When is the area of this region bigger than the area of this one?

When is the area of this region bigger than the area of this one?

Question

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 $a$,

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1. $a > 1$,

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1. $a > \dfrac{6}{5}$,

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1. $a > \left(\dfrac{4}{3}\right)^{3/2}$,

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1. $a > \left(\dfrac{6}{5}\right)^4$.

Suggestion

Drawing a diagram might help to visualise what is happening here.

Can we use calculus to find the areas mentioned?

What do we know about areas below the $x$-axis?

Solution

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).