Solution

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

  1. \(\frac{3}{20}\),

  2. \(\frac{4}{45}\),

  3. \(\frac{7}{13}\),

  4. \(1\).

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

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.