The area bounded by the curves $y^{2} = {4a}^{2}(x - 1)$and lines $x = 1$and $y = 4a$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$On\ solving\ y^{2} = {4a}^{2}(x - 1)and\ y = 4a,\ we\ get\ x = 5$
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$	herefore Required\ area = \int_{1}^{5}{\left( 4a - 2a\sqrt{x - 1} ight)dx}$
$= \left\lbrack 4ax - 2a\frac{(x - 1)^{3/2}}{3/2} ightbrack_{1}^{5}$
$= \frac{16a}{3}\ sq\ units$

Q. The area bounded by the curves $y^{2} = {4a}^{2}(x - 1)$and lines $x = 1$and $y = 4a$ is

Explanation

Reviewed by Dr. Abubacker

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