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TN XII Math
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Integral Calculus and its applications
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Find the area included between the parabola $y^{2}=4ax $ and its latus rectum.
tnstate
class12
bookproblem
ch7
sec-1
exercise7-4
p117
q6
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asked
Apr 27, 2013
by
poojasapani_1
edited
Apr 27, 2013
by
poojasapani_1
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Toolbox:
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 $y^2=4ax$
and latus rectum $=2 \times$ area is first quadrent.
$A= 2 \int \limits_0^a y dx =2 \int \limits _0^a 2 \sqrt a x^{\large\frac{1}{2}} dx$
$\qquad= 4 \sqrt a \bigg [\large\frac{x}{3} $$x^{\large\frac {3}{2}}\bigg]_0^a$
$\qquad= \large\frac{8}{3}$$a^{\large\frac{1}{2}}a^{\large\frac{3}{2}}$
$\qquad=\large\frac{8}{3}$$a^2 sq \;units$
answered
Aug 15, 2013
by
meena.p
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