## Calculus: Early Transcendentals 8th Edition

$\frac{4}{ \pi }+\frac{4}{3}$
Step 1 of 3: Find the point of intersection of the curves $y= \left( sin\frac{ \pi x}{2} \right)$ and $y= x^{2}-2x$ When$x= 0$ , both the curves have value 0. When $x= 2$ , both the curves have value 0. The x-coordinates of the points of intersection are 0 and 2. Step 2 of 3: Sketch the curves. Find the area of the shaded region. Step 3 of 3: The area $A= \int _{0}^{2} \left[ sin \left( \frac{ \pi x}{2} \right) - \left( x^{2}-2x \right) \right] dx$ $A= \int _{0}^{2}sin \left( \frac{ \pi x}{2} \right) dx- \int _{0}^{2} \left( x^{2}-2x \right) dx$ $= \left[ -\frac{2}{ \pi }cos \left( \frac{ \pi x}{2} \right) \right] _{0}^{2}- \left[ \frac{x^{3}}{3}-x^{2} \right] _{0}^{2}$ [By the fundamental theorem of calculus, part 2] $= \left[ -\frac{2}{ \pi }cos \pi +\frac{2}{ \pi }cos \pi \right] - \left[ \frac{8}{3}-4 \right]$ $= \left[ \frac{2}{ \pi }+\frac{2}{ \pi } \right] - \left[ -\frac{4}{3} \right] =\frac{4}{ \pi }+\frac{4}{3}$ $A=\frac{4}{ \pi }+\frac{4}{3}$