Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 10

Sketch the region enclosed by the curves $y = 1 + \sqrt{x}$, $\displaystyle y = \frac{(3+x)}{3}$. Then find the area of the region.


By using vertical strip,
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its points of intersection. Thus,

$
\begin{equation}
\begin{aligned}
1 + \sqrt{x} &= \frac{3+x}{3}\\
\\
3 + 3 \sqrt{x} &= 3+x\\
\\
3\sqrt{x} &= x \\
\\
9x &= x^2\\
\\
x^2 - 9x &= 0\\
\\
x(x-9) &= 0
\end{aligned}
\end{equation}
$

We have, $x = 0$ and $x =9$
Therefore,

$
\begin{equation}
\begin{aligned}
A &= \int^{9}_{0} \left[ (1+\sqrt{x}) - \left( \frac{3+x}{3} \right)\right] dx \\
\\
A &= \int^9_0 \left( \sqrt{x} - \frac{x}{3} \right) dx\\
\\
A &= \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^2}{2(3)} \right]^9_0\\
\\
A &= 4.5 \text{ square units}
\end{aligned}
\end{equation}
$