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)=√(64-x^2)` is the upper half of a circle with radius
`√(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`