Question

Using integration, find the area bounded by the curve ${{x}^{2}}=4y$ and the line $x=4y-2$.

Answer 1

We start solving this problem by first finding the points of intersection of the given curves. Then we find the area of the region bounded by each curve with x-axis between the points of intersection using the integration formula for area $\int\limits_{a}^{b}{f\left( x \right)dx}$. Then we subtract the area of the smaller curve from the area of the larger one to get the area of the required region.

Complete step-by-step solution:
First, let us find the points of intersection of the given curves.
${{x}^{2}}=4y\Rightarrow y=\dfrac{{{x}^{2}}}{4}$
Substituting it in the equation of line
$\begin{aligned} & \Rightarrow x=4y-2 \\\ & \Rightarrow x=4\left( \dfrac{{{x}^{2}}}{4} \right)-2 \\\ & \Rightarrow x={{x}^{2}}-2 \\\ & \Rightarrow {{x}^{2}}-x-2=0 \\\ & \Rightarrow \left( x-2 \right)\left( x+1 \right)=0 \\\ & \Rightarrow x=2,-1 \\\ \end{aligned}$
When x=2, $y=\dfrac{{{x}^{2}}}{4}=\dfrac{{{2}^{2}}}{4}=\dfrac{4}{4}\Rightarrow y=1$
When x=-1, $y=\dfrac{{{x}^{2}}}{4}=\dfrac{{{\left( -1 \right)}^{2}}}{4}=\dfrac{1}{4}\Rightarrow y=\dfrac{1}{4}$
So, points of intersection of the two curves are $\left( 2,1 \right)$ and $\left( -1,\dfrac{1}{4} \right)$.
Now let us look at the plot

We need to find the area of the shaded region. As we see our required area can be written as the difference between the area of line and parabola between x=-1 and x=2.
Let us consider the formula for finding the area of region bounded by a function f(x) between x=a and x=b,
$\int\limits_{a}^{b}{f\left( x \right)dx}$
Using this formula, let us find the area of region bounded by curve ${{x}^{2}}=4y$ and x-axis between x=-1 and x=2.
$\begin{aligned} & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\int\limits_{-1}^{2}{{{x}^{2}}dx} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\left[ \dfrac{{{x}^{3}}}{3} \right]_{-1}^{2} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\left[ \dfrac{{{x}^{3}}}{3} \right]_{-1}^{2} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\left[ \dfrac{{{2}^{3}}}{3}-\dfrac{{{\left( -1 \right)}^{3}}}{3} \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\left[ \dfrac{8}{3}-\dfrac{-1}{3} \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{1}{4}\left[ \dfrac{9}{3} \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{{{x}^{2}}}{4}dx=}\dfrac{3}{4}................\left( 1 \right) \\\ \end{aligned}$
Now, let us find the area of region bounded by curve $x=4y-2$ and x-axis between x=-1 and x=2.
$\begin{aligned} & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\int\limits_{-1}^{2}{\left( x+2 \right)dx} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-1}^{2} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\left[ \left( \dfrac{{{2}^{2}}}{2}+2\left( 2 \right) \right)-\left( \dfrac{{{\left( -1 \right)}^{2}}}{2}+2\left( -1 \right) \right) \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\left[ \left( \dfrac{4}{2}+4 \right)-\left( \dfrac{1}{2}-2 \right) \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\left[ \left( 2+4 \right)-\left( \dfrac{-3}{2} \right) \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\left[ 6+\dfrac{3}{2} \right] \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{1}{4}\times \dfrac{15}{2} \\\ & \Rightarrow \int\limits_{-1}^{2}{\dfrac{x+2}{4}dx=}\dfrac{15}{8}................\left( 2 \right) \\\ \end{aligned}$
So, now we need to subtract the equation (1) from equation (2), to get our required area
$\begin{aligned} & \Rightarrow \dfrac{15}{8}-\dfrac{3}{4} \\\ & \Rightarrow \dfrac{9}{8} \\\ \end{aligned}$
Hence, the area of required region is $\dfrac{9}{8}$ square units.
Hence, the answer is $\dfrac{9}{8}$ square units.

Note: We can solve this by using a formula too. The area bounded by two curves $f\left( x \right)$ and $g\left( x \right)$ between x=a and x=b is $\int\limits_{a}^{b}{\left( f\left( x \right)-g\left( x \right) \right)dx}$. Here our $f\left( x \right)$ is the line and $g\left( x \right)$ is the parabola. Then we get $\int\limits_{-1}^{2}{\left( \dfrac{x+2}{4}-\dfrac{{{x}^{2}}}{4} \right)}dx$, which is same as what we have done separately. So, one need to remember that both are same and should not confuse that both are different.