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2011 Calculus AB free response #4a

Taking derivatives and integrals of strangely defined functions. Created by Sal Khan.

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• At we try to derive the integral evaluated from 0 to x. Would we not want to subtract the lower limit of f(0) from f(x)? Why is g'(x)=2+f(x) and not g'(x)=2+(f(x)-f(0))? This changes the answer since f(0) has a value.
• When you integrate, you get something like F(x) - F(0). F(x) is a function of x, while F(0) is an evaluation of F(x) at x = 0, which is some constant number. When you derive F(x) - F(0), d[F(x)]/dx = f(x), while d[F(0)]/dx = 0, since the derivative of a constant is zero. This equates to f(x) + 0 = f(x). Therefore, g'(x) = 2 + [f(x) - 0] = 2 + f(x).
• Why does the derivative of f(t) is f(x)?
(1 vote)
• because of the first FTC. When you have an integral from a to x of some function with a different variable, in this case f(t) you simply replace it with x to make f(x) when you take the antiderivative.