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I want to find the value of: $$\frac{d}{dx}\int_{\pi}^{x^2} \sin(t) \ \text{dt}$$ I have recently been taught Fundamental Theorem of Calculus Part $1$ (I learnt Part $2$ first I don't know why). I attempted to solve the integral like this: $$\frac{d}{dx}F(x^2)=\frac{d}{dx}\int_{\pi}^{x^2} \sin(t) \ \text{dt}$$ $$\frac{d}{dx}F(x^2)=\sin(x^2)$$ However, the correct answer is $2x\sin(x^2)$. Can someone please explain why?

P.S. If this is a dumb question I apologize. I don't really understand the first part of the theorem, so please forgive me.

Edit: People have told me to use the Chain Rule, but I do not see how $F(x^2)$ is a composition of two functions. I see one function; $F(x^2)$. I think that the only difficulty I am having right now is seeing how $F(x^2)$ is a composition of two functions. I am now worried that I won't get it because then I will not know what to do if the upper bound is not $x$. What if it is $2x$? Do I have to apply a chain rule to that as well? Please help!

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  • $\begingroup$ Chain Rule!${}{}{}{}$ $\endgroup$ – André Nicolas Mar 21 '14 at 18:07
  • $\begingroup$ @André I don't see how this is a composition of two functions please help $\endgroup$ – TrueDefault Mar 21 '14 at 18:08
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    $\begingroup$ Well. You have $x$. By the right upper corner of the $x$ you have a $2$. This is a function: the function that maps each number into its square. And $x^2$ is sorrounded by a pair of parentheses, and there is an $F$. Thisi is the other function. Then you have two functions: first apply the square and then apply $F$. And you want to differentiante $F$ respect to $x$. Since between $F$ and $x$ you have another function (the one that rises a number into its square), you must apply the Chian Rule. Or, if you prefer, the definition of derivative. $\endgroup$ – ajotatxe Mar 22 '14 at 0:17
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You have two functions here $f(x)=\sin(x)$ and $g(x)=x^2$ so $\frac{d}{dx}F(g(x))=2x\sin(x^2)$

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I assume $x>\sqrt\pi$.

You are to differentiate $F(x^2)-F(\pi)$. You have to apply the chain rule:

$$\frac d{dx}(F(x^2)-F(\pi))=2xF'(x^2)=2x\sin(x^2)$$

where $2x$ is the derivative of $x^2$.

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Hint: Let $G(u)=\int_{\pi}^u \sin(t)\,dt$. Then $F(x)=G(x^2)$. Now use the Chain Rule.

Remark: In this particular case you can do the computation without using the Fundamental Theorem, for we know how to integrate $\sin(t)$. We get $F(x)=-\cos(x^2)+\cos(\pi)$. In most problems of this type, you will not be able to find an explicit formula for the antiderivative.

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The Fundamental Theorem of Calculus doesn't say $${d\over dx}\int_a^{x^2}{f(t)\,dt} = f(x^2)\,,$$ it says $${d\over dx}\int_a^{x}{f(t)\,dt} = f(x)\,.$$ You are applying the theorem to a case it doesn't cover, so you are getting the wrong answer.

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  • $\begingroup$ I now understand why it is a composition of two functions, thanks! $\endgroup$ – TrueDefault Mar 21 '14 at 18:30
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if you define $f(u) = \int_\pi^u \sin(t) dt$, you know $f'(u) = \sin(u)$. In this case, your function is $f(x^2)$ and you want $d/dx [ f(x^2)]$.

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