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

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:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + {\text{C}}}$