Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 54

Find the area bounded by the curves $y = 5 \ln x$, $y = x \ln x$



First, we need to get the upper and lower limits of the integral by simply getting the points of intersections of the curves. So,

$
\begin{equation}
\begin{aligned}
5 \ln x &= x \ln x\\
\\
5 \ln x &= x \ln x = 0\\
\\
\ln x( 5- x ) &= 0
\end{aligned}
\end{equation}
$


We have,
$\ln x = 0$ and $ 5- x = 0$
Solving for $x$
$e^{\ln x} = e^0$
$x =1$ and $x =5$

Then, by using vertical strips

$
\begin{equation}
\begin{aligned}
A &= \int^b_a \left( y_{\text{upper}} - y_{\text{lower}} \right) dx\\
\\
A &= \int^5_1 (5 \ln x - x \ln x ) dx\\
\\
A &= \int^5_1 \ln x (5 -x ) dx
\end{aligned}
\end{equation}
$


To evaluate the area, we must use integration by parts, so
If we let $u = \ln x$ and $dv = (5 -x) dx$. Then,
$\displaystyle du = \frac{1}{x} dx \text{ and } v = 5x - \frac{x^2}{2}$

So,

$
\begin{equation}
\begin{aligned}
A = \int^5_1 \ln x (5-x) dx &= uv - \int v du = (\ln x) \left( 5x - \frac{x^2}{2} \right) \int \left( 5x - \frac{x^2}{2} \right) \left( \frac{1}{x} \right) dx\\
\\
&= \ln x \left( 5x - \frac{x^2}{2} \right) - \int \left( 5 - \frac{x}{2} \right) dx\\
\\
&= \ln x \left( 5x - \frac{x^2}{2} \right) - \left(5x - \frac{x^2}{2} \left( \frac{1}{2} \right) \right)
\end{aligned}
\end{equation}
$


Evaluate from $x = 1$ to $x = 5$, we have...
$\displaystyle A = \frac{25}{2} \ln 5 - 14$ square units