Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 12

Sketch the region enclosed by the curves $y = x^2$, $y = 4x - x^2$. 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^2 &= 4x - x^2\\
\\
-2x^2 + 4x &= 0 \\
\\
2x (-x + 2) &= 0
\end{aligned}
\end{equation}
$

we have, $x = 0$ and $x = 2$
Therefore,

$
\begin{equation}
\begin{aligned}
A &= \int^2_0 \left[ \left( 4x - x^2\right) - x^2\right] dx\\
\\
A &= \int^2_0 \left( 4x - 2x^2\right) dx\\
\\
A &= \left[ \frac{4x^2}{2} - \frac{2x^3}{3} \right]^2_0\\
\\
A &= \frac{8}{3} \text{ square units}
\end{aligned}
\end{equation}
$