Blog #28

Well, this week we had exams. (The Calc exam was actually really easy!) But anyway, I’m just going to review cross-sections and substitution since those two things seem to be what I trip up on the most while doing the practice AP’s.

Cross-Sections:

Ex. 1) 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

Next, 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. 2) 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