(a) The integral is equal to the area of the triangle; it has height 20 and base 10, so the area is 20*10/2 = 100.
(b) The integral is equal to the area of the semicircle with radius 10. It's also under the horizontal axis, so the area is negative. The semicircle has area [tex]\frac{\pi10^2}2=50\pi[/tex], so the integral is -50π.
(c) First compute
[tex]\displaystyle\int_{30}^{35}g(x)\,\mathrm dx[/tex]
which is the area of the triangle on the right. It has height and base 5, so its area is 25/2.
Then split up the desired integral as
[tex]\displaystyle\int_0^{35}g(x)\,\mathrm dx=\int_0^{10}g(x)\,\mathrm dx+\int_{10}^{30}g(x)\,\mathrm dx+\int_{30}^{35}g(x)\,\mathrm dx[/tex]
and plug in the integral values you know:
[tex]\displaystyle\int_0^{35}g(x)\,\mathrm dx=100-50\pi+\frac{25}2=\frac{225}2-50\pi[/tex]
