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$

a) \,4{a^2} \\

b) \,8{a^2} \\

c) \,\dfrac{{28}}{3}{a^2} \\

d) \,none\,of\,these \\

$

Answer
Verified

Area $ = \int\limits_a^{4a} {f(x)dx} $

Here, $y = \sqrt {4ax} $

Area $ = \int\limits_a^{4a} {f(x)dx} $

Area $ = \int\limits_a^{4a} {\sqrt {4ax} dx} $

Area $ = \sqrt {4a} \int\limits_a^{4a} {\sqrt x } dx$ { $\sqrt {4a}$ is a constant and can be taken out of integration}

$

= \sqrt {4a} \left[ {\dfrac{{{x^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right]_a^{4a} \\

= \sqrt {4a} \left[ {\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_a^{4a} \\

= \dfrac{2}{3}\sqrt {4a} \left[ {{{\left( {4a} \right)}^{\dfrac{3}{2}}} - {{\left( a \right)}^{\dfrac{3}{2}}}} \right] \\

= \dfrac{2}{3}\sqrt {4a} \left[ {8{{\left( a \right)}^{\dfrac{3}{2}}} - {{\left( a \right)}^{\dfrac{3}{2}}}} \right] \\

= \dfrac{2}{3} \times 7{\left( a \right)^{\dfrac{3}{2}}} \times \sqrt {4a} \\

= \dfrac{{28}}{3}{a^{2\,}}\,\,sq\,units \\

$

Now this is the area of $y = + \sqrt {4ax}$ but not of $y = - \sqrt {4ax}$ which is also a solution of ${y^2} = 4ax$ therefore to find the solution of either we can integrate or we can conclude that by symmetry it is also equal to $ = \dfrac{{28}}{3}{a^{2\,}}\,\,sq\,units$

It can be clearly seen by the plot of ${y^2} = 4ax$

Total Area =$A + A = 2A$

Both graphs are symmetrical therefore total area equals to twice of area of $y = + \sqrt {4ax} $

So total area of ${y^2} = 4ax$ from $x = a\,to\,x = 4a$ is equal to $ = 2 \times \dfrac{{28}}{3}{a^{2\,}}\,\,sq\,units$$ = \dfrac{{56}}{3}{a^2}\,\,sq\,units$