# find the derivative of an integral

Find $f'(x)$ where $f(x)$ is the integral from ${\sqrt{x}}$ to $x$ of $e^x-e^{t^2} dt$

Is there an easy way to do this using the fundamental theorem of calculus because if I try to ingretate w.r.t $t$ then $e^{t^2}$ is a bit of a problem.

• – Harto Saarinen Oct 23 '14 at 20:46
• Perhaps write $\int_{\sqrt x}^x$ as $\int_0^x-\int_0^{\sqrt x}$? For the second integral, let $u=\sqrt x$ and remember that $\frac d{dx} f(u)=f'(u) u'$. – Akiva Weinberger Oct 23 '14 at 20:47
• see this – John Oct 23 '14 at 20:51
• @columbus8myhw can you explain a bit more please? I usually just find antiderivitives and sub in the limits – user108605 Oct 23 '14 at 20:53
• @columbus8myhw Doesn't quite work that way. You still need to evaluate $\int e^{t^2}\,dt$ without using FTC, which doesn't yield a closed form. – Marc Oct 23 '14 at 20:54

## 3 Answers

Apply differentiation under the integral sign, which is a formula derived from FTC: \begin{align*} \frac{d}{dx}\int_{\sqrt x}^x(e^x-e^{t^2})dt&=(e^x-e^{x^2})-(e^x-e^x)(\sqrt x)^\prime+\int_{\sqrt x}^x e^x dt \\&= (e^x-e^{x^2})+e^x(x-\sqrt x) \end{align*}

$\displaystyle\int_{\sqrt{x}}^{x}\left(e^x-e^{t^2}\right)dt=e^x(x-\sqrt{x})-\int_0^{x}e^{t^2}dt+\int_0^{\sqrt{x}}e^{t^2}dt$, so

$\displaystyle\frac{d}{dx}\int_{\sqrt{x}}^{x}\left(e^x-e^{t^2}\right)dt=e^{x}\left(1-\frac{1}{2\sqrt{x}}\right)+e^x(x-\sqrt{x})-e^{x^2}+e^{x}\cdot\frac{1}{2\sqrt{x}}$

$\hspace{.45 in}=e^x(1+x-\sqrt{x})-e^{x^2}$.

Yes there is. Assuming no pathological situations, the derivative of a definite integral is the difference of the integrand evaluated at the two endpoints. So $$\frac{d}{dx} \int_\sqrt{x}^x \left( e^x - e^{t^2} \right) dt = \left. e^x - \right) dt = \left. \left( e^x - e^{t^2} \right) \right|_{t=\sqrt{x}}^{t=x} = e^x - e^{x^2}$$

• Incorrect. Your answer assumes that the integrand is not a function of $x$, but in fact it is here. You need to add the term $\int_{\sqrt{x}}^x \partial ((e^x-e^{t^2})/\partial x) dt$. – Marc Oct 23 '14 at 21:00