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A) $\dfrac{1}{3}$

B) $\dfrac{2}{3}$

C) $1$

D) $\dfrac{1}{6}$

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f(x)$, we can integrate with respect to $x$. No specific interval is given. So we can take the unit interval $(0,1)$.

$A = \int\limits_a^b {(f(x) - g(x))dx} $

We are given the curves ${y^2} = x$ and $y = |x|$.

We have to find the area enclosed between them.

We can rewrite them as follows.

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

And we have,

$|x|$ takes the value $x$ for $x > 0$ and $ - x$ for $x < 0$.

To find the area between the curves,

Consider the interval $(0,1)$.

If we have two curves $y = f(x)$ and $y = g(x)$ such that $f(x) > g(x)$ then the area between them bounded by the

horizontal lines $x = a,x = b$ is given by

$A = \int\limits_a^b {(f(x) - g(x))dx} $

So let $f(x) = \sqrt x $ and $g(x) = |x|$.

In the interval $(0,1)$, we have $\sqrt x > |x| = x$

So substituting we get the area as,

$\Rightarrow$$A = \int\limits_0^1 {(\sqrt x - x)dx} $

This gives,

$\Rightarrow$$A = \int\limits_0^1 {({x^{\dfrac{1}{2}}} - x)dx} $

We know that $\int\limits_0^1 {{x^n}dx} = [\dfrac{{{x^{n + 1}}}}{{n + 1}}]_0^1$

We get,

$\Rightarrow$$A = [\dfrac{{{x^{\dfrac{3}{2}}}}}{{\dfrac{3}{2}}} - \dfrac{{{x^2}}}{2}]_0^1$

Simplifying we have,

$\Rightarrow$$A = [\dfrac{{2{x^{\dfrac{3}{2}}}}}{3} - \dfrac{{{x^2}}}{2}]_0^1$

Substituting the limits we get,

$\Rightarrow$$A = [\dfrac{{2 \times {1^{\dfrac{3}{2}}}}}{3} - \dfrac{{{1^2}}}{2} - (\dfrac{{2 \times {0^{\dfrac{3}{2}}}}}{3} -\dfrac{{{0^2}}}{2})]$

Simplifying we get,

$\Rightarrow$$A = [\dfrac{2}{3} - \dfrac{1}{2} - (0 - 0)]$

$ \Rightarrow A = \dfrac{{4 - 3}}{6}$

So we get,

$\Rightarrow$$A = \dfrac{1}{6}$

That is the area enclosed between the two curves is $\dfrac{1}{6}$.