Monday, February 28, 2011
blogety
ok so this week we reviewed for the ap like the bosses we are so i get to talk about quotient rule. yay me! so you use quotient rule wen finding the derivative of a fraction. you start by taking the bottom and multiplying it by the derivative of the top and then subtracting that from the derivative of the bottom multiplied by the top. all that is then over the bottom squared.
Sunday, February 27, 2011
2/27
So we did APs again this week as well as for the rest of the year to prep for THE BIG ONE.
Basically, the free response consists of certain questions:
1. Derivative work
2. Integral work
3. "Physics" (PVA)
Key words to look for!
Antiderivative, area, and volume all involve integration
If the "graph of f'" is shown in the problem, you may need to predict where f or f'' is
make sure if youre integrating a fraction to see if its a natural log
Its all a review of the first few chapters we covered
keep reviewing and you'll get a 3 and up!
I got a 2 last time and im pretty proud of that
Basically, the free response consists of certain questions:
1. Derivative work
2. Integral work
3. "Physics" (PVA)
Key words to look for!
Antiderivative, area, and volume all involve integration
If the "graph of f'" is shown in the problem, you may need to predict where f or f'' is
make sure if youre integrating a fraction to see if its a natural log
Its all a review of the first few chapters we covered
keep reviewing and you'll get a 3 and up!
I got a 2 last time and im pretty proud of that
Blog #27
Well, more AP tests…so, more review! This time, I’m reviewing Antiderivatives (or Integrals):
First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c.
Second, an antiderivative is just the derivative backwards. You will be given f’(x) and asked to find f(x).
Third, how to find an antiderivative: If your given ax^b (a being a constant and b being an exponent), your formula is (a/b+1)(x^b+1).
*Note: When solving for the antiderivative, always add “+ c” in place of a constant. When solving for an integral, you will be given something like “f(1) = 2”, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
**Note: When given the second derivative, just solve twice.
***Note: When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
****Note: Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
*****Note: Trig functions: You just do the opposite of the derivative.
Ex. 1) S dx = x + c
Ex. 2) S (x+2) dx = (1/1+1)(x^1+1) = x^2/2 + 2x + c
Ex. 3) S (3x^4 – 5x^2 + x) dx = 3x^5/5 – 5x^3/3 + x^2/2 + c
Ex. 4) S sinx dx = -cosx
Ex. 5) S x^1/2 dx = (1/2) + 1 = 3/2 = 2x^(3/2)/3 + c
Ex. 6) S 3x^2 dx = x^3 + c
Ex. 7) Find the integral of f’(x) = 1/x^2 that satisfies the initial f(1) = 0.
S x^-2 dx = (1/-1)(x^-1) = -1/x + c
-1/1 + c = 0
-1 + c = 0
c = 1
-1/x + 1
First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c.
Second, an antiderivative is just the derivative backwards. You will be given f’(x) and asked to find f(x).
Third, how to find an antiderivative: If your given ax^b (a being a constant and b being an exponent), your formula is (a/b+1)(x^b+1).
*Note: When solving for the antiderivative, always add “+ c” in place of a constant. When solving for an integral, you will be given something like “f(1) = 2”, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
**Note: When given the second derivative, just solve twice.
***Note: When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
****Note: Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
*****Note: Trig functions: You just do the opposite of the derivative.
Ex. 1) S dx = x + c
Ex. 2) S (x+2) dx = (1/1+1)(x^1+1) = x^2/2 + 2x + c
Ex. 3) S (3x^4 – 5x^2 + x) dx = 3x^5/5 – 5x^3/3 + x^2/2 + c
Ex. 4) S sinx dx = -cosx
Ex. 5) S x^1/2 dx = (1/2) + 1 = 3/2 = 2x^(3/2)/3 + c
Ex. 6) S 3x^2 dx = x^3 + c
Ex. 7) Find the integral of f’(x) = 1/x^2 that satisfies the initial f(1) = 0.
S x^-2 dx = (1/-1)(x^-1) = -1/x + c
-1/1 + c = 0
-1 + c = 0
c = 1
-1/x + 1
2/27/11
Okay, sooo for some reason for the most recent practice AP's I forgot how to do chain rule.. really weird I know. So I went back and looked and how to do it and I guess i'll explain it now.
CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Chain rule is way different that the product and quotient rule, however it can be used in both of the rules.
Heres an example of just the chain rule by itself:
Ex. (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Example with product rule and chain rule:
Ex. (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Example with quotient rule and chain rule:
Ex. ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1)^3
CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Chain rule is way different that the product and quotient rule, however it can be used in both of the rules.
Heres an example of just the chain rule by itself:
Ex. (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Example with product rule and chain rule:
Ex. (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Example with quotient rule and chain rule:
Ex. ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1)^3
Alaina's blog, 27 February 2011
So most of this week, we focused on our AP tests, which I like because its good practice for the real thing. But I'm having some problems with the APs.
First, sometimes the way the questions on the free response are worded, I don't exactly know what to do. So, for example, when a question tells you that g(x)=f '(x) and h(x)=g(x), and says to find g(x). What are you supposed to do?
So I guess I'll explain explain area if only one equation is given (area between the graph and the x/y axis) and the area between two equations.
a) area under the graph: you integrate the equation and use the Fundamental Theorem of Calculus, plugging in the two given x-values. F(b)-F(a) where b is the larger number and a is the smaller one.
B) area between two graphs: 1st: if the bounds are not given, set the equations equal and solve for the variable.
If you are not allowed a calculator, plug into the formula integral from a to b of top-bottom. Once you integrate, use the Fundamental Theorem of Calculus and plug in end points.
If you are allowed a calculator, graph the two equations in the calculator and use the intersect function.
These concepts are fairly simple, yet I struggle with them when it comes down to working problems involving them on the AP.
So I'm studying my key words sheets for the exammm! Good luck everyone!
First, sometimes the way the questions on the free response are worded, I don't exactly know what to do. So, for example, when a question tells you that g(x)=f '(x) and h(x)=g(x), and says to find g(x). What are you supposed to do?
So I guess I'll explain explain area if only one equation is given (area between the graph and the x/y axis) and the area between two equations.
a) area under the graph: you integrate the equation and use the Fundamental Theorem of Calculus, plugging in the two given x-values. F(b)-F(a) where b is the larger number and a is the smaller one.
B) area between two graphs: 1st: if the bounds are not given, set the equations equal and solve for the variable.
If you are not allowed a calculator, plug into the formula integral from a to b of top-bottom. Once you integrate, use the Fundamental Theorem of Calculus and plug in end points.
If you are allowed a calculator, graph the two equations in the calculator and use the intersect function.
These concepts are fairly simple, yet I struggle with them when it comes down to working problems involving them on the AP.
So I'm studying my key words sheets for the exammm! Good luck everyone!
Stephen Ledbetter Blog Post 2/27/2011
This week went by pretty fast but the days seemed to just drag on and on and on. I'm going to explain how to work on integration by substitution.
If you have the integral of u * v or the integral of u/v, where v is the derivative of u, v will disappear.
Example:
the integral of (x^2 + 1)^2 (2x) dx
u = x^2 + 1
du = 2x dx
plug in to...
the integral of u^2 du
1/3u^3 + c
1/3(x^2 + 1)^3 + c
the integral of 5 cos 5x dx
u = 5x
du = 5 dx
plug in to...
the integral of cos u du
-sin u + c
-sin 5x + c
e^x
d/dx e^(x^3) = e^(x^3) * 3x^2
the integral of e^2x = e^2x/2 = 1/2 e^2x + c
the integral of x^2 e^(x^3) dx
u = x^3
du = 3x^2
1/3 * the integral of 3x^2 e^(x^3) dx
1/3 * the integral of e^u du
1/3 e^u/1 + c
1/3 e^(x^3) + c
blog of chaos
well, let's try to avoid talking about last week for now.....
anyways, almost didn't get this blog done, RTC has faulty hardware, and my internet's out for a while
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
anyways, almost didn't get this blog done, RTC has faulty hardware, and my internet's out for a while
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
Saturday, February 26, 2011
BLOGGG
soooo RAM RAM RAM RAM RAM RAM A BAMMER BAM BAM
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
Then, there's the midpoint rule (MRAM). Midpoint uses rectangles but touches at a midpoint. So if your directions say to find the area bounded by 0 and 1 using 1 rectangle, your midpoint would be 1/2. From there, you repeat the steps from LRAM and RRAM.
2/26/11
Okayyyy, so for this blog I'll just go over some problems that we normally see on the multiple choice no calculator section..
Ex. 1) 1/2 S e^t/2 dt =
*(1/2 times the integral of e^t/2)
*Okay so to solve this integral we must first integrate e..but remember to leave the number, 1/2, out in front until the end
*So to integrate e^t/2, it's going to be e^t/2 / 1/2...where 1/2 is the derivative of the exponent of e
*So you should have this 1/2[e^t/2 / 1/2], which simplifies to:
1/2[2e^t/2]
*And distributing the 1/2 through you get e^t/2 + c...*don't forget the +c because you weren't given bounds for this integral
Ex. 2) At what point on the graph of y=1/2x^2 is the tangent line parallel to the line 2x-4y=3?
*First, since they say "tangent line" you're going to have to derive the 1st equation they gave you, and you should get that y'=x
*And in order to find where the graph of that^ is parallel to the line (2nd equ.) you must find the slope of the 2nd equation and then set that slope and the derivative you just found equal to each other
*So let's find the slope of 2x-4y=3
-first subtract the 2x over to get -4y=-2x+3
-now divide 4 by everything and you should get y=1/2x-3/4, which means your slope is 1/2
*Now set your derivative equal to your slope:
x=1/2 (so that's also your x-value...to find your y-value you must plug the x-value into the original equation y=1/2x^2..so you get that y=1/8
*Since you found your x and y, now you have a point!
(1/2, 1/8) is your final answer
Ex. 3) The average value of cos x on the interval [-3,5] is
*AVERAGE VALUE is the key word in this problem! All you do for this problem is simply plug into a formula..which is this:
1/b-a Sa,b f(x)...where -3 is a and 5 is b
*So plugging in you get: 1/5-(-3) S-3,5 cos x dx
= 1/8 [sinx] l-3,5 ...where the "l" is the bar
Fundamental Theorem of Calculus:
1/8 sin(5) - [1/8 sin(-3)]
(*Since the negative for -3 would become -sin(3) that means the subtraction sign would become a + sign)
*So now you have 1/8 sin(5) + 1/8 sin(3) and that's your final answer..or you could put it as a fraction like this:
(sin5 + sin3)/8
Ex. 1) 1/2 S e^t/2 dt =
*(1/2 times the integral of e^t/2)
*Okay so to solve this integral we must first integrate e..but remember to leave the number, 1/2, out in front until the end
*So to integrate e^t/2, it's going to be e^t/2 / 1/2...where 1/2 is the derivative of the exponent of e
*So you should have this 1/2[e^t/2 / 1/2], which simplifies to:
1/2[2e^t/2]
*And distributing the 1/2 through you get e^t/2 + c...*don't forget the +c because you weren't given bounds for this integral
Ex. 2) At what point on the graph of y=1/2x^2 is the tangent line parallel to the line 2x-4y=3?
*First, since they say "tangent line" you're going to have to derive the 1st equation they gave you, and you should get that y'=x
*And in order to find where the graph of that^ is parallel to the line (2nd equ.) you must find the slope of the 2nd equation and then set that slope and the derivative you just found equal to each other
*So let's find the slope of 2x-4y=3
-first subtract the 2x over to get -4y=-2x+3
-now divide 4 by everything and you should get y=1/2x-3/4, which means your slope is 1/2
*Now set your derivative equal to your slope:
x=1/2 (so that's also your x-value...to find your y-value you must plug the x-value into the original equation y=1/2x^2..so you get that y=1/8
*Since you found your x and y, now you have a point!
(1/2, 1/8) is your final answer
Ex. 3) The average value of cos x on the interval [-3,5] is
*AVERAGE VALUE is the key word in this problem! All you do for this problem is simply plug into a formula..which is this:
1/b-a Sa,b f(x)...where -3 is a and 5 is b
*So plugging in you get: 1/5-(-3) S-3,5 cos x dx
= 1/8 [sinx] l-3,5 ...where the "l" is the bar
Fundamental Theorem of Calculus:
1/8 sin(5) - [1/8 sin(-3)]
(*Since the negative for -3 would become -sin(3) that means the subtraction sign would become a + sign)
*So now you have 1/8 sin(5) + 1/8 sin(3) and that's your final answer..or you could put it as a fraction like this:
(sin5 + sin3)/8
Sunday, February 20, 2011
aps are ca-razy
well, all we've done this week were ap reviews and more practice aps, but at least the last one wasn't as bad as the others
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
blogtiem
so this week we did a bunch of stuff for the AP which would mean that its review tiem. i believe ill go with this weeks popular choice LRAM and RRAM. quick sidenote the average of the two comes to TRAM. anyram here we go. first of all you usually use LRAM and RRAM when you are given several coordinate points and have to find an integral or a missing coordinate point. you start off by dividing the area into sections (usually using the coordinate points given). the y-values become your height and the length between the x-values become the width of each subsection. for LRAM you add all of the y-values together excluding the one furthest to the right and for RRAM you add up all the y-values excluding the one furthest left. you then multiply your answer by the width of your subsections. and shazzam the rabbit was in the hat the whole tiem.
Blog #26
This week, we continued with the practice AP tests, so for this blog, I’m going to go over how to find LRAM and RRAM.
These methods are used to approximate the area of the region bounded by a curve and the x-axis.
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
These methods are used to approximate the area of the region bounded by a curve and the x-axis.
LRAM is the left Reimann Sums. It uses the left points touching the graph.
RRAM is the right Reimann Sums. It uses the right points touching the graph.
In order to get an accurate estimation, you need to do both LRAM and RRAM because one will always be an underestimate and one will always be an overestimate.
Reimann Sums is the method of breaking up the region into smaller, easier shapes and adding the area of each one together. It is commonly used with rectangles.
First, the area of a rectangle is length times width. So, we must find the length and the width of each rectangle, then add all of them together.
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
alaina's blog, 20 Feb 2011
I'm just going to review LRAM, RRAM, MRAM, and TRAP and finding the area between two graphs..
LRAM, RRAM, MRAM and TRAP all circulate around Riemann Sums.
Anyway, LRAM is usually an underestimate; whereas, RRAM is usually an overestimate.
To find LRAM and RRAM, first you need to draw the graph or the number of rectangles within the bounds and figure out your delta X. Then, you need to plug in the numbers at the end of each rectangle within the bounds. This gives you your y-values. From there, for LRAM, you multiply each y-value by delta X except for the last and add them together. For RRAM, you multiply each y-value by delta X except for the first and add them together. These values are over and under estimates.
For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
Then, there's the midpoint rule (MRAM). Midpoint uses rectangles but touches at a midpoint. So if your directions say to find the area bounded by 0 and 1 using 1 rectangle, your midpoint would be 1/2. From there, you repeat the steps from LRAM and RRAM.
LRAM, RRAM, MRAM and TRAP all circulate around Riemann Sums.
Anyway, LRAM is usually an underestimate; whereas, RRAM is usually an overestimate.
To find LRAM and RRAM, first you need to draw the graph or the number of rectangles within the bounds and figure out your delta X. Then, you need to plug in the numbers at the end of each rectangle within the bounds. This gives you your y-values. From there, for LRAM, you multiply each y-value by delta X except for the last and add them together. For RRAM, you multiply each y-value by delta X except for the first and add them together. These values are over and under estimates.
For the Trapezoidal rule, there are two ways to find it. You can add LRAM and RRAM together and multiply by one half, or you can 1/2 (base 1 + base 2)H where bases are y-values and h is like delta X. This method gets you closer to the true area.
Then, there's the midpoint rule (MRAM). Midpoint uses rectangles but touches at a midpoint. So if your directions say to find the area bounded by 0 and 1 using 1 rectangle, your midpoint would be 1/2. From there, you repeat the steps from LRAM and RRAM.
2/20/11
I think i'll go over definite integrals since I saw a couple of problems like that on the practice AP's. Sooo, the definite integral of a function represents the area between the curve and the x-axis. *Definite integrals can also be calculated by counting squares and Riemann Sums. Basically what you're doing is drawing the graph of the function, putting your points, drawing however many rectangles there asking for, looking to see which rectangle (right or left) touches the graph, or goes over, and plugging in..Or finding the area.
Here's an example:
What are 2 types of Riemann sums that we could use to find the area of the region bounded by the graph of f(x)=1/2^2 and the x-axis between x=0 and x=3. Find the area ofthis region using all methods.
*So switching this to an integral we get the bounds of 3 as the upper, and 0 as the lower bound 1/2x^2=1/6x^3 bar 3 at top and 0 at the bottom.
*Now that we know 1/6x^3 is our formula were gonna plug in the bounds 3 and 0. 1/6(3)^3-1/6(0)^3=9/2 (this is usuing the fundamental theorm of calculus)
**Now draw the LRRAM
We know the function is a parabola, and the x axis goes up to 3, with 3 triangles, and the left point of the rectangles are touching the graph(which means it does NOT go over)
*Now plug in the 3 rectangle points into the original function (1/2x^2)
We're only going to plug in 0,1, and 2 because 3 is not touching the graph.
*1/2(0)^2=0
1/2(1)^2=1/2
1/2(2)^2=2
*Add them all together to get 5/2
**Now draw the RRAM
It's drawn the exact same except we know the rectangles go over, and when we plug in points were going to plug in 1,2, and 3.
Here's an example:
What are 2 types of Riemann sums that we could use to find the area of the region bounded by the graph of f(x)=1/2^2 and the x-axis between x=0 and x=3. Find the area ofthis region using all methods.
*So switching this to an integral we get the bounds of 3 as the upper, and 0 as the lower bound 1/2x^2=1/6x^3 bar 3 at top and 0 at the bottom.
*Now that we know 1/6x^3 is our formula were gonna plug in the bounds 3 and 0. 1/6(3)^3-1/6(0)^3=9/2 (this is usuing the fundamental theorm of calculus)
**Now draw the LRRAM
We know the function is a parabola, and the x axis goes up to 3, with 3 triangles, and the left point of the rectangles are touching the graph(which means it does NOT go over)
*Now plug in the 3 rectangle points into the original function (1/2x^2)
We're only going to plug in 0,1, and 2 because 3 is not touching the graph.
*1/2(0)^2=0
1/2(1)^2=1/2
1/2(2)^2=2
*Add them all together to get 5/2
**Now draw the RRAM
It's drawn the exact same except we know the rectangles go over, and when we plug in points were going to plug in 1,2, and 3.
2/20
So we've been doing practice AP's lately and I got a 2 on the last one which is good by my standards. I learned that the free response is a point fest when it comes to just showing that you know how to set it up.
for example, if youre confused by a certain problem: find the key words in it (most likely a derivative or an integral) then set it up (free point). If you know what to do after setting up, carry it out for 1 to 2 more points.
The calculator portion is where I'm sort of weak at at the moment, along with the free response, but after some binder assignments and extra practice I can easily get a 3 :D
for example, if youre confused by a certain problem: find the key words in it (most likely a derivative or an integral) then set it up (free point). If you know what to do after setting up, carry it out for 1 to 2 more points.
The calculator portion is where I'm sort of weak at at the moment, along with the free response, but after some binder assignments and extra practice I can easily get a 3 :D
Saturday, February 19, 2011
2/19/11
So this week we continued with practice AP's; I'll just give a few example problems like those on the multiple choice section..
Ex. 1) If f(x) = -x^3 + x + 1/x, then f'(-1) =
*So first let's determine what we're given in this problem.
*They give you the original equation, and they're asking for f'(-1)..which means they want you to take the derivative and then plug in -1 for x after.
*So first let's take the derivative and you should have this:
-3x^2 + 1 - 1/x^2
*Now all you do is plug in -1 for x:
-3(-1)^2 + 1 - 1/(-1)^2
= -3
Ex. 2) d/dx cos^2(x^3)
*So when you first take a look at this problem, you see d/dx, which means all you have to do is differentiate it
*So since this is a chain rule, first you have to derive the outside, cos^2, so you should first get this:
2(cos(x^3))
*Buttt, you're not done yet; next you have to derive cos and then recopy the rest of the equation. So now you should have this:
2(cos(x^3)) * sin(x^3)
*Last, multiply that by the derivative of the inside, x^3, which is 3x^2
*So now you have 2(cos(x^3)) * sin(x^3) * 3x^2
*Now simplifying all of that, your final answer should be:
6x^2cos(x^3)sin(x^3)
Ex. 3) The area of the region enclosed by the graph of y = x^2 + 1 and the line y = 5 is....
*Alright they're asking for AREA so that means you're going to use some kind of INTEGRAL..
*First, let's figure out what our graphs look like...y=x^2+1 is a parabola with vertex (0,1) and y=5 is a horizontal line at the point (0,5)
*So looking at the two graphs on the same coordinate plane, the line y=5 is on the top and the parabola is on the bottom, and you're looking for the area in between.
So you're going to have to use the formula: S top-bottom
*Plugging into the formula you get:
S 5-x^2-1
= S 4-x^2
Now integrating that^ you get:
4x - 1/3x^3....but we still have to find the bounds!
So to do that all you do is take the two equations and set them equal and solve for x:
x^2 + 1 = 5
x^2 = 4
x = -2,2 ...and those are your bounds-- -2 being the lower bound and 2 being the upper bound
*So now you have this:
[4x - 1/3x^3] I-2,2 ...the "I" being the bar (meaning you didn't plug in yet)
*Now Fundamental Theorem of Calculus!
4(2)-1/3(2)^3-[4(-2)-1/3(-2)^3]
Simplifying that you should end up with 32/3 and that's your area
Ex. 1) If f(x) = -x^3 + x + 1/x, then f'(-1) =
*So first let's determine what we're given in this problem.
*They give you the original equation, and they're asking for f'(-1)..which means they want you to take the derivative and then plug in -1 for x after.
*So first let's take the derivative and you should have this:
-3x^2 + 1 - 1/x^2
*Now all you do is plug in -1 for x:
-3(-1)^2 + 1 - 1/(-1)^2
= -3
Ex. 2) d/dx cos^2(x^3)
*So when you first take a look at this problem, you see d/dx, which means all you have to do is differentiate it
*So since this is a chain rule, first you have to derive the outside, cos^2, so you should first get this:
2(cos(x^3))
*Buttt, you're not done yet; next you have to derive cos and then recopy the rest of the equation. So now you should have this:
2(cos(x^3)) * sin(x^3)
*Last, multiply that by the derivative of the inside, x^3, which is 3x^2
*So now you have 2(cos(x^3)) * sin(x^3) * 3x^2
*Now simplifying all of that, your final answer should be:
6x^2cos(x^3)sin(x^3)
Ex. 3) The area of the region enclosed by the graph of y = x^2 + 1 and the line y = 5 is....
*Alright they're asking for AREA so that means you're going to use some kind of INTEGRAL..
*First, let's figure out what our graphs look like...y=x^2+1 is a parabola with vertex (0,1) and y=5 is a horizontal line at the point (0,5)
*So looking at the two graphs on the same coordinate plane, the line y=5 is on the top and the parabola is on the bottom, and you're looking for the area in between.
So you're going to have to use the formula: S top-bottom
*Plugging into the formula you get:
S 5-x^2-1
= S 4-x^2
Now integrating that^ you get:
4x - 1/3x^3....but we still have to find the bounds!
So to do that all you do is take the two equations and set them equal and solve for x:
x^2 + 1 = 5
x^2 = 4
x = -2,2 ...and those are your bounds-- -2 being the lower bound and 2 being the upper bound
*So now you have this:
[4x - 1/3x^3] I-2,2 ...the "I" being the bar (meaning you didn't plug in yet)
*Now Fundamental Theorem of Calculus!
4(2)-1/3(2)^3-[4(-2)-1/3(-2)^3]
Simplifying that you should end up with 32/3 and that's your area
Sunday, February 13, 2011
blog? blog. *serious face* BLOG.
duuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuude, we've learned nothing new lately, which is awesome
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
Blog #25
Well, this week we just went over the AP practice test, so I’m just going to review a few things that were on the AP tests.
First, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Ex. 1) Find the area of the shape between the boundaries given.
2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Next, Cross-Sections:
Ex. 2) Find the volume of a solid with triangular perpendicular cross-section for: y = 1 – x/2 & y = -1 + x/2.
To solve a cross-section problem, you simply do what we’ve been doing with area and volume this whole time, just with the area formula for whatever shape they give you. In this case, that would be a triangle.
The area of a triangle: ½ bh
So, you simply take the integral of base times height (the first equation times the second equation) and multiply it by ½. Oh, and don’t forget the dx.
½ S(1 – x/2)(-1 + x/2)dx = ½ S(-1 - x²/4 + x)dx
For this particular problem, they did not give you any bounds to work with, so you just leave the answer as an equation.
For other problems like this but with different shapes, use these area formulas:
Triangles >>> S ½ bh
Rectangles >>> S lw
Semi-cirlces >>> S ½ π r²
Cirles >>> S π r²
Squares >>> S s²
Finally, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 3) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
First, the Fundamental Theorem of Calculus:
This theorem is very simple; it just states that “bSa f(x) dx = F(b) – F(a) = area where F(x) is the integrated function”.
Ex. 1) Find the area of the shape between the boundaries given.
2S0 (x + 1) dx
F(x) = ½ x² + x
Area = (½ (2)² + (2)) – (½ (0)² + (0)) = 4 – 0
= 4
Next, Cross-Sections:
Ex. 2) Find the volume of a solid with triangular perpendicular cross-section for: y = 1 – x/2 & y = -1 + x/2.
To solve a cross-section problem, you simply do what we’ve been doing with area and volume this whole time, just with the area formula for whatever shape they give you. In this case, that would be a triangle.
The area of a triangle: ½ bh
So, you simply take the integral of base times height (the first equation times the second equation) and multiply it by ½. Oh, and don’t forget the dx.
½ S(1 – x/2)(-1 + x/2)dx = ½ S(-1 - x²/4 + x)dx
For this particular problem, they did not give you any bounds to work with, so you just leave the answer as an equation.
For other problems like this but with different shapes, use these area formulas:
Triangles >>> S ½ bh
Rectangles >>> S lw
Semi-cirlces >>> S ½ π r²
Cirles >>> S π r²
Squares >>> S s²
Finally, Substitution:
This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Ex. 3) S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
2/13/11
Alright, sooo were still doing AP stuff.. with nothing new learned I think i'm just gonna review some old topics. I know questions similar to these showed up on the practice AP's we took, and I forgot how to do them. So i'll explain it and give some examples.
EXAMPLES OF SLOPE:
1. Find the slope m of the line tangent to the graph of the function g(x) = 5 - x^2 at the point (2,1).
*To do this, you must find the derivative, then plug in your x-value.
Derivative = -2x
Plug in: -2 (2) = -4
The slope is -4
2. Find the slope of the graph of the function at the given value.
f(x) = 2 (4x + 6)^2 when x = 5
*Find the derivative and then plug in.
Derivative:
2 * (2 (4x + 6)) * 4
16(4x + 6)
16(4(5) + 6)
16(20 + 6)
16 (26) = 416
416 is the slope of the graph.
EXAMPLE: Mean Value Theorem
1. Check if the function is continuous.
2. Check for differentiability or non-differentiability.
3. Slope
f ' (c) = f(b) - f(a) / b - a
* It means on an interval [a,b]. There must be some x-value where the derivative = the slope between a & b.
slope of a secant line = slope of a tangent line between a & b
1. Given f(x) = 5 - 4/x, find all values of c in the open interval (1,4) such that
f ' (c) = f(4) - f(1) / 4-1
yes it is both continuous and differentiable
= 4-1/4-1 = 1
5 - 4x^-1
4/x^2 = 1
4 = x^2
x = +/- 2
c = +/- 2
*By the mean value theorem, the function is continuous and differentiable, therefore some value c=2 where f ' (x) = the slope between (1,4).
EXAMPLES OF SLOPE:
1. Find the slope m of the line tangent to the graph of the function g(x) = 5 - x^2 at the point (2,1).
*To do this, you must find the derivative, then plug in your x-value.
Derivative = -2x
Plug in: -2 (2) = -4
The slope is -4
2. Find the slope of the graph of the function at the given value.
f(x) = 2 (4x + 6)^2 when x = 5
*Find the derivative and then plug in.
Derivative:
2 * (2 (4x + 6)) * 4
16(4x + 6)
16(4(5) + 6)
16(20 + 6)
16 (26) = 416
416 is the slope of the graph.
EXAMPLE: Mean Value Theorem
1. Check if the function is continuous.
2. Check for differentiability or non-differentiability.
3. Slope
f ' (c) = f(b) - f(a) / b - a
* It means on an interval [a,b]. There must be some x-value where the derivative = the slope between a & b.
slope of a secant line = slope of a tangent line between a & b
1. Given f(x) = 5 - 4/x, find all values of c in the open interval (1,4) such that
f ' (c) = f(4) - f(1) / 4-1
yes it is both continuous and differentiable
= 4-1/4-1 = 1
5 - 4x^-1
4/x^2 = 1
4 = x^2
x = +/- 2
c = +/- 2
*By the mean value theorem, the function is continuous and differentiable, therefore some value c=2 where f ' (x) = the slope between (1,4).
Saturday, February 12, 2011
2/12/11
Soooooo...time for another blog I suppose. About what? Nothing specific really..All this week we just took practice AP's and did work with that. But anyway, I think I'll just give a few examples of some recent things we learned..
Ex. 1) S sin(3x+5) dx ...(integrate)
*Okay first you should notice that with this problem there's two things going on at once..kind of like a chain rule; so, that tells you that you're going to have to use substitution to solve this
*So your u is going to be what's inside the function, which is 3x+5 and your du will be 3 dx (because 3 is the derivative of your u)
*Now you're going to replace 3x+5 with u in your equation to solve...butttttt, you're going to have to add a 1/3 on the outside to even it out (since your derivative isn't actually in the original problem) So you should get this:
1/3 S sin(u) du
*Now you're going to integrate sin(u)....and you should get -cos(u)
So now you have this:
1/3[-cos(u)] + c
*Now all you have left to do is plug back in for u like this:
= 1/3[-cos(3x+5)] + c
Ex. 2) For y = 3x^2+2x+5 at the point (4,0), what is the slope of the normal line?
*First, since they're asking for slope you know that you're going to have to take the derivative of the equation they gave you.
*So once you take the derivative you should get 6x+2
*Since they didn't ask for an equation, all they want is the slope (just the number)..and they gave you an x-value so you're going to plug that (4) into your derivative (6x+2) like this:
6(4)+2
= 26...so that's your slope
*Buttttt, since they're asking for the slope of the NORMAL LINE, that means to get your final answer you'll have to take the negative reciprocal of the slope you just found
*So the slope of the normal line is -1/26
Ex. 3) Find the average value for f(x)=5x^2+6x+2 on the interval [0,2].
*Sooooo when you first look at this problem I bet the word "average" stands out to you..it probably makes you think like average rate of change or average velocity or something which means that you'll have to take the derivative....wellllll, not in this case. Average VALUE is totally different. It involves an integral, but it's super easy because all you have to do is use this simple formula:
1/b-a Sa,b f(x) ....(where its the integral of f(x) from bounds a-b multiplied by 1/b-a
*So plugging in the information they gave us into the formula you should get this:
1/2 S0,2 5x^2+6x+2 dx
*Now all you do is integrate it! And you should end up with this:
1/2[5/3x^3 + 3x^2 +2x] I2,0 ...(pretend that "I" is the bar that means you haven't plugged in yet, haha)
*So now do the Fundamental Theorem of Calculus:
1/2 [5/3(2)^3 + 3(2)^2 + 2(2)] - [0]
**^don't forget, the 1/2 is still out front because that's in the formula
*So once you simplify everything you should get this for your final answer:
44/3
Ex. 1) S sin(3x+5) dx ...(integrate)
*Okay first you should notice that with this problem there's two things going on at once..kind of like a chain rule; so, that tells you that you're going to have to use substitution to solve this
*So your u is going to be what's inside the function, which is 3x+5 and your du will be 3 dx (because 3 is the derivative of your u)
*Now you're going to replace 3x+5 with u in your equation to solve...butttttt, you're going to have to add a 1/3 on the outside to even it out (since your derivative isn't actually in the original problem) So you should get this:
1/3 S sin(u) du
*Now you're going to integrate sin(u)....and you should get -cos(u)
So now you have this:
1/3[-cos(u)] + c
*Now all you have left to do is plug back in for u like this:
= 1/3[-cos(3x+5)] + c
Ex. 2) For y = 3x^2+2x+5 at the point (4,0), what is the slope of the normal line?
*First, since they're asking for slope you know that you're going to have to take the derivative of the equation they gave you.
*So once you take the derivative you should get 6x+2
*Since they didn't ask for an equation, all they want is the slope (just the number)..and they gave you an x-value so you're going to plug that (4) into your derivative (6x+2) like this:
6(4)+2
= 26...so that's your slope
*Buttttt, since they're asking for the slope of the NORMAL LINE, that means to get your final answer you'll have to take the negative reciprocal of the slope you just found
*So the slope of the normal line is -1/26
Ex. 3) Find the average value for f(x)=5x^2+6x+2 on the interval [0,2].
*Sooooo when you first look at this problem I bet the word "average" stands out to you..it probably makes you think like average rate of change or average velocity or something which means that you'll have to take the derivative....wellllll, not in this case. Average VALUE is totally different. It involves an integral, but it's super easy because all you have to do is use this simple formula:
1/b-a Sa,b f(x) ....(where its the integral of f(x) from bounds a-b multiplied by 1/b-a
*So plugging in the information they gave us into the formula you should get this:
1/2 S0,2 5x^2+6x+2 dx
*Now all you do is integrate it! And you should end up with this:
1/2[5/3x^3 + 3x^2 +2x] I2,0 ...(pretend that "I" is the bar that means you haven't plugged in yet, haha)
*So now do the Fundamental Theorem of Calculus:
1/2 [5/3(2)^3 + 3(2)^2 + 2(2)] - [0]
**^don't forget, the 1/2 is still out front because that's in the formula
*So once you simplify everything you should get this for your final answer:
44/3
Sunday, February 6, 2011
umad bloggity bloggers?
duuuuuuuuuuuuuuuuuuuuuuuuude, the super bowl was boring
anyway, we didn't learn a single new thing this week, so what am I supposed to blog about?
something: !
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
anyway, we didn't learn a single new thing this week, so what am I supposed to blog about?
something: !
Ex. Find the area of the region bounded by the curve f(x) = x² and the x-axis between x=0 and x=3. Use 3 rectangles.
First of all, x=0 and x=3 are your bounds. So to rewrite this equation, it would be: 3S0 x² dx. (*”S” is the integral symbol.)
Next, what is the width? To find it, subtract your upper and lower bounds, then divide by the number of rectangles: (3-0)/3 = 1
Now, to find the lengths, plug in to the original equation every number between 0 and 3 that is 1 width away. The numbers will be 0, 1, 2, and 3. To plug in: (0)² = 0; (1)² = 1; (2)² = 4; (3)² = 9
Finally, for LRAM, multiply the width by the first three lengths (all of the left points) and add them together: (1)(0) + (1)(1) + (1)(4) = 5
For RRAM, multiply the width by the last three lengths (all of the right points) and add them together: (1)(1) + (1)(4) + (1)(9) = 14
Your answer: 5 < area < 14
BLOGbro
This week we started taking practice AP testing and we also went over PVA too. PVA: position, velocity, and acceleration. Displacement or position is the integral of velocity; and distance is the absolute value of the integral of velocity. Most of the PVA questions will have more than one part to them. It's good to realize what the question is asking for (as far as distance/displacement).
Here's an example:
Chad accelerates his car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after he started accelerating.
A: Find the time(s) at which v = 0
Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
In reality these problems are really easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals.
Here's an example:
Chad accelerates his car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after he started accelerating.
A: Find the time(s) at which v = 0
Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
In reality these problems are really easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals.
Blog #24
This week, we mainly just did AP practice tests. Besides that, we learned how to integrate using substitution. This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S u x v
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
This week, we also learned the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
S u x v
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
This week, we also learned the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
2/6/11
Soooo this week we started taking practice AP testing..well just the multipule choice part, and actually it is really hard. I know the information, but for some reason I just can't work the problems out..hopefully with practice I will get better at taking the test. Anyways, since we started taking the AP test we haven't learned anything new. I pretty much went over everything besides PVA, so i'll explain how to do that. PVA: position, velocity, and acceleration. The week before we took the AP test we learned about these functions and how to apply integrals with them. For instance, displacement or position is the integral of velocity; and distance is the absolute value of the integral of velocity. Most of the PVA questions will have more than one part to them. It's good to realize what the question is asking for (as far as distance/displacement).
Here's an example:
1.) Meg accelerates her car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after she started accelerating.
A: Find the time(s) at which v = 0
*Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Notice they give you the time interval [1,9]..which is going to be your bounds for your integral
*Also notice there asking for the displacement, you're going to take the integral of the velocity function with bounds 1,9
*So plugging into that formula you should get this:
S1,9 t^1/2-2 dt
Now taking the integral of that equation you should end up with:
2/3t^3/2 - 2t
Now use the Fundamental Theorem of Calculus to plug in 9 and 1:
2/3(9)^3/2 - 2(9) - [2/3(1)^3/2 - 2(1)]
= 4/3 ft
is displacement
C: Find the total distance she travels for the time interval [1,9].
*So now since they're asking for distance, all we're going to do is use the distance formula..which says to take the absolute value of the integral of velocity
*So plugging that into your calculator with bounds [1,9] you should end up with 4
*So 4 ft is your total distance
These problems are reallllly easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals. It's pretty much everything rolled into one. And that's alllll :)
Here's an example:
1.) Meg accelerates her car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after she started accelerating.
A: Find the time(s) at which v = 0
*Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4
B: Find her net displacement for the time interval [1,9].
*Notice they give you the time interval [1,9]..which is going to be your bounds for your integral
*Also notice there asking for the displacement, you're going to take the integral of the velocity function with bounds 1,9
*So plugging into that formula you should get this:
S1,9 t^1/2-2 dt
Now taking the integral of that equation you should end up with:
2/3t^3/2 - 2t
Now use the Fundamental Theorem of Calculus to plug in 9 and 1:
2/3(9)^3/2 - 2(9) - [2/3(1)^3/2 - 2(1)]
= 4/3 ft
is displacement
C: Find the total distance she travels for the time interval [1,9].
*So now since they're asking for distance, all we're going to do is use the distance formula..which says to take the absolute value of the integral of velocity
*So plugging that into your calculator with bounds [1,9] you should end up with 4
*So 4 ft is your total distance
These problems are reallllly easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals. It's pretty much everything rolled into one. And that's alllll :)
2/6/11
Sooo..starting off this week we learned the Second Fundamental Theorem of Calculus. All that is is the derivative of an integral. So it would be written in the form: d/dx S1,2x x^2 dx...where d/dx infers derivative; 1,2x are the bounds; and x^2 is the equation. In order to solve a problem like this, all you do is take f(x) (which would be the equation), plug in the top bound for x and multiply that by the derivative of the top bound. *Also, in order for this method to work, the bottom bound has to be a number. Nexxttttt, we learned integration by Substitution..you use this method when you have an equation that looks like it would be a product rule or a quotient rule. To use substitution, one part of the equation is going to be the derivative of the other part..or close to it. It's a weird process to explain so I'll just give an example haha...And also this week we learned how to integrate e..and all you do for that is basically the opposite that you would do to take its derivative. First you recopy the e part, then instead of multiplying by the exponent's derivative, you divide by it. So here are a few examples of what I just talked about:
Ex. 1) d/dx S6,2x^4 2x+1 dx......where 6,2x^4 are the bounds and 2x+1 is the equation
*So first you're going to take the equation (f) and plug in the top bound for x..so you should get this:
2(2x^4)+1
*Now multiply that by the derivative of 2x^4...which is 8x^3
*So you should have 8x^3(4x^4+1)
Ex. 2) S(6x)(3x^2+1) dx
*Now since this looks like it would be a product rule, you have to check to see if you can use substitution to solve it
*6x is the derivative of 3x^2+1..so yes you can use substitution
*To use this method, you're going to replace one of the terms with u to make it simpler to solve
*Since 6x is the derivative of 3x^2+1, u becomes 3x^2+1 And du, the derivative of u, would be 6x dx..(<
S u du (and the 6x is not included because it's going to disappear)
*So once you integrate u, you should end up with this:
1/2u^2 + c ...*don't forget the +c because there are no bounds
*Now all you do is plug back in for u. So you're final equation should look like this:
1/2(3x^2+1) + c
Ex. 3) S e^4x^2
*To integrate this, you first recopy the equation:
= e^4x^2
*Now all you do is divide by the derivative of e's exponent...which is 8x
*So you should get this for your answer:
e^4x^2/8x
**Well this week was good overall. I understood everything; I just think I could use more practice on the substitution problems because those are kinda tricky.
Ex. 1) d/dx S6,2x^4 2x+1 dx......where 6,2x^4 are the bounds and 2x+1 is the equation
*So first you're going to take the equation (f) and plug in the top bound for x..so you should get this:
2(2x^4)+1
*Now multiply that by the derivative of 2x^4...which is 8x^3
*So you should have 8x^3(4x^4+1)
Ex. 2) S(6x)(3x^2+1) dx
*Now since this looks like it would be a product rule, you have to check to see if you can use substitution to solve it
*6x is the derivative of 3x^2+1..so yes you can use substitution
*To use this method, you're going to replace one of the terms with u to make it simpler to solve
*Since 6x is the derivative of 3x^2+1, u becomes 3x^2+1 And du, the derivative of u, would be 6x dx..(<
S u du (and the 6x is not included because it's going to disappear)
*So once you integrate u, you should end up with this:
1/2u^2 + c ...*don't forget the +c because there are no bounds
*Now all you do is plug back in for u. So you're final equation should look like this:
1/2(3x^2+1) + c
Ex. 3) S e^4x^2
*To integrate this, you first recopy the equation:
= e^4x^2
*Now all you do is divide by the derivative of e's exponent...which is 8x
*So you should get this for your answer:
e^4x^2/8x
**Well this week was good overall. I understood everything; I just think I could use more practice on the substitution problems because those are kinda tricky.
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