Review question

# When is the area of this region bigger than the area of this one? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8350

## Solution

Consider two functions $f(x) = a - x^2,$ $g(x) = x^4 - a.$

For precisely which values of $a > 0$ is the area of the region bounded by the $x$-axis and the curve $y = f(x)$ bigger than the area of the region bounded by the $x$-axis and curve $y = g(x)$?

1. all values of $a$,
2. $a > 1$,
3. $a > \dfrac{6}{5}$,
4. $a > \left(\dfrac{4}{3}\right)^{3/2}$,
5. $a > \left(\dfrac{6}{5}\right)^4$.

The area of the region bounded by the $x$-axis and the curve $y = f(x)$ is $\int_{-a^{1/2}}^{a^{1/2}} a - x^2 \:dx = \left[ax-\dfrac{x^3}{3}\right]_{-a^{1/2}}^{a^{1/2}} = \dfrac{4a^{3/2}}{3}.$

On the other hand the area of the region bounded by the $x$-axis and the curve $y = g(x)$ is $-\int_{-a^{1/4}}^{a^{1/4}} x^4 - a \:dx = -\left[\dfrac{x^5}{5}-ax\right]_{-a^{1/4}}^{a^{1/4}} = \dfrac{8a^{5/4}}{5}.$

Since $5/4<3/2$ it follows that when $a$ is small, $\dfrac{8a^{5/4}}{5} > \dfrac{4a^{3/2}}{3}$, but when $a$ is large, $\dfrac{8a^{5/4}}{5} < \dfrac{4a^{3/2}}{3}$.

So when are these two areas equal? When $\dfrac{8a^{5/4}}{5} = \dfrac{4a^{3/2}}{3} \iff a^{1/4}=\dfrac{6}{5} \iff a = \left(\dfrac{6}{5}\right)^4,$ and so the answer is (e).