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# 2011 Calculus AB free response #6c

Average value of a piecewise-defined function on an interval. Created by Sal Khan.

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• At the end, shouldn't we distribute the `1/2` over the expression in the brackets to get `(13/8)-cos(-1)-((e^-4)/8)?` The solution feels unfinished to me without taking this final step. • Why didn't Sal simplify cos(-1)? That's zero and that term drops. Distribute the 1/2 and = 13/8-(e^-4)/8 • Hi blahdee327. I hope you don't mind that I changed your comment to a question and did some editing to it.
Are you sure that cos(-1) = 0? If you graph the function y=cos(x), you'll see that that is not the case. Indeed, the function y = cos(x) is equal to zero at - π/2 which does not equal -1.
You may have gotten confused because cos(0) = 1.
• the exam of AP calculus the word "BC" stands off for integral calculus right¿ • At , Sal takes the definite integral from 0 to 1 of the function f(x). The function f(x)at x=0 is defined as 1-2sin(x). Everywhere else on the interval, the function f(x) is defined as e^-4x. Even though we determined in part (a) that the function f(x) is continuous at x=0, how can we take a definite integral of a function over an interval that contains the dividing line between two parts of a piecemeal function. Is this action defined in such a case? 