Question

The area bounded by \[{y^2} = 2x + 1\;\] and \[x - y - 1 = 0\;\] is
A) $\dfrac{4}{3}$
B) $\dfrac{8}{3}$
C) $\dfrac{{16}}{3}$
D) None of these

Answer 1

Hint:In this question, we have given the equation of the curve and the line. We have to find the area bounded by them. Firstly, calculate the coordinates of the curve and the line using their equation put $x = 0$ then put $y = 0$. Also, calculate the coordinates of the point of intersection by using any of the methods (Substitution, elimination, and graphical method). Then, plot all the coordinates in the graph. To calculate the area covert each equation in terms of $y$ and integrate their difference with respect to $dy$ from $ - 1$ to $4$ 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 is \[{y^2} = 2x + 1\;\] and the equation of line is \[x - y - 1 = 0\;\],
Therefore, the coordinates of the given curve \[{y^2} = 2x + 1\;\] are $\left( { - \dfrac{1}{2},0} \right)$ and $\left( {0,1} \right)$
(At $x = 0$, we get $y = 1$
At $y = 0$, we get $x = - \dfrac{1}{2}$)
Similarly, the coordinate of the line \[x - y - 1 = 0\;\] are $\left( {0, - 1} \right)$ and $\left( {1,0} \right)$
(At $x = 0$, we get $y = - 1$
At $y = 0$, we get $x = 1$)
Now, to calculate the point of intersection of the given curve and the line
Put $x = y + 1$ in the equation of the curve\[{y^2} = 2x + 1\;\]
It will be, \[{y^2} = 2\left( {y + 1} \right) + 1\;\]
\[{y^2} - 2y - 3 = 0\]
Now, calculating the roots of the required quadratic equation
\[{y^2} - 3y + y - 3 = 0\]
\[y\left( {y - 3} \right) + 1\left( {y - 3} \right) = 0\]
So, we get $y = - 1$ and $y = 3$
Put the required values in $x = y + 1$
At $y = - 1$, $x = 0$ and at $y = 3$, we get $x = 4$
Therefore, the point of intersection of the curve and the line are $\left( {0, - 1} \right)$ and $\left( {4,3} \right)$
Graph of the required points is attached below i.e., figure 1;

Figure 1: A graph plotting the coordinates of the given curve and the line
As we know that, $x = y + 1$ and $x = \dfrac{{{y^2} + 1}}{2}$
For the area of the shaded portion integrate the difference of the above equations with respect to $dy$ from $ - 1$ to $3$.
It will be,
\[A = \int\limits_{ - 1}^3 {\left\{ {\left( {y + 1} \right) - \dfrac{1}{2}\left( {{y^2} - 1} \right)} \right\}dy} \]
\[ = \int\limits_{ - 1}^3 {\left\{ {y + 1 - \dfrac{{{y^2}}}{2} + \dfrac{1}{2}} \right\}dy} \]
\[ = \int\limits_{ - 1}^3 {\left\{ {y - \dfrac{{{y^2}}}{2} + \dfrac{3}{2}} \right\}dy} \]
As we know that, $\int {cdx = cx} $ where $c$ is the constant and $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
\[ = \left[ {\dfrac{{{y^2}}}{2} - \dfrac{{{y^3}}}{6} + \dfrac{{3y}}{2}} \right]_{ - 1}^3\]
Resolving the limits, we get
\[ = \dfrac{{{{\left( 3 \right)}^2}}}{2} - \dfrac{{{{\left( 3 \right)}^3}}}{6} + \dfrac{{3\left( 3 \right)}}{2} - \left( {\dfrac{{{{\left( { - 1} \right)}^2}}}{2} - \dfrac{{{{\left( { - 1} \right)}^3}}}{6} + \dfrac{{3\left( { - 1} \right)}}{2}} \right)\]
\[ = \dfrac{9}{2} - \dfrac{{27}}{6} + \dfrac{9}{2} - \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{3}{2}\]
L.C.M. of $2$ and $6$ is $6$
\[ = \dfrac{{27 - 27 + 27 - 3 - 1 + 9}}{6}\]
\[ = \dfrac{{32}}{6}\]
\[ = \dfrac{{16}}{3}\]

Therefore, the correct option is (C).

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.