Differentiating a definite integral

By the fundamental theorem of calculus :

$\frac{\mathrm{d}I}{\mathrm{d}a} = 4 - 2^{a^2}$ $4 = 2^{a^2}$ $2 = a^2$ $a = \pm\sqrt{2}$

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We can also show this in the usual way of solving a definite integral:

$f(x) = 4 - 2^{x^2}$ $I(a) = \int_0^a f(x) \, \mathrm{d}x = F(a) - F(0)$

where $F(x)$ is an antiderivative of $f(x)$ , i.e. $F'(x) = f(x)$

$\frac{\mathrm{d}I}{\mathrm{d}a} = \frac{\mathrm{d}}{\mathrm{d}a} F(a) - \frac{\mathrm{d}}{\mathrm{d}a} F(0) = f(a) - 0 = 4 - 2^{a^2}$

I present a geometrical and intuitive approach.

This shows a sketch of the graph. As we progress along the $x$ -axis from $0$ in the positive direction, $I$ is increasing because we are taking a greater area above the $x$ axis into account in the integral. However, when we reach a zero of the function, the contribution of the area to the integral becomes negative and hence $I$ begins to decrease.

Therefore, there is a local maximum where $4-2^{a^2}=0\iff 2^{a^2}=4\impliedby \color{#20A900}{\boxed{a=\sqrt 2}}$ .