Question

The area enclosed by the curve ${x^2}y = 36$, the $x - axis$ and the lines $x = 6$ and $x = 9$ is
A. $6$
B. $1$
C. $4$
D. $2$

Answer 1

Hint: In this question, we are given the equation of the curve i.e., ${x^2}y = 36$ and the lines. We have to find the area bounded by them. Firstly, plot the graph of the following. Then, for the area integrate the equation of the curve in terms of $x$ with respect to $dx$ from $x = 6$ to $x = 9$. Solve it further using integration and logarithm formulas and resolve the limits.

Formula Used: Integration formula –
$\int {cdx = cx} $ where $c$ is the constant
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $

Complete step by step solution: Given that,
Equation of the curve ${x^2}y = 36$ and the equation of the lines $x = 6$ to $x = 9$.
Equation of the curve also written as $y = \dfrac{{36}}{{{x^2}}}$.
Graph of the given equations is attached below (figure 1);


        Figure 1: A Graph contains the plot of the given curve and the lines

Now, as you can see bounded area is under the equation of the curve from the equation of both the lines.
Therefore, the area of the bounded region will be the integration of the equation of the given curve in terms of $x$ i.e., $y = \dfrac{{36}}{{{x^2}}}$ with respect to $dx$ from $x = 6$ to $x = 9$.
Thus, the area of the required region will be
$A = \int\limits_6^9 {\left( {\dfrac{{36}}{{{x^2}}}} \right)dx} $
As we know that, $\int {cdx = cx} $ where $c$ is the constant and $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
It implies that,
$A = 36\left[ {\dfrac{{ - 1}}{x}} \right]_6^9$
On resolving the limits, we get $A = 36\left[ { - \dfrac{1}{9} - \left( { - \dfrac{1}{6}} \right)} \right]$
On solving, we get $A = 2$ square units
Hence, Option (D) is the correct answer i.e., $2$ sq. unit

Option ‘D’ is correct

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.