Q. 26 - 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.) 

${\int }_1^{3}2x\sqrt{x^2+1}$

Step-by-Step Solution

Answer

The required value is $\frac{2}{3}{\left(10\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(2\right)}^{\frac{3}{2}}$.

Step 1. Given Information

We are given the definite integral ${\int }_1^{3}2x\sqrt{x^2+1}$ 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:  

$$\begin{gathered} {\int }_1^{3}2x\sqrt{x^2+1} \\ ={\left[\frac{2}{3}{\left(x^2+1\right)}^{\frac{3}{2}}\right]}_1^3 \\ =\left[\frac{2}{3}{\left(3^2+1\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(1+1\right)}^{\frac{3}{2}}\right] \\ =\left[\frac{2}{3}{\left(10\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(2\right)}^{\frac{3}{2}}\right] \end{gathered}$$

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 $\frac{2}{3}{\left(10\right)}^{\frac{3}{2}}-\frac{2}{3}{\left(2\right)}^{\frac{3}{2}}$ square units.