Question

The graph of the function f(x) is shown below.

Answer 1

The integral of $f(x)$ over the interval $[-a,a]$ corresponds to the area of a semicircle of radius $a$, so

$\displaystyle \int_{-a}^a f(x) \, dx = \frac{\pi a^2}2$

The integral over $[a,3a]$ corresponds to the negative area of a trapezoid with height $a$ and base lengths $2a$ and $a$, so

$\displaystyle \int_a^{3a} f(x) \, dx = -\frac{a(2a+a)}2 = -\frac{3a^2}2$

Then combining the integrals, we have

$\displaystyle \int_{-a}^{3a} f(x) \, dx = \frac{\pi a^2}2 + \left(-\frac{3a^2}2\right) = \frac{(\pi-3)a^2}2$

When $a=9$, the integral evaluates to

$\displaystyle \int_{-9}^{27} f(x) \, dx = \frac{(\pi-3)\times81}2 \approx \boxed{5.7345}$[/tex]