Prove that $\forall a>0$ $\exists c\in [0,1]$ such that $$\int_0^a e^{x^2} dx =\frac{c}{a}(e^{a^2}-1)$$ I initially tried to use Cauchy's Mean Value Theorem, but I wasn't successful. The fact that we have $e^{a^2}-1$, which is precisely $e^{x^2} \bigg |_{0}^{a}$, made me think about IBP, but this didn't work out either.


2 Answers 2


For $a\ge 0,$ define

$$f(a) = a\int_0^a e^{x^2}\,dx + 1 - e^{a^2}.$$

Then $f(0)=0$ and

$$f'(a) = \int_0^a e^{x^2}\,dx + ae^{a^2} - 2ae^{a^2} = \int_0^a e^{x^2}\,dx - ae^{a^2}.$$

Clearly the last expression is negative for $a>0.$ Thus $f$ is strictly decreasing on $[0,\infty).$ Hence $f(a)<0$ for $a>0.$ This implies

$$\tag 1 \int_0^a e^{x^2}\,dx < \frac{e^{a^2}-1}{a},\,a>0.$$

It follows that the left side of $(1)$ equals $c$ times the right side of $(1)$ for some $c\in (0,1).$

  • $\begingroup$ Very elegant (+1) $\endgroup$
    – an4s
    Feb 18, 2020 at 21:30
  • $\begingroup$ @an4s Thank you. $\endgroup$
    – zhw.
    Feb 19, 2020 at 16:04

Express $c$ as follows,

$$c =\frac {\int_0^a e^{x^2} dx }{\frac1a(e^{a^2}-1)} =\frac {\int_0^a e^{x^2} dx }{\int_0^ae^{ax}dx} =\frac {\int_0^1 h(t) e^{a^2t}dt }{\int_0^1e^{a^2t}dt}$$

where $h(t)=e^{-a^2t(1-t)}$ and the substitute $x = at$ is made in the last step. For $0\le t \le 1$ and $a>0$, we have

$$0<e^{-a^2t(1-t)} \le 1$$

which leads to

$$0<\int_0^1 h(t) e^{a^2t}dt \le \int_0^1e^{a^2t}dt$$

and, hence,

$$0 < c \le 1$$


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