Answer
$ \frac{4}{ \pi }+\frac{4}{3} $
Work Step by Step
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} $
