## Thomas' Calculus 13th Edition

a. $\frac{\pi}{4}-\frac{3}{2}$ b. $\frac{\pi}{2}$
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 $\int_{-1}^{1}(3x+ \sqrt {1-x^2})dx=\frac{3}{2}(1)^2-\frac{3}{2}(-1)^2+\frac{1}{2}\pi (1)^2=\frac{\pi}{2}$