Tuesday, October 14, 2014

Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 22

Evaluate $\displaystyle \int^9_4 \frac{\ln y}{\sqrt{y}} dy$ by using Integration by parts.
If we let $u = \ln y$ and $\displaystyle dv = \frac{dy}{\sqrt{y}}$,then
$\displaystyle du = \frac{1}{y} dy \text{ and } v = \int \frac{dy}{\sqrt{y}} = \frac{y^{-\frac{1}{2}+1} }{-\frac{1}{2}+1} = 2\sqrt{y}$

So,

$
\begin{equation}
\begin{aligned}
\int^9_4 \frac{\ln y}{\sqrt{y}} dy = uv - \int vdu &= 2 \sqrt{y} \ln y - \int \frac{2\sqrt{y}dy}{y}\\
\\
&= 2 \sqrt{y} \ln y - \int 2 y^{\left( \frac{1}{2}-1\right)} dy\\
\\
&= 2 \sqrt{y} \ln y - 2 \int y^{-\frac{1}{2}} dy\\
\\
&= 2 \sqrt{y} \ln y - 2 (2 \sqrt{y})\\
\\
&= \left[ 2 \sqrt{y} (\ln y -2) \right]^9_4\\
\\
&= \left[ 2\sqrt{9} (\ln(9)-2) \right] - \left[ 2\sqrt{4}(\ln 4-2) \right]\\
\\
&= 6 \ln 9 - 12 - 4 \ln 4 + 8\\
\\
&= 6 \ln 9 - 4 \ln 4 - 4\\
\\
&= \ln 9^6 - \ln 4^4 - 4\\
\\
&= \ln \left( \frac{9^6}{4^4} \right) - 4\\
\\
&= \ln \left( \frac{531441}{256} \right) - 4

\end{aligned}
\end{equation}
$

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...