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## anonymous 5 years ago use part 1 of the fundamental theorem of calculus to find the derivative of: Int. from x to 4 of sin(t^3)dt

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1. anonymous

you have to start with integrating with respect to t can you do that?

2. anonymous

You want to differentiate with respect to x, right? You don't have to integrate a sin(t^3). Use the idea that differentiation is the inverse of integration. $d/dx[\int\limits_{x}^{4}f'(t)dt]=d/dx[f(4)-f(x)]$ I let f'(t) = sin(t^3). Since sin(t^3) is a continuous function, it must have an antiderivative (we just don't know what it is, nor do we care). We then plug 4 into that antiderivative and subtract f(x) from it. $d/dx[f(4)-f(x)]=0-f'(x)*1=-f'(x)$ Plug x in for t in f'(t) to get your answer: -sin(x^3)

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