# Prove that $\forall a>0$ $\exists c\in [0,1]$ such that $\int_0^a e^{x^2} dx =\frac{c}{a}(e^{a^2}-1)$

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.

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 $$c\in (0,1).$$

• Very elegant (+1)
– an4s
Feb 18, 2020 at 21:30
• @an4s Thank you.
– 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

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