Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 22

Sketch the region enclosed by the curves $\displaystyle y = \sin \left( \frac{\pi x}{2} \right)$, $ y = x$. Then find the area of the region.


By using vertical strips
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its point of intersection. Thus
$\displaystyle y = \sin \left( \frac{\pi x}{2} \right) = x$

By trial and error, we have $x = 1 $ and $x = -1$
Notice that the orientation between the graphs at $x < 0 $ differs from the graph at $x > 0 $. Let $A_1$ and $A_2$ be the area of the left most part and right most part respectively. So,


$
\begin{equation}
\begin{aligned}
A_1 &= \int^0_{-1} \left[ \left(x - \sin \left( \frac{\pi x}{2} \right) \right) \right] dx\\
\\
A_1 &= 0.1366 \text{ square units}
\end{aligned}
\end{equation}
$

For the area of the right most part.


$
\begin{equation}
\begin{aligned}
A_2 &= \int^1_0 \left[ \sin \left( \frac{\pi x}{2} \right) - x\right] dx \\
\\
A_2 &= 0.1366 \text{ square units}
\end{aligned}
\end{equation}
$

Therefore, the total area is $A_1 + A_2 = 0.1366 + 0.1366 = 0.2732$ square units.