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I'm confused about problems involving differentiation with respect to the limit of an integral, I just want to check that my understanding is correct.

For example, are the following statements correct? :

$$ \frac{d}{dx}\int_0^xs^2ds=x^2 $$

$$ \frac{d}{ds}\int_0^xs^2ds=\int_0^x2s~ds $$

and by the product rule: $$ \frac{d}{dx}\int_0^x~x~s^2ds=\int_0^xs^2~ds+x^3 $$

Git Gud
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WeakLearner
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  • In case of doubts you may go back to the bases. Your first answer appears right, the second doesn't make sense to me (integration variables are 'dummy' and the answer should be $0$), the third should be right (you may too put $x$ out of the integral). – Raymond Manzoni Oct 21 '14 at 11:52
  • I agree with Raymond Manzoni, the 2nd integral is confusing. – SDiv Oct 21 '14 at 11:56

2 Answers2

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In general;

$$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f(t) \mathrm{d}t = f(x)$$

$$\frac{\mathrm{d}}{\mathrm{d}t}\int_0^x f(t) \mathrm{d}t = \color{red}{0}$$

Now let's spend some time on this as this is where you did a mistake. Note that $$y(?)=\int_0^x f(t) \mathrm{d}t$$ is a function of $x$!($?=x$)Why? Remember how you deal with definite integrals. You find an antiderivative, then substract the lower bound from the upper. Formalizing this, let's denote $F$ an antiderivative of $f$. Then $$\int_a^b f(x) \mathrm{d}x=F(b)-F(a)$$

If you do this with yours, what do you get? $F(x)-F(a)$. What does this mean? This means the result is a function of $x$. So what right? Well, what happens when you differentiate a function with respect to something it is not related? You treat it as a constant. What happens when you differentiate a constant? Well you get $0$. So, $$\frac{\mathrm{d}}{\mathrm{d}s} \int_0^x f(s) \mathrm{d}s =\frac{\mathrm{d}}{\mathrm{d}s} \left ( F(x)-F(0) \right ) =0$$

As for the third, your approach is dead-on. I don't know how you treated $x$ though. Since you're integrating wrt to $s$ you can treat $x$ as a constant and "pull it out of the integral".

UserX
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There is a nice Wikipedia page on this: Differentiation under the integral sign.

Direct from that page we have $$\frac{\text{d}}{\text{d}x}\left( \int_{a(x)}^{b(x)}f(x,t)\text{d}t \right ) = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}f_x(x,t)\text{d}t.$$

Then for $f(x,t) = t^2$ we have $$\frac{\text{d}}{\text{d}x}\left( \int_{0}^{x}t^2\text{d}t \right ) = f(x,x)(1)-f(x,0)(0)+\int_{0}^{x}0\text{d}t\\=x^2(1) - 0 + 0 = x^2.$$

For $f(x,t)=xt^2$ we have $$\frac{\text{d}}{\text{d}x}\left( \int_{0}^{x}xt^2\text{d}t \right ) = f(x,x)(1)-f(x,0)(0)+\int_{0}^{x}t^2\text{d}t\\=x^3(1) - 0 + x^3/3 = \frac{4}{3}x^3.$$

SDiv
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