Question

Evaluating integral by net area

Answer 1

The graph of $f(x)=2$ is a horizontal line. So the area under $f(x)$ on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.

The graph of $g(x)=\sqrt{64-x^2}$ is the upper half of a circle with radius $\sqrt{64}=8$. It's symmetric about $x=0$, so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is $\frac{\pi 8^2}4=16\pi$.

Then use the fact that the definite integral is distributive over sums, meaning

$\displaystyle\int_0^8f(x)+g(x)\,\mathrm dx=\int_0^8f(x)\,\mathrm dx+\int_0^8g(x)\,\mathrm dx=16+16\pi$