Answer
a. $\frac{\pi}{4}-\frac{3}{2}$
b. $\frac{\pi}{2}$
Work Step by Step
a. We need two formulas: $\int_{a}^b xdx=\frac{1}{2}(b)^2-\frac{1}{2}(a)^2, \int_{-1}^0 xdx=\frac{1}{2}(0)^2-\frac{1}{2}(-1)^2=-\frac{1}{2}$ representing the area under the line $y=x$
Since $y= \sqrt {1-x^2}$ represents a semicircle as shown in the figure, we have
$\int_{-1}^0 \sqrt {1-x^2}dx=\frac{1}{4}\pi (1)^2=\frac{\pi}{4}$
(quarter of the unit circle area).
Thus
$\int_{-1}^0(3x+ \sqrt {1-x^2})dx=-\frac{3}{2}+\frac{\pi}{4}$
b. Repeat part (a), except that the second area is half of a circle; we have
$\int_{-1}^{1}(3x+ \sqrt {1-x^2})dx=\frac{3}{2}(1)^2-\frac{3}{2}(-1)^2+\frac{1}{2}\pi (1)^2=\frac{\pi}{2}$
