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.)

${\int }_1^3\frac{\sqrt{x}-1}{\sqrt{x}}dx$

Short answer

The required value is$4-2\sqrt{3}$.

Step-by-step solution

Step 1. Given Information   

We are given the definite integral ${\int }_1^3\frac{\sqrt{x}-1}{\sqrt{x}}dx$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:

$$\begin{gathered} {\int }_1^3\frac{\sqrt{x}-1}{\sqrt{x}}dx \\ ={\int }_1^3\left(1-\frac{1}{\sqrt{x}}\right)dx \\ ={\int }_1^{3}dx-{\int }_1^3\frac{1}{\sqrt{x}}dx \\ ={\left[x\right]}_1^3-{\int }_1^3{x}^{-\frac{1}{2}}dx \\ =\left(3-1\right)-{\left[\frac{x^{{\displaystyle \frac{1}{2}}}}{\frac{1}{2}}\right]}_1^3 \\ =2-2\left({\left(3\right)}^{\frac{1}{2}}-{\left(1\right)}^{\frac{1}{2}}\right) \\ =2-2\sqrt{3}+2 \\ =4-2\sqrt{3} \end{gathered}$$

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 $4-2\sqrt{3}$square units.