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What is $\frac{d}{dx}\int_0^{x^2}f(t) dt$

My understanding is that if $F(t)$ is antiderivative of $f(t)$, then it should be $F(x^2)-F(0)$, but it is $f(x^2)(2x)$

The entire problem is as follow:

Find $f(4)$ if $\int_0^{x^2}f(t)\, dt=x\cos (\pi x)$

and the solution is

$\frac{d}{dx}\int_0^{x^2}f(t) \,dt=\cos \pi x -\pi x \sin \pi x\Rightarrow f(x^2)x=\cos \pi x -\pi x \sin \pi x \Rightarrow f(x^2)=\frac{\cos \pi x -\pi x \sin \pi x}{x}$

Thus $x=2\Rightarrow f(4)=1/4$

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  • $\begingroup$ Apparently it isn't a case of the Fundamental Theorm of calculus $\endgroup$ – Goldname Jan 20 '17 at 6:31
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Let $u=x^2$ and apply chain rule. $$\frac{d}{dx}\int_0^{u}f(t) dt = f(u) \frac{du}{dx}=f(x^2)\cdot2x.$$

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You were correct in assessing the situation in terms of an antiderivative of $f$, but you didn't follow through. If $F$ is an antiderivative of $f$, by the Fundamental Theorem of Calculus, $$ F(x^2)-F(0) = \int_0^{x^2}f(t)\,dt. $$ To take the derivative, we apply the chain rule since we have to differentiate the composite $F(x^2)$. Since $F(0)$ is a constant, we have $$ F'(x^2)(x^2)' = \frac{d}{dx}\int_0^{x^2}f(t)\,dt. $$ $F$ is an antiderivative of $f$, so $F'(x^2) = f(x^2)$. Putting it all together, $$ \frac{d}{dx}\int_0^{x^2}f(t)\,dt = 2x\,f(x^2). $$

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  • $\begingroup$ how do we know that F(0) is constant? $\endgroup$ – Vikram Jan 20 '17 at 7:13
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    $\begingroup$ @Vikram, $F\colon\Bbb R\to \Bbb R$ implies that $F(0) \in\Bbb R$. All numbers are constant. $\endgroup$ – Ennar Jan 20 '17 at 14:40
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As you said, if $f$ is continuous, then $\int_0^{x^2}f(t)dt=F(x^2)-F(0)$, so by taking the derivative on both sides using the chain rule one gets:

$$\frac{\partial}{\partial x}\int_0^{x^2}f(t)dt=F'(x^2)\cdot 2x=2xf(x^2)$$

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  • $\begingroup$ You mean $F'(x^2)\cdot 2x$. $\endgroup$ – Ennar Jan 20 '17 at 14:39
  • $\begingroup$ Yes, you are right. Corrected. $\endgroup$ – Momo Jan 20 '17 at 15:19
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You can use this general formula : $$\frac {d}{dx} \int_{u(x)}^{v(x)} f(t) dt = f(v(x))v'(x)-f(u(x))u'(x).$$

Note that $u(x)$ and $v(x)$ must be differentiable which is your case.

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Another (very slightly different) approach.

Start by setting $u=x^2$, and define a function $$F(u) = \int_0^u f(t)\space dt$$ The second fundamental theorem of calculus tells us that $\frac{dF}{du} =f(u) = f(x^2)$. But that's not what we need to compute. We are asked to find $\frac{dF}{dx}$. (Notice the different variable in the bottom of the derivative.)

Fortunately the chain rule comes to our rescue. We have $$\frac{dF}{dx} = \frac{dF}{du} \cdot \frac{du}{dx}$$ The second factor on the right tells you where the unexpected $2x$ in the solution comes from.

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