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Draw the curve with the given data where \[y = 4\] and let line AB represents \[y = 4\].

Also, $ y = {x^2} $

It can be re-written as –

Let the curve AOB represents $ {x^2} = y $

Now, the area of AOBA $ = 2 \times $ Area BONB

Place the values in the above expression –

Area $ = 2\int_0^4 x dy $ ..... (A)

We are given that –

$ {x^2} = y $

Take square-root on both the sides of the equation –

$ \sqrt {{x^2}} = \sqrt y $

Square and square-root cancel each other on the left hand side of the equation. Also, the square of positive and the negative term gives the positive term.

$ \therefore x = \pm \sqrt y $

Since BONB lies in first quadrant, we have $ x = + \sqrt y $

Place the above value in the equation (A)

Area of AOBA $ = 2\int_0^4 {xdy} $

Place value for “x”

Area of AOBA $ = 2\int_0^4 {\sqrt y dy} $

It can be written as using the law of exponent.

Area of AOBA $ = 2\int_0^4 {{y^{\dfrac{1}{2}}}dy} $

Use the identity: $ \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $ in the above equation.

$ \Rightarrow A = 2\left[ {\dfrac{{{y^{\dfrac{1}{2} + 1}}}}{{\dfrac{1}{2} + 1}}} \right]_0^4 $

Simplify the above equation.

$ \Rightarrow A = 2\left[ {\dfrac{{{y^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}}} \right]_0^4 $

Place the limit in the above equation. Lower limit is subtracted from the upper limit.

$ \Rightarrow A = 2 \times \dfrac{2}{3}\left[ {{4^{\dfrac{3}{2}}} - 0} \right] $

Simplify the above equation.

$ \Rightarrow A = \dfrac{4}{3}\left[ {{{(2)}^{2 \times }}^{\dfrac{3}{2}}} \right] $

Common factors from the numerator and the denominator cancels each other. Therefore remove from the numerator and the denominator.

$ \Rightarrow A = \dfrac{4}{3}\left[ {{{(2)}^3}} \right] $

Simplify the above equation –

$ \Rightarrow A = \dfrac{{32}}{3} $

This is the required solution.

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