# The area of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is \begin{array}{1 1}(A)\;\frac{3}{8}sq.units & (B)\;\frac{5}{8}sq.units\\(C)\;\frac{7}{8}sq.units & (D)\;\frac{9}{8} sq.units \end{array}

Toolbox:
• The area enclosed by the curve $y=f(x)$,the $x$-axis and the ordinates $x=a$ and $x=b$ is given by $\int_a^b ydx.$
• $\int x^n dx=\large\frac{x^{n+1}}{n+1}$$+c. Step 1: The given curves are x^2=4y and the line x=4y-2 Now let us find the points of intersection of the curves can be obtained by solving the two equations. x^2=4y and 4y=x+2 Hence x^2=x+2 or x^2-x-2=0 On factorizing we get, (x-2)(x+1)=0 Hence the points of intersection are (2,1)(-1,\large\frac{1}{4}) Step 2: Hence the required area A=\int_{-1}^2(y_2-y_1)dx. Where y_2=\large\frac{x+2}{4} and y_1=\sqrt{\large\frac{x^2}{4}} A=\int_{-1}^2\begin{bmatrix}\big(\large\frac{x+2}{4}\big)-\frac{x}{2}\end{bmatrix}$$dx$
$A=\large\frac{1}{4}\int_{-1}^2\normalsize(x+2) dx-\large\frac{1}{2}\int_{-1}^2\normalsize xdx.$
Step 3:
On integrating we get,
$A=\large\frac{1}{4}\begin{bmatrix}\large\frac{x^2}{2}+\normalsize 2x\end{bmatrix}_{-1}^2-\large\frac{1}{2}\begin{bmatrix}\large\frac{x^2}{2}\end{bmatrix}_{-1}^2$
$\;\;\;=\large\frac{1}{8}$$\big(x^2\big)_{-1}^2+\large\frac{1}{2}$$\big(x\big)_{-1}^2-\large\frac{1}{4}$$\big(x^2\big)_{-1}^2 On applying limits we get, A=\large\frac{1}{8}$$[(4-1)]+\large\frac{1}{2}$$[2+1]-\large\frac{1}{4}$$[4-1].$
$\;\;=\large\frac{3}{8}+\frac{3}{2}-\frac{3}{4}=\frac{9}{8}$sq.units.
Hence D is the correct option.
answered May 6, 2013
edited May 6, 2013