Single Variable Calculus

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

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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} $

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