Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 56

Find the area bounded by the curves $y = \arctan 3x$ and $\displaystyle y = \frac{1}{2} x$ by approximating the $x$-coordiantes of the points of intersections.



Based from the graph, we can estimate the $x$-coordinates of the points of intersections as $ x \approx - 2.90$ and $x \approx 2.90$. Since both graphs are symmetric to the origin, we can simply evaluate the half region and multiply it by two toget the area of the entire region. So,


$
\begin{equation}
\begin{aligned}
A &= 2 \int^{2.90}_0 \left(y_{\text{upper}} - y_{\text{lower}} \right) dx\\
\\
A &= 2 \int^{2.90}_0 \left(\arctan (3x) - \frac{x}{2}\right) dx\\
\\
A &= 2 \left[ \int^{2.90}_0 \arctan (3x) dx - \int^{2.90}_0 \frac{x}{2} dx \right]\\
\\
A &= 2.7953 \text{ square units}
\end{aligned}
\end{equation}
$