Question

Find the area bounded between the curve
${x}^{2} = y \: and \: x - y + 2 = 0$
​

Answer 1

Answer:

Correct option is

B

3

4

sq.units

Curve y=2x−x

2

cuts x-axis at x=0,2

So, points are (0,0),(2,0)

Area of bounded region is

A=∫

0

2

(2x−x

2

)dx

=[x

2

−

3

x

3

]

0

2

=[2

2

−

3

2

3

]−[0

2

−

3

0

3

]

=

3

4

Answer 2

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

Given curves are

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

and

$\rm :\longmapsto\:x - y + 2 = 0$

can be rewritten as

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

Now, Step 1 Point of intersection of Curve 1 and 2

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

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

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

$\rm :\longmapsto\:x(x - 2) + 1(x - 2) = 0$

$\rm :\longmapsto\:(x - 2) (x+ 1)= 0$

$\bf\implies \:x = 2 \: \: or \: \: x = - 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 4 \\ \\ \sf - 1 & \sf 1 \end{array}} \\ \end{gathered}$

Step - 2 :- Curve Sketching

See the attachment

Step - 3

Required area bounded between these two curves is

$\rm \:  =  \: \: \displaystyle\int_{-1}^2(x + 2) \: dx \: - \: \displaystyle\int_{-1}^2 {x}^{2} \: dx \:$

We know,

$\boxed{ \tt{ \: \displaystyle\int \: {x}^{n} \: dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

So, using this identity, we get

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

$\rm \:  =  \: \bigg[\dfrac{4 - 1}{2} + 2(2 + 1)\bigg] - \bigg[\dfrac{8 + 1}{3} \bigg]$

$\rm \:  =  \: \bigg[\dfrac{3}{2} + 2(2 + 1)\bigg] - \bigg[\dfrac{9}{3} \bigg]$

$\rm \:  =  \: \bigg[\dfrac{3}{2} + 6\bigg] - \bigg[3 \bigg]$

$\rm \:  =  \:\dfrac{3}{2} + 3$

$\rm \:  =  \:\dfrac{3 + 6}{2}$

$\rm \:  =  \:\dfrac{9}{2} \: square \: units$

More to know :-

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