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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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The area of the region bounded by the curve $x=2y+3$ and the y-lines $y=1$ and $y=-1$ is \[(A)\;4 sq.units\quad(B)\;\frac{3}{2} sq. units\quad(C)\;6 sq.units\quad(D)8 sq.units\]

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  • The area enclosed by the curve $x=f(y)$,the $y$-axis and the abscissae $y=c$ and $y=d$ is given by $\int_c^d xdy.$
  • $\int y^ndx=\large\frac{y^{n+1}}{n+1}$$+c$
Step 1:
Given curve is $x=2y+3$ between the lines $y=1$ and $y=-1$
The required area is the shaded portion shown in the fig.
Step 2:
Area of the shaded portion is $A=\int_{-1}^1 xdy=\int_{-1}^1(2y+3)dy$
On integrating we get,
$\;\;\;=\begin{bmatrix}2\big(\large\frac{y^2}{2}\big)\normalsize +3y\end{bmatrix}_{-1}^1=\begin{bmatrix}y^2+3y\end{bmatrix}_{-1}^1$
On applying limits we get,
$[(1-1)+3(1+1)]=6$ sq.units.
Hence the correct option is C.
answered May 7, 2013 by sreemathi.v
edited May 8, 2013 by sreemathi.v

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