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Question:

Let $$F(x) =\int_{x^3}^{5}(cos^2t-te^t)dt $$

Find $F'(x)$

We were not explicitly taught about this during the semester but from what I can gather from online readings is that

$$F'(x)=F'(5)-F'(x^3)$$

Therefore,

$$F'(x)=(cos^2(5)-5e^5)-(cos^2(x^3)-x^3e^{x^3})$$

Is this correct? Thank you for any help!

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1 Answer

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You're on the right track. This is an application of the fundamental theorem of calculus. In general,

$$\frac{d}{dx}\int_{f(x)}^{g(x)} h(t)\,dt = h(g(x))g'(x)-h(f(x))f'(x).$$

In your case, $f(x) = x^3$, $g(x) = 5$ and $h(x) = \cos^2(x)-xe^x$.

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The way to see that the equality above holds is to note that if $H$ is the antiderivative of $h$,

$$\int h(t)\,dt = H(t)+C$$

Then

$$\int_{f(x)}^{g(x)} h(t)\,dt = H(g(x))-H(f(x)).$$

Taking a derivative gives

$$\frac{d}{dx}\int_{f(x)}^{g(x)} h(t)\,dt = \frac{d}{dx}\left(H(g(x))-H(f(x))\right) = H'(g(x))g'(x)-H'(f(x))f'(x).$$

However since $H$ is the antiderivative of $h$, $H' = h$ so we get that

$$\frac{d}{dx}\int_{f(x)}^{g(x)} h(t)\,dt = h(g(x))g'(x) - h(f(x))f'(x)$$

as desired.

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