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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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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}

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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 by sreemathi.v
edited May 6, 2013 by sreemathi.v
 

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