Chapter A6, Problem 54E

To determine

To Evaluate: the integrals.

Answer to Problem 54E

The result is $0.636$

Explanation of Solution

Given:

  ${\displaystyle \underset{0}{\overset{\ln 2}{\int }}4e^{-\theta }sinh\theta d\theta }$

Calculation:

Therefore the integral can now be evaluated as,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\ln 2}{\int }}4e^{-\theta }sinh\theta .d\theta }={\displaystyle \underset{0}{\overset{\ln 2}{\int }}4e^{-\theta }\left(\frac{e^{\theta }-e^{-\theta }}{2}\right).d\theta }\\ ={\displaystyle \underset{0}{\overset{\ln 2}{\int }}\left(1-e^{2\theta }\right)}.d\theta \\ =2{\left[\frac{e^{-2\theta }}{2}+\theta \right]}_0^{\ln 2}\\ =2\left[\left(\frac{e^{-2\ln 2}}{2}+\ln 2\right)-\left(\frac{e^{-0}}{2}-0\right)\right]\\ =2\left(\frac{1}{8}+\ln 2-\frac{1}{2}\right)\\ =\frac{1}{4}-1+2\ln 2\\ \approx 0.636\end{array}$

Conclusion:

The result is $0.636$