Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises.

${\int }_0^1{\left(3x-1\right)}^{\frac{-2}{3}}dx$

Short answer

The value of the integral is ${\int }_0^1{\left(3x-1\right)}^{\frac{-2}{3}}dx=1+\sqrt[3]{2}$

Step-by-step solution

Step 1. Given information. 

The given integral is the following.

${\int }_0^1{\left(3x-1\right)}^{\frac{-2}{3}}dx$

Step 2. Value of integral. 

The function $f\left(x\right)={\left(3x-1\right)}^{\frac{-2}{3}}$has a vertical asymptote at $x=\frac{1}{3}.$

solve this improper integral by splitting at localid="1648875492622" $x=\frac{1}{3}.$

localid="1648876694124" $$\begin{gathered} {\int }_0^1{\left(3x-1\right)}^{\frac{-2}{3}}dx={\int }_0^{\frac{1}{3}}{\left(3x-1\right)}^{\frac{-2}{3}}dx+{\int }_{\frac{1}{3}}^1{\left(3x-1\right)}^{\frac{-2}{3}}dx \\ =\underset{b\to {\frac{1}{3}}^{-}}{\lim }{\int }_0^b{\left(3x-1\right)}^{\frac{-2}{3}}dx+\underset{a\to {\frac{1}{3}}^{+}}{\lim }{\int }_a^1{\left(3x-1\right)}^{\frac{-2}{3}}dx \\ =\underset{b\to {\frac{1}{3}}^{-}}{\lim }{\left[{\left(3x-1\right)}^{\frac{1}{3}}\right]}_0^b+\underset{a\to {\frac{1}{3}}^{+}}{\lim }{\left[{\left(3x-1\right)}^{\frac{1}{3}}\right]}_a^1 \\ =\underset{b\to {\frac{1}{3}}^{-}}{\lim }\left({\left(3b-1\right)}^{\frac{1}{3}}-\sqrt[3]{-1}\right)+\underset{a\to {\frac{1}{3}}^{+}}{\lim }\left(\sqrt[3]{2}-{\left(3a-1\right)}^{\frac{1}{3}}\right) \\ =0+1+\sqrt[3]{2}-0 \\ =1+\sqrt[3]{2} \end{gathered}$$

So the value of the integral is localid="1648876733115" ${\int }_0^1{\left(3x-1\right)}^{\frac{-2}{3}}dx=1+\sqrt[3]{2}.$