Review question

# When does this definite integral have a minimum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7236

## 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.