Chapter 6, Problem 29RE

To determine

To calculate: the integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{2}{y+1}}dy$ .

Answer to Problem 29RE

  $2\ln 3$

Explanation of Solution

Given information: The integral ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{2}{y+1}}dy$

Formula used: Use integration formula ${\displaystyle \int \frac{1}{x}dx}=\ln x+c$

Calculation: Use property ${\displaystyle \int af\left(x\right)dx}=a{\displaystyle \int f\left(x\right)dx}$

  ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{2}{y+1}}dy=2{\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{y+1}}dy$

Substitute $y+1=t$

  $\Rightarrow dy=dt$

And limit of integration is $t=1$ to $t=3$

Hence,

  $\begin{array}{c}2{\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{y+1}}dy=2{\displaystyle \underset{1}{\overset{3}{\int }}\frac{1}{t}}dt\\ =2{\left[\ln \left(t\right)\right]}_1^3\\ =2\left[\ln \left(3\right)-\ln \left(1\right)\right]\left(\text{Substitute the upper and lower limit}\right)\\ =2\left[\ln \left(3\right)-0\right]\left(\text{Since, }\ln \left(1\right)=0\right)\\ =2\ln 3\end{array}$

Therefore, ${\displaystyle \underset{0}{\overset{2}{\int }}\frac{2}{y+1}}dy=2\ln 3$