When does this definite integral have a minimum?

When does this definite integral have a minimum?

Question

The smallest value of $I(a) = \displaystyle\int_0^1 \! (x^2 - a)^2 \, \mathrm{d}x,$ as $a$ varies, is

1. $\frac{3}{20}$,

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1. $\frac{4}{45}$,

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1. $\frac{7}{13}$,

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1. $1$.

Interactive graph

The smallest value of $I(a) = \displaystyle\int_0^1 \! (x^2 - a)^2 \, \mathrm{d}x,$ as $a$ varies, is…

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/ppMqEM5G/width/600/height/400/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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The orange curve here is $y = (x^2-a)^2$, while the white region is $I(a)$

Solution

$$\begin{align*} I(a) \hspace{2mm} & = \displaystyle\int_0^1 \! (x^2 - a)^2 \, \mathrm{d}x\\ & = \displaystyle\int_0^1 \! (x^4 - 2a x^2 + a^2) \, \mathrm{d}x \\ & = \left[ \frac{x^5}{5} - \frac{2ax^3}{3} + a^2 x\right]_0^1 \\ & = \frac{1}{5} - \frac{2a}{3} + a^2 \\ & = \frac{1}{5}+ \left(a - \frac{1}{3}\right)^2 -\frac{1}{9}\\ & = \left(a - \frac{1}{3}\right)^2 + \frac{4}{45}. \end{align*}$$

So the smallest possible value of $I(a)$ is $\frac{4}{45}$ (which occurs when $a=\frac{1}{3}$).

Hence the answer is (b).

Alternatively we can differentiate $I(a) = \frac{1}{5} - \frac{2a}{3} + a^2$ with respect to $a$, and put this equal to zero to get the same result.