The integral of [tex]f(x)[/tex] over the interval [tex][-a,a][/tex] corresponds to the area of a semicircle of radius [tex]a[/tex], so
[tex]\displaystyle \int_{-a}^a f(x) \, dx = \frac{\pi a^2}2[/tex]
The integral over [tex][a,3a][/tex] corresponds to the negative area of a trapezoid with height [tex]a[/tex] and base lengths [tex]2a[/tex] and [tex]a[/tex], so
[tex]\displaystyle \int_a^{3a} f(x) \, dx = -\frac{a(2a+a)}2 = -\frac{3a^2}2[/tex]
Then combining the integrals, we have
[tex]\displaystyle \int_{-a}^{3a} f(x) \, dx = \frac{\pi a^2}2 + \left(-\frac{3a^2}2\right) = \frac{(\pi-3)a^2}2[/tex]
When [tex]a=9[/tex], the integral evaluates to
[tex]\displaystyle \int_{-9}^{27} f(x) \, dx = \frac{(\pi-3)\times81}2 \approx \boxed{5.7345}[/tex][/tex]
