2
$\begingroup$

Find $f'(0)$ and $f'(\sqrt{3})$ when $f(x)=\int_{x}^{x^3} e^{t^2} dt$

I know that $\int_{}^{} e^{t^2} dt$ doesn't return elementary functions, but I am not sure how to solve this. Can anyone give a little hint?

$\endgroup$
4
$\begingroup$

Recall by the Fundamental Theorem of Calculus, $$\frac{d}{dx}\int_a^xB(t)dt=B(x),\space\text{for $a\in \mathbb{R}$}$$

So if we have $$I=\int_{g(x)}^{h(x)}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt+\int_{g(x)}^{a}B(t)\space dt=\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt$$

We want to differentiate $I$, namely $$\frac{dI}{dx}=\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt-\int_{a}^{g(x)}B(t)\space dt \right)=\underbrace{\frac{d}{dx}\left(\int_{a}^{h(x)}B(t)\space dt\right)}_{\frac{dI_1}{dx}}-\underbrace{\frac{d}{dx}\left(\int_{a}^{g(x)}B(t)\space dt\right)}_{\frac{dI_2}{dx}}$$

So $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}$$

Calculate separately:

$$\text{For $\frac{dI_1}{dx}$, let $u=h(x)$ so that $\frac{du}{dx}=\frac{dh}{dx}$}$$

Hence $$\frac{dI_1}{dx}=\frac{d}{du}\left(\int_{h(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(h(x))\cdot h'(x)$$

And similarly, $$\text{For $\frac{dI_2}{dx}$, let $u=g(x)$ so that $\frac{du}{dx}=\frac{dg}{dx}$}$$

Hence $$\frac{dI_2}{dx}=\frac{d}{du}\left(\int_{g(a)}^u B(t)\space dt\right)\cdot\frac{du}{dx}=B(u)\cdot\frac{du}{dx}=B(g(x))\cdot g'(x)$$ $$$$

We conclude $$\frac{dI}{dx}=\frac{dI_1}{dx}-\frac{dI_2}{dx}=B(h(x))\cdot h'(x)-B(g(x))\cdot g'(x)$$ or that $$$$

$$\frac{d}{dx}\left(f(x)\right)=\frac{d}{dx}\left(\int_{g(x)}^{h(x)}B(t)\space dt\right)=h'(x)B(h(x))-g'(x)B(g(x))$$

Note in your case,

  • $f(x)=\int_x^{x^3}e^{t^2}dt$
  • $h(x)=x^3$
  • $g(x)=x$
  • $B(t)=e^{t^2}$

So really, you're not trying to find an antiderivative of $y_1=e^{x^2}$, but a derivative of $y_2=\int_{g(x)}^{h(x)}e^{t^2}dt$.

$\endgroup$
0
$\begingroup$

Hint:

$\int_{u}^{v} f'(t)\mathrm dt=f(v)-f(u)$, we get that $\left(\int_{u}^{v}f'(t)\mathrm dt\right)'=f'(v)v'-f'(u)u'$. Here $u$ and $v$ are functions of $x$. Using this fact which is popularly called Leibniz Integral Rule, you can get the required result by setting $f(t)=e^{t^2}$, $u(x)=x^2$ and $v(x)=x^3$ and you'll be done in no time.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.