Question

The value of $a$, \[\left( {a > 0} \right)\;\] for which the area bounded by the curves $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$, $y = 0$, $x = a$ and $x = 2a$ has the least value is
A. $2$
B. $\sqrt 2 $
C. ${\left( 2 \right)^{\dfrac{1}{3}}}$
D. $1$

Answer 1

Hint: In this question, we are given the equation of curve $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$ which lies in the $x - axis$ from $x = a$ and $x = 2a$. We have to find the value of $a$ at which the area will be least. Also, \[\left( {a > 0} \right)\;\] (given). So, calculate the area by integrating the equation of the curve from $a$ to $2a$. Then, differentiate the required answer with respect to $a$ and equate it to $0$. Solve it further.

Formula Used: Integration formula –
$\int {f\left( y \right)dx = \dfrac{{\int {f\left( y \right)dx} }}{{\left( {\dfrac{{dy}}{{dx}}} \right)}}} $
Differentiation formula –
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step Solution:
Given that,
Equation of the curve is $y = \dfrac{x}{6} + \dfrac{1}{{{x^2}}}$ from $x = a$ to $x = 2a$
Therefore, the area bounded by the curve will be the integration of the equation of the curve from $x = a$ to $x = 2a$
Therefore, Area $ = \int\limits_a^{2a} {\dfrac{x}{6} + \dfrac{1}{{{x^2}}}} dx$
$ = \left[ {\dfrac{{{x^2}}}{{12}} - \dfrac{1}{x}} \right]_a^{2a}$
Now, resolving the limits
Thus, area $\left( A \right) = \dfrac{{{a^2}}}{4} + \dfrac{1}{{2a}}$
Differentiate the required area with respect to $a$,
We get,
$\dfrac{{dA}}{{da}} = \dfrac{{2a}}{4} + \dfrac{1}{{2{a^2}}}$
Also, written as $\dfrac{{dA}}{{da}} = \dfrac{a}{2} + \dfrac{1}{{2{a^2}}}$
Now, for minimum value
Put $\dfrac{{dA}}{{da}} = 0$
It implies that,
$\dfrac{a}{2} + \dfrac{1}{{2a}} = 0$
L.C.M of $2$ and $2a$ is $2a$
Therefore, $\dfrac{{{a^2} - 1}}{{2a}} = 0$
We get, $a = \pm 1$
According to the question, $a$ should be greater than $0$. So, $a = - 1$ will not exist.
Thus, at $a = 1$ the area of the given curve will be least.

Therefore, the correct option is (D).

Note:Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.