The area enclosed between the ${y^2} = x$ and $y = |x|$ is
A) $\dfrac{1}{3}$
B) $\dfrac{2}{3}$
C) $1$
D) $\dfrac{1}{6}$
Answer
177.3k+ views
Hint: The area between two curves can be found using definite integral. Since the curves are expressed in terms of $y =
f(x)$, we can integrate with respect to $x$. No specific interval is given. So we can take the unit interval $(0,1)$.
Formula used: 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} $
Complete step-by-step solution:
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 $\dfrac{1}{6}$.
Therefore the answer is option D.
Note: We took the value of $|x|$ as $x$ since the values are positive in the unit interval. Also, we have for positive numbers less than one, its root exceeds the number. So, we get the function $f(x)$ greater than the function $g(x)$. If in the question any interval is specified, we have to change the range of $x$.
f(x)$, we can integrate with respect to $x$. No specific interval is given. So we can take the unit interval $(0,1)$.
Formula used: 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} $
Complete step-by-step solution:
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 $\dfrac{1}{6}$.
Therefore the answer is option D.
Note: We took the value of $|x|$ as $x$ since the values are positive in the unit interval. Also, we have for positive numbers less than one, its root exceeds the number. So, we get the function $f(x)$ greater than the function $g(x)$. If in the question any interval is specified, we have to change the range of $x$.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
