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Find the area of the region bounded by the parabola $y^{2}=4x$ and the line $2x-y=4$

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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 $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
$y^2=4x(i)$
$2x=y+4 (ii)$
$y^2 =2(y+4)$
$y^2-2y-8=0$
$(y-4)(8+2)=0$
$=>y=-4,-2$
and $x=4,1$
The points of intersection are $(4,4),(1,-2)$
Required area= Area to left of line - area of left of area
$\qquad=\large\frac{1}{2} $$ \int \limits_{-2}^ 4(y+4) dy- \large\frac{1}{4}$$\int \limits _{-2}^4y^2 dy$
On integrating we get,
$\qquad=\large\frac{1}{2}$$ \bigg[4y+\large\frac{y^2}{2}$$\bigg]_{-2}^{4}-\large\frac{1}{4}\bigg[\large\frac{y^3}{3}\bigg]_{-2}^4$
Applying the limits we get
$\qquad= \large\frac{1}{2}$$[(16+8)+(8-2)]-\large\frac{1}{12}$$(64+8)$
$\qquad=\large\frac{1}{2}$$(30)-\large\frac{1}{12}$$(72)$
$\qquad= 15-6 $
$\qquad=9$ sq.units
answered Aug 16, 2013 by meena.p
edited Nov 25, 2014 by sreemathi.v
 

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