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# 2011 Calculus AB free response #3 (a & b)

Equation of a tangent line and area between curves. Created by Sal Khan.

## Want to join the conversation?

• Don't you have to write units squared right after the -1/8 + 1/pi because you are finding the area?
• Unless units are given, you don't have to add a generic 'units^2' to your answer on the AP exam. That said it doesn't count against you if you do add it.
• Where do these trig function properties of derivatives/antiderivatives come from? I'm having trouble grasping how to take the derivative of functions in which x is being multiplied by a constant, such as sin(2x), cos(3x/4) etc.
I perused the obvious vids on here, but didn't see any mention of these types of derivatives/antiderivatives.
• I found this on wolfram alpha but I'm still not sure what is going on in the u substitution step.

Possible intermediate steps:
integral sin(pi x) dx
For the integrand sin(pi x), substitute u = pi x and du = pi dx:
= 1/pi integral sin(u) du
The integral of sin(u) is -cos(u):
= -(cos(u))/pi+constant
Substitute back for u = pi x:
= -(cos(pi x))/pi+constant

Is this saying to multiply the entire integral by pi/pi? So just using a little trick of multiplying it by 1 and hence not changing anything?

(1 vote)
• at how do you know which function is above/below?
• Usually the best way is to make a graph of the region and determine the answer by looking at the graph. In this case they've given you the graph, so you just have to recognize which line looks like part of the sine function (sin(πx) looks like sin(x) except for greater frequency) and which line looks like x^3 (again, there's a coefficient, but that doesn't change the general shape of the function).

If you're unsure, you can figure the slope of each function at the left boundary (x = 0), and the one with the higher slope would be the one that's above the other. In a pinch you can try to find a point within the range where you can calculate the value of both functions. By far the most valuable skill here, though, is to understand the graphs.
• @ Why are we evaluating/integrating from 0 to 1/2 instead of 0 to 1?
• Because we are integrating with respect to x, and the x values in consideration range from 0 to 1/2. We could change the integration to be with respect to y, in which case you would integrate from 0 to 1.
• I see that the derivative of -1/pi(cos-pi-x) is sin-pi-x, but going the other way is not overtly obvious. What is the operation for taking the integral of sin-pi-x
• If you know how to do u-subs, then make u=pi*x, and du=pi*dx. If you know what to do from here, great. Otherwise, let me know.