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If $F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt$

a.) The integral of $F(x)$ is $3x^2\sqrt{x^2+x^4}$.

b.) The derivative of $F(x)$ does not exist.

c.) $F'(x)=3x^2\sqrt{x^6+x^{12}}$.

I can't seem to find the answer. I found that $$F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt= \frac{(x^6+1)^{3/2}}{3}-\frac{5 \sqrt{5}}{3}$$

and $$\frac{d}{dx}F(x)=\int_2^{x^3}\sqrt{t^2+t^4}dt=3x^5\sqrt{x^6+1}$$

so, to me the answer is not provided. Any help would be much appreciated.

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    $\begingroup$ Please think about why you think you found a closed form expression for $F$. That integral has no nice answer - the problem is really about the FTC. What you've written suggests that there's some algebra you got wrong, independent of calculus. $\endgroup$ – Ethan Bolker Feb 21 at 23:37
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Let's say $G(x)$ is an antiderivative of $\sqrt{x^2+x^4}$. (we know about its existence from the fundamental theorem of calculus). So now from Newton-Leibniz formula we have $F(x)=G(x^3)-G(2)$. Now if we differentiate by $x$ using the chain rule we get:

$F'(x)=3x^2G'(x^3)=3x^2\sqrt{x^6+x^{12}}$

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  • $\begingroup$ Thank you Mark! $\endgroup$ – Project 2501 Feb 21 at 18:17
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Hint: If $F(x)= \int_a^{g(x)} f(t) dt$ , then $$F’(x) = f(g(x))g’(x)$$

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  • $\begingroup$ Thank you Jose! $\endgroup$ – Project 2501 Feb 21 at 18:17
  • $\begingroup$ Your welcome, notice that with this method is not necessary to calculate an undefined integral, which is useful because is not always possible (see $\int e^{-x^2} dx$) @RyanPennell $\endgroup$ – Ahlfkushevich Feb 21 at 18:57
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By the way, aside from the good answers provided by Mark and JoseSquare, note that you actually did compute $F(x)$ and $F'(x)$ correctly! The result you obtained for $F'(x)$ is equivalent to the expression given in choice (c).

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  • $\begingroup$ I believe the OP's answer for $F'(x)$ is only the same as choice (c) when $x\ge0$... $\endgroup$ – A. Howells Feb 22 at 2:08

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