Question

Find the area bounded by the curve ${x^2} = 4y$ and the line $x = 4y - 2$.
A $\dfrac{5}{4}$
B $\dfrac{9}{8}$
C $\dfrac{3}{4}$
D $\dfrac{7}{8}$

Answer 1

Hint: In this question we need to find the area bounded by a parabola ${x^2} = 4y$ and a line $x = 4y - 2$, for that firstly we will draw a graph and plot the parabola and the line then we will be finding the point of intersection. After that for finding the area we will be finding the integration of both the parabola and the line with respect to dx.

Complete step by step answer:

We have been provided with a parabola ${x^2} = 4y$ and a straight line $x = 4y - 2$,

Now we will be finding the point of intersection of the parabola and straight line,

For that we will keep the value of 4y in $x = 4y - 2$,

So, the equation becomes: $x = {x^2} - 2$,

Solving it further, ${x^2} - x - 2 = 0$,

Now for finding the roots we will be using splitting the middle term method: ${x^2} - 2x + x - 2 = 0$,

Now we will be simplifying it by taking the terms common: $x(x - 2) + 1(x - 2) = 0$,

Now take (x-2) common, we will get: $(x - 2)(x + 1) = 0$,

So, the roots would be: $x = - 1,2$.

Now substitute this value in the equation of the parabola: ${{x}^{2}}=4y$, we will get,

For $x=-1,y=\dfrac{1}{4}$ and for $x=2,y=1$.

So, we will be using these points to find the point of intersection,

Now plot the graph for the parabola ${{x}^{2}}=4y$ and line $x=4y-2$,

We will get the point of intersection as: $\left( -1,\dfrac{1}{4} \right)$ and (2, 1).

Now we will be finding the area bounded by the parabola ${x^2} = 4y$ and the line $x = 4y - 2$,

From the figure we can say that the area will be: A= $\smallint $ (equation of line) dx- $\smallint $(equation of parabola) dx

So, the area of the shaded region: $A = \int\limits_{ - 1}^2 {\left( {\dfrac{{x + 2}}{4} - \dfrac{{{x^2}}}{4}} \right)dx} $,

Now we will be doing the integration using the formula: $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $,

So, the integration would be: $\dfrac{1}{4}\left[ {\int\limits_{ - 1}^2 {(x + 2} )dx - \int\limits_{ - 1}^2 {({x^2})} dx} \right]$,

Simplifying it further: $\dfrac{1}{4}{\left[ {\dfrac{{{x^2}}}{2} + 2x - \dfrac{{{x^3}}}{3}} \right]^2}_{ - 1}$,

Now we will be putting the limits, we will get: $\dfrac{1}{4}\left[ {\left( {\dfrac{4}{2} + 4 - \dfrac{8}{3}} \right) - \left( {\dfrac{1}{2} - 2 + \dfrac{1}{3}} \right)} \right]$

Simplifying it further: $\dfrac{1}{4}\left[ {8 - \dfrac{1}{2} - 3} \right]$,

So, the area of the bounded region comes out to be: $\dfrac{9}{8}$ sq. units.

So, the correct answer is “Option B”.

Note: While solving this question, do plot the parabola and the line on the graph otherwise the chances of mistakes would increase. For integration do not use this formula $\int {{x^n}dx = \log x} $ as this formula is just applicable for n=1.