Question

Evaluating more integrals

Answer 1

(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 $\frac{\pi10^2}2=50\pi$, so the integral is -50π.

(c) First compute

$\displaystyle\int_{30}^{35}g(x)\,\mathrm dx$

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

$\displaystyle\int_0^{35}g(x)\,\mathrm dx=\int_0^{10}g(x)\,\mathrm dx+\int_{10}^{30}g(x)\,\mathrm dx+\int_{30}^{35}g(x)\,\mathrm dx$

and plug in the integral values you know:

$\displaystyle\int_0^{35}g(x)\,\mathrm dx=100-50\pi+\frac{25}2=\frac{225}2-50\pi$