Q. 23 - Calculus

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.) 

${\int }_2^5\frac{1}{\sqrt{x^5}}dx$

Step-by-Step Solution

Answer

The required value is $\frac{-2}{3}{\left(5\right)}^{-\frac{3}{2}}+\frac{2}{3}{\left(2\right)}^{-\frac{3}{2}}$.

Step 1. Given Information

We are given the definite integral ${\int }_2^5\frac{1}{\sqrt{x^5}}dx$ and we need to use the Fundamental Theorem of Calculus to find the exact value of the integral.     

Step 2. Finding the integral

The required value is:  

$$\begin{gathered} {\int }_2^5\frac{1}{\sqrt{x^5}}dx \\ ={\int }_2^5{x}^{-\frac{5}{2}}dx \\ ={\left[\frac{x^{-\frac{5}{2}+1}}{-\frac{5}{2}+1}\right]}_2^5 \\ ={\left[\frac{x^{-\frac{3}{2}}}{-{\displaystyle \frac{3}{2}}}\right]}_2^5 \\ =-\frac{2}{3}{\left(5\right)}^{-\frac{3}{2}}+\frac{2}{3}{\left(2\right)}^{-\frac{3}{2}} \end{gathered}$$

Step 3. Rechecking solution

The required graph is:  

After plotting the graph we can see that the area under the graph is exactly same as the area obtained from the definite integral. The area under the graph is $\frac{-2}{3}{\left(5\right)}^{-\frac{3}{2}}+\frac{2}{3}{\left(2\right)}^{-\frac{3}{2}}$ square units.