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Find the area of the region bounded by the lines $x-2y-12=0 $ and $y$-axis,$y=-1$ and $y=-3$

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  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
Area bounded by $x-2y-12=0$ and $y-axis,y=-1,y=-3$
$x-2y-12=0=>x=2y+12$
$ A=\int \limits _{-3}^{-1} x dy$
$\quad=\int \limits _{-3}^{-1} (2y+12)dy$
$\quad=y^2+12y \bigg]_{-3}^{-1}$
$\quad=(1-12)-(9-36)$
$\quad=-11+27$
$\quad=16 sq.units$
answered Aug 15, 2013 by meena.p
 

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