Question

How do you evaluate the integral $\int{\dfrac{1}{{{\left( x-1 \right)}^{\dfrac{2}{3}}}}}$ from [0,2]?

Answer 1

Hint: In this question, we have to find the value of a definite integral. Thus, we will apply the integration formula and the basic mathematical rules to get the solution. First, we will rewrite the given integral in the form of$\int\limits_{0}^{t}{{{x}^{-m}}dx}$ . Then, we will apply the integration formula $\int\limits_{0}^{t}{{{x}^{-m}}dx}=\left[ \dfrac{{{x}^{-m+1}}}{-m+1} \right]_{0}^{t}$ in the new integral. After that, we will substitute the value of limits in x using the limit formula $\int\limits_{a}^{b}{f(x)dx}=f(b)-f(a)$ . In the end, we will make the necessary calculations to get the required result for the solution.

Complete step by step solution:
According to the problem, we have to find the value of definite integral.
Thus, we will apply the integration formula and the basic mathematical rules to get the solution.
The definite integral given to us is $\int{\dfrac{1}{{{\left( x-1 \right)}^{\dfrac{2}{3}}}}}$ from [0,2] ------------- (1)
First, we will rewrite the value of expression (1), we get
$\Rightarrow \int\limits_{0}^{2}{{{\left( x-1 \right)}^{-\dfrac{2}{3}}}dx}$
Now, we will apply the formula $\int\limits_{0}^{2}{{{x}^{-m}}dx}=\left[ \dfrac{{{x}^{-m+1}}}{-m+1} \right]_{0}^{2}$ in the above integral, we get
$\Rightarrow \left[ \dfrac{{{\left( x-1 \right)}^{-\dfrac{2}{3}+1}}}{-\dfrac{2}{3}+1} \right]_{0}^{2}$
Now, we will take the least common multiple of the denominator in the above expression, we get
$\Rightarrow \left[ \dfrac{{{\left( x-1 \right)}^{\dfrac{-2+3}{3}}}}{\dfrac{-2+3}{3}} \right]_{0}^{2}$
On further simplify the above expression, we get
$\Rightarrow \left[ \dfrac{{{\left( x-1 \right)}^{\dfrac{1}{3}}}}{\dfrac{1}{3}} \right]_{0}^{2}$
$\Rightarrow \left[ 3{{\left( x-1 \right)}^{\dfrac{1}{3}}} \right]_{0}^{2}$
Now, we will apply the limits in place of x using the formula $\int\limits_{a}^{b}{f(x)dx}=f(b)-f(a)$ , we get
$\Rightarrow 3{{\left( 2-1 \right)}^{\dfrac{1}{3}}}-3{{\left( 0-1 \right)}^{\dfrac{1}{3}}}$
On further solving the above expression, we get
$\Rightarrow 3{{\left( 1 \right)}^{\dfrac{1}{3}}}-3{{\left( -1 \right)}^{\dfrac{1}{3}}}$
Therefore, we get
$\Rightarrow 3-3{{\left( -1 \right)}^{\dfrac{1}{3}}}$
Therefore, the value of integral $\int{\dfrac{1}{{{\left( x-1 \right)}^{\dfrac{2}{3}}}}}$ from [0,2] is $3-3{{\left( -1 \right)}^{\dfrac{1}{3}}}$ .

Note:
While solving this problem, do mention all the formulas you are using to avoid confusion and mathematical error. Do not forget to solve the limits given in the problem, to get an accurate answer.