Question

find the area of region $\rm \left\{(x, y) \mid x^{2} \leq y \leq x\right\}$
$\\ \rule{300pt}{0.1pt}$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given region is

$\rm \left\{(x, y) \mid x^{2} \leq y \leq x\right\}$

So, given curves are

$\rm \: {x}^{2} = y - - - (1)$

and

$\rm \: y = x - - - (2)$

Step :- 1 Point of intersection of two curves

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

$\rm \: {x}^{2} = x$

$\rm \: {x}^{2} - x = 0$

$\rm \: x(x - 1) = 0$

$\rm\implies \:x = 0 \: \: \: or \: \: \: x = 1$

Hᴇɴᴄᴇ,

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

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

Step :- 2 Curve Sketching

The curve $x^2 = y$ 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

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

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

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

$\rm \: = \: \dfrac{1}{2} - \dfrac{1}{3}$

$\rm \: = \: \dfrac{3 - 2}{6}$

$\rm \: = \: \dfrac{1}{6} \: square \: units$

$\rule{190pt}{2pt}$

Additional Information :-

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

Answer 2

$\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =$

Given region is

$\begin{gathered}\rm \left\{(x, y) \mid x^{2} \leq y \leq x\right\} \\ \end{gathered} </p><p>{(x,y)∣x </p><p>2</p><p> ≤y≤x}$

So, given curves are

$\begin{gathered}\rm \: {x}^{2} = y - - - (1) \\ \end{gathered} </p><p>x </p><p>2</p><p> =y−−−(1)$

and

$\begin{gathered}\rm \: y = x - - - (2) \\ \end{gathered} </p><p>y=x−−−(2)</p><p>$

Step :- 1 Point of intersection of two curves

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

$\begin{gathered}\rm \: {x}^{2} = x \\ \end{gathered} </p><p>x </p><p>2</p><p> =x$

$\begin{gathered}\rm \: {x}^{2} - x = 0\\ \end{gathered} </p><p>x </p><p>2</p><p> −x=0$

$\begin{gathered}\rm \: x(x - 1) = 0\\ \end{gathered} </p><p>x(x−1)=0$

$\begin{gathered}\rm\implies \:x = 0 \: \: \: or \: \: \: x = 1 \\ \end{gathered} </p><p>⟹x=0orx=1$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf 1 & \sf 1 \end{array}} \\ \end{gathered}\end{gathered} </p><p>x</p><p></p><p> </p><p>0</p><p>1</p><p></p><p> </p><p>y</p><p></p><p> </p><p>0</p><p>1</p><p>$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

$\begin{gathered}\rm \: = \: \displaystyle\int_{0}^{1}\rm [y_{line} - y_{parabola}] \: dx \\ \end{gathered} </p><p>=∫ </p><p>0</p><p>1</p><p></p><p> [y </p><p>line</p><p></p><p> −y </p><p>parabola</p><p></p><p> ]dx$

$\begin{gathered}\rm \: = \: \displaystyle\int_{0}^{1}\rm [x - {x}^{2} ] \: dx \\ \end{gathered} </p><p>=∫ </p><p>0</p><p>1</p><p></p><p> [x−x </p><p>2</p><p> ]dx</p><p>$

$\begin{gathered}\rm \: = \: \bigg[\dfrac{ {x}^{2} }{2} - \dfrac{ {x}^{3} }{3} \bigg] _{0}^{1} \\ \end{gathered} </p><p>=[ </p><p>2</p><p>x </p><p>2</p><p> </p><p></p><p> − </p><p>3</p><p>x </p><p>3</p><p> </p><p></p><p> ] </p><p>0</p><p>1$

$\begin{gathered}\rm \: = \: \dfrac{1}{2} - \dfrac{1}{3} \\ \end{gathered} </p><p>= </p><p>2</p><p>1</p><p></p><p> − </p><p>3</p><p>1</p><p>$

$\begin{gathered}\rm \: = \: \dfrac{3 - 2}{6} \\ \end{gathered} </p><p>= </p><p>6</p><p>3−2$

$\begin{gathered}\rm \: = \: \dfrac{1}{6} \: square \: units\\ \end{gathered} </p><p>= </p><p>6</p><p>1</p><p></p><p> squareunits</p><p></p><p> </p><p></p><p>\rule{190pt}{2pt}$

Additional Information :-

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