Q. 72 - Calculus

Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

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

Step-by-Step Solution

Answer

The solution of the given integral is ${\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{2}{3}\left[2-\sqrt{10}\right]$.

Step 1. Given Information

Solving the given integrals. 

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

Step 2. Using the substitution method.

Let

$$\begin{gathered} u=1-3x \\ \frac{du}{dx}=-3 \\ du=-3dx \\ -\frac{1}{3}du=dx \end{gathered}$$

Step 3. We will now write the limits of integration in terms of the new variable u .

When $x=-3$, we have

$$\begin{gathered} u=1-3x \\ u=1-3(-3) \\ u=1+9 \\ u=10 \end{gathered}$$

When $x=-1$, we have

$$\begin{gathered} u=1-3x \\ u=1-3(-1) \\ u=1+3 \\ u=4 \end{gathered}$$

Step 4. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}{\int }_{10}^4\frac{1}{\sqrt{u}}du \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}{\int }_{10}^4\frac{1}{u^{1/2}}du \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}{\int }_{10}^4{u}^{-1/2}du \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}{\left[\frac{u^{-1/2+1}}{-1/2+1}\right]}_{10}^4 \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}{\left[\frac{u^{1/2}}{1/2}\right]}_{10}^4 \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{1}{3}·2{\left[u^{1/2}\right]}_{10}^4 \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{2}{3}{\left[\sqrt{u}\right]}_{10}^4 \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{2}{3}\left[\sqrt{4}-\sqrt{10}\right] \\ {\int }_{-3}^{-1}\frac{1}{\sqrt{1-3x}}dx=-\frac{2}{3}\left[2-\sqrt{10}\right] \end{gathered}$$