Question

Evaluate the integral\(\int\limits_1^3 {{r^3}\ln rdr} \)

Short answer

We will use theintegration by part method to solve this given integral

\(\int {\mu d\nu = \mu \nu } - \int {\nu .} d\mu \)

Step-by-step solution

given data

Given = \({\rm I} = \int\limits_1^3 {{r^3}\ln rdr} \)

Let \(\mu = {r^3}\)\(d\mu = 3{r^3}dr\)

\(\int {d\nu = \int {\ln rdr} } \) \(\nu = \frac{1}{r}\)

use integration by part

\(\int\limits_1^3 {{r^3}\ln rdr} = {\left( {{r^3}\left( {\frac{1}{r}} \right)} \right)_1} - \int {\frac{1}{r}} \left( {3{r^2}dr} \right)\)

\(\begin{aligned}{l} = \left( {{r^2}} \right)_1^3 - 3\int\limits_1^3 {{r^2}dr} \\ = \left( {{3^2} - {1^2}} \right) - \frac{3}{3}\left( {{r^3}} \right)_1^3\\ = \left( {9 - 1} \right) - \left( {27 - 1} \right)\\ = 8 - 26\\ = - 18\end{aligned}\)

Hence, \(\int\limits_1^3 {{r^3}\ln rdr} = - 18\)