Question

The value of integral \[\int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {x - {x^3}} \right)}^{\dfrac{1}{3}}}}}{{{x^4}}}} dx\]
A. \[6\]
B. \[0\]
C. \[3\]
D. \[4\]

Answer 1

Hint: We will solve this question by doing a simple integration formula which states that for a given function
\[f\left( x \right)\] the integration will be equals to as below:
\[\int {f\left( x \right)dx = {\text{F}}\left( x \right) + {\text{C}}} \], if
\[{\text{F}}'\left( x \right) = f\left( x \right)\]. Here,
\[\int {} \]is the integral symbol,
\[f\left( x \right)\] is called integrand,
\[x\] will be our variable,
\[dx\] is known as the differential of the variable
\[x\], and \[{\text{C}}\] is called the constant.

Complete step-by-step solution:
Step 1: By taking \[{x^3}\] as common from the numerator of the expression
\[\int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {x - {x^3}} \right)}^{\dfrac{1}{3}}}}}{{{x^4}}}} dx\], we get:
\[ \Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {{x^3}} \right)}^{\dfrac{1}{3}}}{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^4}}}} dx\]
By dividing the powers of the term \[{\left( {{x^3}} \right)^{\dfrac{1}{3}}}\] in the above expression we get:
\[ \Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{x{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^4}}}} dx\]
By doing division w.r.t to
\[x\] in the above expression we get:
\[ \Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx\]
Step 2: We will assume that
\[t = \dfrac{1}{{{x^2}}} - 1\], so by differentiating it w.r.t
\[x\] we get:
\[ \Rightarrow \dfrac{{dt}}{{dx}} = \dfrac{{ - 2}}{{{x^3}}}\]
\[\left( {\because \dfrac{1}{{{x^2}}} - 1 = \dfrac{{1 - {x^2}}}{{{x^2}}}} \right)\]
By bringing \[x\] terms at one side of the above expression we get:
\[ \Rightarrow \dfrac{{ - dt}}{2} = \dfrac{{dx}}{{{x^3}}}\]
Now, as per the given information from the question, the value of
\[x\] varies from \[\dfrac{1}{3}\] to \[1\], so by substituting these values in the expression
\[t = \dfrac{1}{{{x^2}}} - 1\] we get:
At \[x = \dfrac{1}{3}\]:
\[ \Rightarrow t = \dfrac{1}{{{{\left( {\dfrac{1}{3}} \right)}^2}}} - 1\]
By simplifying the above expression, we get:
\[ \Rightarrow t = 9 - 1\]
By doing the final subtraction, we get:
\[ \Rightarrow t = 8\]
Similarly, At \[x = 1\]:
\[ \Rightarrow t = \dfrac{1}{{{1^2}}} - 1\]
By simplifying the above expression, we get:\[ \Rightarrow t = 1 - 1\]
By doing the final subtraction, we get:\[ \Rightarrow t = 0\]
So, we can say that when \[x\] varies from \[\dfrac{1}{3}\] to \[1\], the value \[t\] varies from \[0\] to \[8\].
Step 3: By substituting the value of \[t\] in the expression
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx\], we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2}\int\limits_0^8 {{t^{\dfrac{1}{3}}}} dt\]
By doing integration in the RHS side of the expression we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2}{\left[ {\dfrac{{{t^{\dfrac{1}{3} + 1}}}}{{\dfrac{1}{3} + 1}}} \right]^8}_0\]
By adding the powers inside the brackets of the RHS side of the above expression we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}}dx=\dfrac{1}{2}{\left[{\dfrac{{{t^{\dfrac{4}{3}}}}}{{\dfrac{4}{3}}}} \right]^8}_0\]
By bringing \[3\] from the denominator to the numerator side of the RHS side of the above expression we get:
\[ \Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2}{\left[ {\dfrac{{3{t^{\dfrac{4}{3}}}}}{4}} \right]^8}_0\]
By putting the limits \[t\] in the RHS side of the above expression, we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = - \dfrac{1}{2}\left( {0 - \dfrac{{3{{\left( 8 \right)}^{\dfrac{4}{3}}}}}{4}} \right)\]
By simplifying the above expression, we can write it as below:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2} \times \left( {\dfrac{3}{4}} \right) \times {\left( 8 \right)^{\dfrac{4}{3}}}\]
By writing \[8 = {2^3}\] in the above expression we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2} \times \left( {\dfrac{3}{4}} \right) \times {\left( {{2^3}} \right)^{\dfrac{4}{3}}}\]
By replacing
\[{\left( {{2^3}} \right)^{\dfrac{4}{3}}} = {2^4}\], we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2} \times \left( {\dfrac{3}{4}} \right) \times {2^4}\]
By simplifying the above expression, we get:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = \dfrac{1}{2} \times \left( {\dfrac{3}{4}} \right) \times 2 \times 2 \times 2 \times 2\]
By solving the above expression, we get the required answer as below:
\[\Rightarrow \int\limits_{\dfrac{1}{3}}^1 {\dfrac{{{{\left( {\dfrac{1}{{{x^2}}} - 1} \right)}^{\dfrac{1}{3}}}}}{{{x^3}}}} dx = 6\]

\[\because \] Option A is the correct answer.

Note: Students need to be very careful while solving the integration of powers. For example, the integration of \[{x^n}\] will be equals to as below:
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + {\text{C}}} \]