Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_0^4\frac{1}{{(2x+1)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx$

Short answer

${\int }_0^4\frac{1}{{(2x+1)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx=\frac{26}{81}$.

Step-by-step solution

Step 1. Given information.

A definite integral is given as${\int }_0^4\frac{1}{{(2x+1)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx$.

Step 2. Using the Fundamental theorem of Calculus.

We get

$$\begin{gathered} {\int }_0^4\frac{1}{{(2x+1)}^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dx \\ ={\int }_0^4{(2x+1)}^{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx \\ =\frac{1}{2}{\left[\frac{{(2x+1)}^{-\frac{5}{2}+1}}{-\frac{5}{2}+1}\right]}_0^4 \\ =\frac{1}{2}{\left[\frac{{(2x+1)}^{-\frac{3}{2}}}{-\frac{3}{2}}\right]}_0^4 \\ =\frac{1}{2}(-\frac{2}{3}){\left[\frac{1}{{(2x+1)}^{\frac{3}{2}}}\right]}_0^4 \\ =-\frac{1}{3}(\frac{1}{9^{\frac{3}{2}}}-1) \\ =-\frac{1}{3}(\frac{1}{27}-1) \\ =-\frac{1}{3}(-\frac{26}{27}) \\ =\frac{26}{81} \end{gathered}$$

So the exact value of the given definite integral is$\frac{26}{81}$.

Step 3. The graph to verify the answer is

The solution is area under graph which is

$$\begin{gathered} a\approx 0.320987 \\ \approx \frac{26}{81} \end{gathered}$$