Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 8

Sketch the region enclosed by the curves $y = x^2 - 2x$, $y = x+4$. Then find the area of the region.



By using a vertical strip,

$
\begin{equation}
\begin{aligned}
A &= \int^{x_2}_{x_1} (y_{\text{upper}} y_{\text{lower}}) dx\\
\\
\end{aligned}
\end{equation}
$


In order to get the value of the upper and lower limits, we equate the two functions to get its points of intersection, so...

$
\begin{equation}
\begin{aligned}
x^2 - 2x = x + 4\\
\\
x^2 - 3x - 4 = 0
\end{aligned}
\end{equation}
$

By using quadratic formula, we get
$x = 4 $ and $ x = -1$

So, we have

$
\begin{equation}
\begin{aligned}
A &= \int^4_{-1} \left((x+4) - (x^2 - 2x) \right) dx\\
\\
A &= \int^4_{-1} \left( -x^2 + 3x + 4 \right) dx\\
\\
A &= \left[ \frac{-x^3}{3} + \frac{3x^2}{2} + 4x \right]^4_{-1}\\
\\
A &= 20.8333 \text{ square units}
\end{aligned}
\end{equation}
$