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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: $$ \frac{d}{dx}\int_0^x~x~s^2ds=\int_0^xs^2~ds+x^3 $$

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  • $\begingroup$ 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). $\endgroup$ Commented Oct 21, 2014 at 11:52
  • $\begingroup$ I agree with Raymond Manzoni, the 2nd integral is confusing. $\endgroup$
    – SDiv
    Commented Oct 21, 2014 at 11:56

2 Answers 2

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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".

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    $\begingroup$ I think it should be "Let $F$ be an antiderivative of $f$. $\endgroup$ Commented May 2, 2016 at 2:11
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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 $$\frac{\text{d}}{\text{d}x}\left( \int_{0}^{x}xt^2\text{d}t \right ) = f(x,x)(1)-f(x,0)(0)+\int_{0}^{x}t^2\text{d}t\\=x^3(1) - 0 + x^3/3 = \frac{4}{3}x^3.$$

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  • $\begingroup$ The wikipedia page is way more complete and thorough than the accepted answer, thanks for sharing it. $\endgroup$ Commented Oct 22, 2023 at 12:10

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