2000 Calculus AB Free-Response: Question 1 Part A

In my last post, Learn from the past AP Tests: 2000, I linked to the 2000 AP Calculus AB Free-Response questions. Today, I will start to go through the solutions, starting with question number one, part A.

Question number one, part A asks for the area of a region, R, between two functions (also known as curves), which I will call f(x) and g(x). f(x)=e^(-(x^2)) and g(x)=1 - cos(x)

To solve, we must first find the area under the top curve, f(x). To do this, we must integrate f(x) (Also known as taking the antiderivative. The antiderivative of f(x) can be written as F(x).). So F(x)= -(1/2)e^(-(x^2)). The easiest way to do that is u substitution.

Now, lets integrate g(x). G(x)= x - sin(x).

Now that we know the two antiderivatives, we need to find out wher f(x) intersect with g(x); so where f(x) = g(x). Remember, this is a calculator portion, so you can simply plug this into a calculator (In fact, you can solve this question in a much easier manner using your calculator. I will show you how in my next post.). The results show the intersection, B, is at .94194.

Lets plug it all in! The integral of f(x) - g(x) from A=0 to B=.94194, can also be written as [F(x) - G(x)]₀ ^.94194, which can be written as F(.94194)-F(0) - (G(.94194) - G(0)).

It all equals .591.

If you have any questions, please leave a comment.