AP Calculus AB Help - FRQ 2009 #6

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<p>I have been practicing free responce questions and am stuck on 2009 #6, part B. Here is a link to the question/answer: <a href="http://apcentral.collegeboard.com/apc/public/repository/ap09_calculus_ab_sgs.pdf%5B/url%5D"&gt;http://apcentral.collegeboard.com/apc/public/repository/ap09_calculus_ab_sgs.pdf&lt;/a&gt;&lt;/p>

<p>I can integrate from 0 to 4, but cannot figure out how the markers integrated from 0 to -4. I cannot figure out where the 5-(8-2Pi) expression comes from. Any help is appreciated.</p>

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<p>haha, I took this last year and remembering this problem. For a split second I was wondering how to do that then realized what to do and said to myself, “Wooow…”</p>

<p>Just do area under the curve. Take the area of the rectangle minus the area of the semicircle.</p>

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<p>Awesome! Thanks a lot.</p>

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<p>The integral from 0 to -4 is the negative of the integral from -4 to 0. </p>

<p>Therefore, you take the area of the rectangle whose base is 4 and whose height is 2 to get 8. You then subtract the area of the semi-circle to get the area of the two pieces that you want the value of. The area of the semi-circle is 1/2 pi (2^2) or 2pi. This gives you 8-2pi. However, it is -8+2pi since you take the negative since you had to switch the limits of integration. Then, you add your f(0), which is 5, to get 2pi -3.</p>