The graph of [tex]f(x)=2[/tex] is a horizontal line. So the area under [tex]f(x)[/tex] on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.
The graph of [tex]g(x)=\sqrt{64-x^2}[/tex] is the upper half of a circle with radius [tex]\sqrt{64}=8[/tex]. It's symmetric about [tex]x=0[/tex], so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is [tex]\frac{\pi 8^2}4=16\pi[/tex].
Then use the fact that the definite integral is distributive over sums, meaning
[tex]\displaystyle\int_0^8f(x)+g(x)\,\mathrm dx=\int_0^8f(x)\,\mathrm dx+\int_0^8g(x)\,\mathrm dx=16+16\pi[/tex]
