Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 16

Sketch the region enclosed by the curves $y = x^3 - x$, $y = 3x$. 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

$
\begin{equation}
\begin{aligned}
x^3 - x &= 3x\\
\\
x^3 - 4x &= 0\\
\\
x\left( x^2 - 4 \right) &= 0
\end{aligned}
\end{equation}
$


we have, $x = 0$ and $x = 2$ and $x = -2$
Notice that the orientation between the graphs at $x < 0$ is different with $x > 0 $. Let $A$ be the area at $x < 0 $ and $A_2$ be the area at $x > 0$. So...

$
\begin{equation}
\begin{aligned}
A_1 &= \int^0_{-2} \left[ \left( x^3 - x \right) -3x\right] dx\\
\\
A_1 &= \int^0_{-2} \left[ x^3 - 4x \right] dx\\
\\
A_1 &= \left[ \frac{x^4}{4} - \frac{4x^2}{2} \right]^0_{-2}\\
\\
A_1 &= 4 \text{ square units}
\end{aligned}
\end{equation}
$


Similarly,

$
\begin{equation}
\begin{aligned}
A_2 &= \int^2_0 \left[ 3x - (x^3 - x) \right] dx\\
\\
A_2 &= \int^2_0 \left[ 4x - x^3 \right] dx\\
\\
A_2 &= \left[ \frac{4x^2}{2} - \frac{x^4}{4} \right]^2_0\\
\\
A_2 &= 4 \text{ square units}
\end{aligned}
\end{equation}
$


Therefore, the total area is $A_1 + A_2 = 4 + 4 = 8$ square units.