Question

Multiple-Choice Integration, Picture Included, Please Include Work

Answer 1

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx$

Recall that a circle of radius 2 centered at the origin has equation

$x^2+y^2=4\implies y=\pm\sqrt{4-x^2}$

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius $r$ is $\pi r^2$, it follows that the area of a quarter-circle would be $\dfrac{\pi r^2}4$.

You have $r=2$, so the definite integral is equal to $\dfrac{2^2\pi}4=\pi$.

Another way to verify this is to actually compute the integral. Let $x=2\sin u$, so that $\mathrm dx=2\cos u\,\mathrm du$. Now

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du$

Recall the half-angle identity for cosine:

$\cos^2u=\dfrac{1+\cos2u}2$

This means the integral is equivalent to

$\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi$