0

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.

user108605
  • 621
  • See e.g. http://math.stackexchange.com/questions/6155/derivative-of-integral – 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
  • @user108605 https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign – Marc Oct 23 '14 at 20:54
  • @Marc I'm fairly certain that you don't need to evaluate it! You just need to know that an antiderivative exists (elementary or not), and then you can use the FTC—specifically, $\frac d{dx} \int_0^xf(t)dt=f(x)$. – Akiva Weinberger Oct 23 '14 at 21:00
  • @columbus8myhw I was under the impression that you are trying to symbolically integrate the function. In any case, when you do apply the FTC, you'll find out that you don't really need to integrate $\sqrt x$ anyway as it has a zero coefficient. Some care is, however, necessary, since the integrand is a function of $x$ as well as $t$. – Marc Oct 23 '14 at 21:03

3 Answers3

2

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*}

Marc
  • 965
  • 4
  • 13
1

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

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

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

user84413
  • 27,211
-1

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 $$\frac{d}{dx} \int_\sqrt{x}^x \left( e^x - e^{t^2} \right) dt = \left. e^x - \right) dt = \left. \left( e^x - e^{t^2} \right) \right|_{t=\sqrt{x}}^{t=x} = e^x - e^{x^2} $$

Mark Fischler
  • 41,743
  • 1
    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