Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given region is

[tex]\rm \left\{(x, y) \mid x^{2} \leq y \leq x\right\} \\ [/tex]

So, given curves are

[tex]\rm \: {x}^{2} = y - - - (1) \\ [/tex]

and

[tex]\rm \: y = x - - - (2) \\ [/tex]

Step :- 1 Point of intersection of two curves

On equating equation (1) and (2), we get

[tex]\rm \: {x}^{2} = x \\ [/tex]

[tex]\rm \: {x}^{2} - x = 0\\ [/tex]

[tex]\rm \: x(x - 1) = 0\\ [/tex]

[tex]\rm\implies \:x = 0 \: \: \: or \: \: \: x = 1 \\ [/tex]

Hᴇɴᴄᴇ,

➢ Pair of points of the intersection of two curves are shown in the below table.

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}[/tex]

Step :- 2 Curve Sketching

The curve [tex] x^2 = y[/tex] represents the upper parabola having vertex at (0, 0) and axis along y - axis.

The curve y = x is a line passes through (0, 0) and intersects the other curve at (0, 0) and (1, 1)

See the attachment.

Step : - 3 Required Area

The required area between the two curves is

[tex]\rm \: = \: \displaystyle\int_{0}^{1}\rm [y_{line} - y_{parabola}] \: dx \\ [/tex]

[tex]\rm \: = \: \displaystyle\int_{0}^{1}\rm [x - {x}^{2} ] \: dx \\ [/tex]

[tex]\rm \: = \: \bigg[\dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} \bigg] _{0}^{1} \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{2} - \dfrac{1}{3} \\ [/tex]

[tex]\rm \: = \: \dfrac{3 - 2}{6} \\ [/tex]

[tex]\rm \: = \: \dfrac{1}{6} \: square \: units\\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]

[tex]\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} = [/tex]

Given region is

[tex]\begin{gathered}\rm \left\{(x, y) \mid x^{2} \leq y \leq x\right\} \\ \end{gathered}

{(x,y)∣x

2

≤y≤x}[/tex]

So, given curves are

[tex]\begin{gathered}\rm \: {x}^{2} = y - - - (1) \\ \end{gathered}

x

2

=y−−−(1)[/tex]

and

[tex]\begin{gathered}\rm \: y = x - - - (2) \\ \end{gathered}

y=x−−−(2)

[/tex]

Step :- 1 Point of intersection of two curves

On equating equation (1) and (2), we get

[tex]\begin{gathered}\rm \: {x}^{2} = x \\ \end{gathered}

x

2

=x[/tex]

[tex]\begin{gathered}\rm \: {x}^{2} - x = 0\\ \end{gathered}

x

2

−x=0[/tex]

[tex]\begin{gathered}\rm \: x(x - 1) = 0\\ \end{gathered}

x(x−1)=0[/tex]

[tex]\begin{gathered}\rm\implies \:x = 0 \: \: \: or \: \: \: x = 1 \\ \end{gathered}

⟹x=0orx=1[/tex]

Hᴇɴᴄᴇ,

➢ Pair of points of the intersection of two curves are shown in the below table.

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}\end{gathered}

x

0

1

y

0

1

[/tex]Step :- 2 Curve Sketching

The curve x^2 = yx

2

=y represents the upper parabola having vertex at (0, 0) and axis along y - axis.

The curve y = x is a line passes through (0, 0) and intersects the other curve at (0, 0) and (1, 1)

See the attachment.

Step : - 3 Required Area

The required area between the two curves is

[tex]\begin{gathered}\rm \: = \: \displaystyle\int_{0}^{1}\rm [y_{line} - y_{parabola}] \: dx \\ \end{gathered}

=∫

0

1

[y

line

−y

parabola

]dx[/tex]

[tex]\begin{gathered}\rm \: = \: \displaystyle\int_{0}^{1}\rm [x - {x}^{2} ] \: dx \\ \end{gathered}

=∫

0

1

[x−x

2

]dx

[/tex]

[tex]\begin{gathered}\rm \: = \: \bigg[\dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} \bigg] _{0}^{1} \\ \end{gathered}

=[

2

x

2

−

3

x

3

]

0

1[/tex]

[tex]\begin{gathered}\rm \: = \: \dfrac{1}{2} - \dfrac{1}{3} \\ \end{gathered}

=

2

1

−

3

1

[/tex]

[tex]\begin{gathered}\rm \: = \: \dfrac{3 - 2}{6} \\ \end{gathered}

=

6

3−2[/tex]

[tex]\begin{gathered}\rm \: = \: \dfrac{1}{6} \: square \: units\\ \end{gathered}

=

6

1

squareunits

\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c[/tex]