Question

Find the area of the region bounded by the parabola ${{y}^{2}}=2x+1$ and the line $x-y=1$

Answer 1

For solving this question, first we will plot the given curves on the same $x-y$ plane. Then, we will find the desired region whose area is asked in the question. After that, we will solve the given equations and find the coordinates of the intersection points. Then, we will take an elementary horizontal strip of width $dy$ and try to write its length in terms of the variable $y$ . Then, we will write the area of the elementary in terms of $y$ and $dy$ by multiplying its length and width. And finally, we will integrate the area of the elementary strip with suitable limits to get the total area of the given region.

Complete step-by-step answer:
Given,
We have to find the area of the region bounded by the parabola ${{y}^{2}}=2x+1$ and the line $x-y=1$ .
Now, before we proceed we should plot the curve ${{y}^{2}}=2x+1$ and the line $x-y=1$ on the same $x-y$ plane. For more clarity, look at the figure given below:

In the above figure, we have to find the area of the region ABC.
Now, for the coordinates of point A we should put $y=0$ in the equation ${{y}^{2}}=2x+1$ . Then,
$\begin{aligned} & {{y}_{A}}^{2}=2{{x}_{A}}+1 \\\ & \Rightarrow 0=2{{x}_{A}}+1 \\\ & \Rightarrow -2{{x}_{A}}=1 \\\ & \Rightarrow {{x}_{A}}=-0.5 \\\ \end{aligned}$
Now, from the above result, we can say that coordinates of the point $A\equiv (-0.5,0)$ . And from the figure, we can say that for the coordinates of points C and B we should equate the equations ${{y}^{2}}=2x+1$ and $x-y=1$ . Then,
$\begin{aligned} & x-y=1 \\\ & \Rightarrow x=1+y \\\ & \Rightarrow 2x=2+2y \\\ & \Rightarrow 2x+1=3+2y \\\ \end{aligned}$
Now, we put $2x+1={{y}^{2}}$ in the above equation. Then,
$\begin{aligned} & 2x+1=3+2y \\\ & \Rightarrow {{y}^{2}}=3+2y \\\ & \Rightarrow {{y}^{2}}-2y-3=0 \\\ & \Rightarrow {{y}^{2}}-3y+y-3=0 \\\ & \Rightarrow y\left( y-3 \right)+\left( y-3 \right)=0 \\\ & \Rightarrow \left( y+1 \right)\left( y-3 \right)=0 \\\ & \Rightarrow y=-1,3 \\\ \end{aligned}$
Now, from the above result and the figure, we can say that to get the coordinates of point C we should put $y=-1$ and to get the coordinates of point B we should put $y=3$ in the equation $x-y=1$ . Then,
$\begin{aligned} & {{x}_{C}}-{{y}_{C}}=1 \\\ & \Rightarrow {{x}_{C}}-\left( -1 \right)=1 \\\ & \Rightarrow {{x}_{C}}+1=1 \\\ & \Rightarrow {{x}_{C}}=0 \\\ & {{x}_{B}}-{{y}_{B}}=1 \\\ & \Rightarrow {{x}_{B}}-3=1 \\\ & \Rightarrow {{x}_{B}}=4 \\\ \end{aligned}$
Now, from the above result, we conclude that, coordinates of points $C\equiv (0,-1)$ and $B\equiv (4,3)$ .
Now, we take an elementary horizontal strip at $y$ of width $dy$ . For more clarity, look at the figure given below:

Now, to find the length of the elementary strip, we should subtract the ${{x}_{right}}=y+1$ and ${{x}_{left}}=\dfrac{\left( {{y}^{2}}-1 \right)}{2}$ . Then,
Length of the elementary strip $={{x}_{right}}-{{x}_{left}}=y+1-\dfrac{\left( {{y}^{2}}-1 \right)}{2}$ .
Now, as we know that width of the elementary strip is $dy$ . So, the area of the elementary strip will be length multiplied by width. Then,
Area of the elementary strip $=dA=\left( y+1-\dfrac{\left( {{y}^{2}}-1 \right)}{2} \right)dy$ .
Now, to get the total area of the region ABC, we should add the area of such elementary strips from $y=-1$ to $y=3$ so, to get the desired area we should integrate the expression $\left( y+1-\dfrac{\left( {{y}^{2}}-1 \right)}{2} \right)dy$ from $y=-1$ to $y=3$ . Then,
Area of the desired region $=\int\limits_{-1}^{3}{\left( y+1-\dfrac{\left( {{y}^{2}}-1 \right)}{2} \right)dy}$ .
Now, we will use the formula $\int{{{y}^{n}}dy=\dfrac{{{y}^{n+1}}}{n+1}+c}$ to integrate the above integral. Then,

& \int\limits_{-1}^{3}{\left( y+1-\dfrac{\left( {{y}^{2}}-1 \right)}{2} \right)dy} \\\ & \Rightarrow \int\limits_{-1}^{3}{\left( y+1-\dfrac{{{y}^{2}}}{2}+\dfrac{1}{2} \right)dy} \\\ & \Rightarrow \int\limits_{-1}^{3}{\left( y+\dfrac{3}{2}-\dfrac{{{y}^{2}}}{2} \right)}dy \\\ & \Rightarrow \left[ \dfrac{{{y}^{2}}}{2}+\dfrac{3y}{2}-\dfrac{{{y}^{3}}}{6} \right]_{-1}^{3} \\\ & \Rightarrow \left[ \left( \dfrac{{{3}^{2}}}{2}+\dfrac{3\times 3}{2}-\dfrac{{{3}^{3}}}{6} \right)-\left( \dfrac{{{\left( -1 \right)}^{2}}}{2}+\dfrac{3\times \left( -1 \right)}{2}-\dfrac{{{\left( -1 \right)}^{3}}}{6} \right) \right] \\\ & \Rightarrow \left[ \left( \dfrac{9}{2}+\dfrac{9}{2}-\dfrac{27}{6} \right)-\left( \dfrac{1}{2}-\dfrac{3}{2}+\dfrac{1}{6} \right) \right] \\\ & \Rightarrow \left[ \left( 9-\dfrac{9}{2} \right)-\left( \dfrac{4}{6}-\dfrac{3}{2} \right) \right] \\\ & \Rightarrow \left[ \left( \dfrac{9}{2} \right)-\left( -\dfrac{5}{6} \right) \right] \\\ & \Rightarrow \dfrac{9}{2}+\dfrac{5}{6} \\\ & \Rightarrow \dfrac{16}{3} \\\ \end{aligned}$$ Now, from the above result, we conclude that the area of the region ABC will be $\dfrac{16}{3}\text{ sq}\text{.units}$ . Thus, the area of the region bounded by the parabola ${{y}^{2}}=2x+1$ and the line $x-y=1$ will be equal to $\dfrac{16}{3}\text{ sq}\text{.units}\approx 5.333\text{ sq}\text{.units}$ . **Note:** Here, the student should first plot the given curves carefully and then find the desired region whose area is asked in the question and proceed in a stepwise manner. Then, we should be careful while writing the dimensions of the elementary strip and for that, we should take help from the plot of the given curves. Moreover, for easy calculation, we should take horizontal elementary strips, and we should take upper and lower limits correctly, to get the correct answer and whenever we get stuck at some point we should see the plot of the given curves and use the basic concepts of integral calculus.