Find the area bounded by the curves $y^{2} = 4 a^{2} (x - 1)$ and the line $x = 1, y = 4 a$ .

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Solution

$(4 a)^{2} = 4 a (x - 1)$
$\Rightarrow x = 5$
$y^{2} = 4 a^{2} (x - 1), x = 1, y = 4 a$
Reqd asce $= \int_{1}^{5} (4 a - 2 a \sqrt{x - 1}) d x$
$= 40 \times (5 - 1) - 20 {[\frac{(x - 1)^{3 / 2}}{3 / 2}]}^{5}$
$= 40 \times 4 - \frac{2 a \times 2}{3} \times 8 = \frac{48 a - 32 a}{3}$
$= \frac{160}{3}$

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