# Ncert Solutions For Class 12 Maths Ex 8.3

## Ncert Solutions For Class 12 Maths Chapter 8 Ex 8.3

Q.1: Find the area enclosed by the curve whose equations are: y = 2x2, x = 2, x = 3 and x-axis.

Sol:

The equation y = 2x2 represents a parabola symmetrical about y-axis.

Now, the Area of region enclosed by the curve ABCDA [y = 2x2]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{2}^{3}y\;dx=\int_{2}^{3}2x^{2}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{2x^{3}}{3} \right ]_{2}^{3}=\left [ 18-\frac{16}{3} \right ]=\frac{38}{3}}$$ unit2

Therefore, the Area of shaded region $$=\frac{38}{3}$$ unit2

Q.2: Find the area enclosed by the curve whose equations are: y = 5x4, x = 3, x = 7 and x-axis.

Sol:

The equation y = 2x2 represents a quartic parabola symmetrical about y-axis.

Now, the Area of region enclosed by the curve ABCDA [y = 5x4]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{7}y\;dx=\int_{3}^{7}5x^{4}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{5x^{5}}{5} \right ]_{3}^{7}=\left [ 16807-243 \right ]=16564}$$ unit2

Therefore, the Area of shaded region = 16807 unit2

Q.3: Find the area enclosed between the curve y2 = 3x and line y = 6x.

Sol:

Equation y2 = 3x represents a parabola, symmetrical about x-axis.

Now, substituting the Equation of line y = 6x in the equation of parabola:

(6x)2 = 3x $$\Rightarrow x = \frac{1}{12}$$ which gives y = $$\frac{1}{2}$$

Hence the coordinates of point A are $$(\frac{1}{12},\frac{1}{2})$$

The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

Now, the Area enclosed by the curve, OMABO [y2 = 3x]:

Since, y2 = 3x

Therefore, y = $$\sqrt{3x}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{12}} y\;dx = \int_{0}^{\frac{1}{12}}\sqrt{3x}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{3}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{1}{12}}=\sqrt{3}\times \frac{2}{3}\times \left ( \frac{1}{12} \right )^{\frac{3}{2}}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\sqrt{3} \times \frac{2}{3}\times \frac{1}{24\sqrt{3}}=\frac{1}{36}}$$ unit2

Therefore, the Area enclosed by the curve OMABO = $$\frac{1}{36}$$unit2

Now, the Area enclosed by the curve OAMO [y = 6x]:

$$\\\boldsymbol{\Rightarrow }$$ $$Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude$$

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{48}}$$unit2

Therefore, the Area enclosed by the curve OAMO = $$\frac{1}{48}$$unit2

Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{36}-\frac{1}{48} = \frac{1}{144}$$unit2

Therefore, the Area enclosed by the curve OABO = $$\frac{1}{144}$$unit2

Q.4 Find the area enclosed by the curve y = 2x2 and the lines y = 1, y = 3 and the y-axis.

Sol:

Equation y = 2x2 represents a parabola symmetrical about y-axis.

The Area of the region bounded by the curve y = 2x2, y = 1, and y = 3, is the Area enclosed by the curve AA’B’BA.

Now, the Area of region AA’B’BA = 2 (Area of region ABNMA)

Since, 2x2 = y

Therefore, x = $$\sqrt{\frac{y}{2}}$$

Thus, the Area of region bounded by the curve ABNMA [y = 2x2]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{3}x\;dy=\int_{1}^{3}\sqrt{\frac{y}{2}}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{\sqrt{2}}\times {\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{3}=\frac{\sqrt{2}}{3}\times [(3^{\frac{3}{2}})-(1)^{\frac{3}{2}}}}]\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{\sqrt{2}}{3}\times (3\sqrt{3}-1) = \frac{\sqrt{2}}{3}(3\sqrt{3}-1)}$$unit2

Therefore, the Area of region bounded by the curve ABNMA = $$\frac{\sqrt{2}}{3}(3\sqrt{3}-1)$$unit2

Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)$$= \frac{2\sqrt{2}}{3}(3\sqrt{3}-1)$$unit2

Q.5: Find the area enclosed between the curve y2 = 9ax and line y = mx.

Sol:

Equation y2 = 9ax represents a parabola, symmetrical about x-axis.

Now, substituting the Equation of line y = mx in the equation of parabola:

9ax = (mx)2 i.e x = $$\frac{9a}{m^{2}}$$ which gives y = $$\frac{9a}{m}$$

Hence the co-ordinates of point A are $$\left ( \frac{9a}{m^{2}},\frac{9a}{m}\right )$$

The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

Now, the Area enclosed by the curve OMABO [y2 = 9ax]:

Since, y2 = 9ax

Therefore, y = $$3\sqrt{ax}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{9a}{m^{2}}} y\;dx = \int_{0}^{\frac{9a}{m^{2}}}3\sqrt{ax}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{3\sqrt{a}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{9\;a}{m^{2}}}=3\sqrt{a}\times \frac{2}{3}\times \left ( \frac{9\;a}{m^{2}} \right )^{\frac{3}{2}}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{2\sqrt{a} \times 27\times a\sqrt{a}\times \frac{1}{m^{3}}=\frac{54\;a^{2}}{m^{3}}}$$ unit2

Therefore, the Area enclosed by the curve OMABO $$=\frac{54\;a^{2}}{m^{3}}$$ unit2

Now, the Area enclosed by the curve OAMO [y = mx]:

$$\\\boldsymbol{\Rightarrow }$$ $$Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude$$

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{9\;a}{m^{2}}\times \frac{9\;a}{m}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{81\;a^{2}}{2\;m^{3}}}$$ unit2

Therefore, the Area enclosed by the curve OAMO =$$\frac{81\;a^{2}}{2\;m^{3}}$$ unit2

Now, the Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO

i.e. $$\frac{54\;a^{2}}{m^{3}}-\frac{81\;a^{2}}{2\;m^{3}} = \frac{27\;a^{2}}{2\;m^{3}}$$ unit2

Therefore, the Area enclosed by the curve OABO = $$\frac{27\;a^{2}}{2\;m^{3}}$$unit2

Q.6: Find the area bounded by the curve whose equation is 3x2 = 4y and the line 2y – 12 = 3x.

Sol:

Equation 3x2 = 4y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola 3x2 = 4y and the line 2y -12 = 3x is the Area enclosed under the curve ABC0A.

Since, the parabola 3x2 = 4y and the line 3x = 2y – 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:

Since, $$\;x=\frac{2y-12}{3}$$

$$\\\boldsymbol{\Rightarrow }$$ $$3\left ( \frac{2y-12}{3} \right )^{2}=4y\;\;\;i.e. \;\;\;(2y-12)^{2}=12y$$

$$\\\boldsymbol{\Rightarrow }$$ 4y2 +144 – 48y – 12 y = 0

$$\\\boldsymbol{\Rightarrow }$$ y2 – 15y + 36 = 0

By splitting the middle term Method solutions of this quadratic equation are:

y2 – (12+3)y + 36 = 0 $$\Rightarrow$$ y(y – 12) –3(y – 12) = 0

$$\Rightarrow$$ (y – 3) (y – 12) = 0

Therefore, y = 12 and y = 3 which gives x = 4 and x = -2 respectively.

Hence, the co-ordinates of point A and point C are (-2, 3) and (4, 12) respectively.

Since, 3x2 = 4y

Therefore, y = $$\frac{3x^{2}}{4}$$

The Area of region bounded by the curve ABCOA = [Area of region aACba] – [Area of region OCbO + Area of region OAaO ]

The Area enclosed by the curve aACBa [2y – 12 = 3x]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{4} y\;dx \Rightarrow \int_{-2}^{4}\frac{3x+12}{2}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\left | \frac{3x^{2}}{2}+ 12x \right |_{-2}^{4}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\left [ 24+48-6-(-24) \right ]=45}$$ unit2

The Area enclosed by the curve OAaO [3x2 = 4y]:

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} \frac{3x^{2}}{4}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{-2}^{0}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{3}{4}(0-\frac{-8}{3}) \right |=2}$$ unit2

The Area enclosed by the curve OCbO [3x2 = 4y]:

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{4}\;y\;dx\;\Rightarrow \int_{0}^{4} \frac{3x^{2}}{4}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{0}^{4}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{3}{4}(\frac{64}{3}-0) \right |=16}$$ unit2

Now, the Area of region bounded by the curve ABCOA = [ Area of region aACba ] – [ Area of region OCbO + Area of region OAaO ]

$$\\\boldsymbol{\Rightarrow }$$ 45 – [2 + 16] = 27 unit2

Therefore, the Area of shaded region ABCOA = 27 unit2

Q.7: Find the area enclosed by the curves {(x , y) : 6y x2 and y = |x|}

Sol:

Equation x2 = 6y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by the curve x2 = 6y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)

Now, Area of region OAEO = OABO – OEABO

Since, x2 = 6y

Therefore, $$y=\frac{x^{2}}{6}$$

Now, the Area of region bounded by the curve OEABO:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}\frac{x^{2}}{6}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{6}\left | \frac{x^{3}}{3} \right |_{0}^{6}=12}$$ unit2

Therefore, the Area of region bounded by the curve OEABO = 12 unit2

Now, the Area of region bounded by the curve OABO:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}x\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{6}=18}$$ unit2

Therefore, the Area of the region bounded by the curve OABO = 18 unit2

Now, Area of region OAEO = Area of region (OABO – OEABO)

$$\\\boldsymbol{\Rightarrow }$$ 18 – 12 = 6 unit2

Therefore, the total Area of shaded region = 2 × 6 = 12 unit2

Q.8: Find the area enclosed by the sides of a triangle whose vertices have coordinates (3, 0) (5, 8) and (7, 5).

Sol:

Form the above figure:

Let, A (3, 0), B (5, 8) and C (7, 5) be the vertices of triangle ABC.

Now, the equation of line AB:

Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$

$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-3)\times \left(\frac{8-0}{5-3}\right)$$

$$\boldsymbol{\Rightarrow }$$ 2y = 8x – 24

$$\boldsymbol{\Rightarrow }$$ y=4x-12

The Equation of line BC:

Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$

$$\boldsymbol{\Rightarrow }$$ $$(y-8)=(x-5)\times \left(\frac{5-8}{7-5}\right)$$

$$\boldsymbol{\Rightarrow }$$ 2y – 16 = -3x + 15

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\Rightarrow y=\frac{31-3x}{2}}$$

The Equation of line AC:

Since, $$(y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)$$

$$\boldsymbol{\Rightarrow }$$ $$(y-0)=(x-3)\times \left(\frac{5-0}{7-3}\right)$$

$$\boldsymbol{\Rightarrow }$$ 4y=5x-15

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\Rightarrow y=\frac{5x-15}{4}}$$

Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.

The Area under the curve ABMA [y = 4x – 12]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}(4x-12)\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{4x^{2}}{2}-12x\right ]_{3}^{5}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[50-60]-[18-36]=8}$$unit2

Therefore, Area under the curve ABMA = 8 unit2

The Area under the curve MBCN [2y + 3x = 31]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{5}^{7}y\;dx\;=\;\int_{5}^{7}\frac{31-3x}{2}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times \left [ 31x-\frac{3x^{2}}{2}\right ]_{5}^{7}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[\frac{217}{2}-\frac{147}{4}]-[\frac{155}{2}-\frac{75}{4}]=13}$$ unit2

Therefore, Area under the curve MBCN = 13 unit2

The Area under the curve ACNA [4y = 5x – 15]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{3}^{7}y\;dx\;=\;\int_{3}^{7}\left ( \frac{5x-15}{4} \right )dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{5x^{2}}{8}-\frac{15x}{4} \right ]_{3}^{7}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{245}{8}-\frac{105}{4} \right ]-\left [ \frac{45}{8}-\frac{45}{4} \right ]=10}$$ unit2

Therefore, Area under the curve ACNA = 10 unit2

Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA.

Therefore, the Area of triangle ABC = 8 + 13 – 10 = 11 unit2

Q.9: Find the area enclosed by the sides of a triangle whose equations are: 2x – 4 = y, – 2y = -3x + 6 and x – 3y = -5.

Sol:

From the above figure:

The Equation of line AB: 3y = x + 5 . . . . . . (1)

The Equation of line BC: y = 4 – 2x . . . . . . (2)

The Equation of line AC: 2y = 3x – 6 . . . . . . (3)

From equation (1) and equation (2):

3(4 – 2x) = x + 5 i.e. x = 1, which gives y = 2

Therefore, the coordinates of point B are (1, 2)

From equation (2) and equation (3):

2(4 – 2x) = 3x – 6 i.e. x = 2 which gives y = 0

Therefore, the coordinates of point C are (2, 0).

From equation (1) and equation (3):

2y = 3(3y – 5) – 6 i.e. y = 3 which gives x = 4

Therefore, the coordinates of point A are (4, 3).

Now, the Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA

The Area under the curve ABMNA [3y = x + 5]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{4}y\;dx\;=\;\int_{1}^{4}\frac{x+5}{3}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{2}}{6}+\frac{5x}{3} \right ]_{1}^{4}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{16}{6}+\frac{20}{3}-\frac{1}{6}-\frac{5}{3}=\frac{15}{2}}$$ unit2

Therefore, the Area under the curve ABMNA $$=\frac{15}{2}$$ unit2

The Area under the curve BMCB [y = 4 – 2x]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{1}^{2}y\;dx\;=\;\int_{1}^{2}(4-2x)\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 4x-x^{2} \right ]_{1}^{2}}$$

$$\\\boldsymbol{\Rightarrow }$$ [8 – 4] – [4 – 1] =1 unit2

Therefore, the Area under the curve MBCN = 1 unit2

The Area under the curve ACNA [2y = 3x – 6]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{2}^{4}y\;dx\;=\;\int_{2}^{4}\frac{3x-6}{2}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{3x^{2}}{4}-\frac{6x}{2} \right ]_{2}^{4}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{3\times 16}{4}-\frac{6\times 4}{2} \right ]- \left [ \frac{3\times 4}{4}-\frac{6\times 2}{2} \right ]=3}$$ unit2

Therefore, the Area under the curve ACNA = 3 unit2

The Area of triangle ABC = Area enclosed by the curve ABMNA Area enclosed by the curve BMCB – Area enclosed by the curve ACNA

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{15}{2}-1-3 = 3.5$$ unit2

Therefore, the Area of triangle ABC = 3.5 unit2

Q.10: Find the area enclosed by the curve 2x2 = y and the line y = 2x + 12 and x – axis.

Sol:

Equation 2x2 = y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola 2x2 = y and the line y = 2x + 12 and x-axis is the Area enclosed under the curve ABCOA.

Since, the parabola 2x2 = y and the line y = 2x + 12 intersect each other at points A and C, hence the coordinates of points A and C are given by:

Since, y = 2x + 12

$$\\\boldsymbol{\Rightarrow }$$ 2x2 = (2x+12)

$$\\\boldsymbol{\Rightarrow }$$ x2 – x – 6 = 0

By splitting the middle term Method solutions of this quadratic equation are:

x2 – (3 – 2)x – 6 = 0 $$\Rightarrow$$ x(x – 3) +2(x – 3) = 0

$$\Rightarrow$$ (x – 3) (x + 2) = 0

Therefore, x = 3 and x = -2 which gives y = 18 and y = 8 respectively.

Hence, the co-ordinates of point E and point A are (3, 18) and (-2, 8) respectively.

Since, 2x2 = y

Therefore, y = 2x2

The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]

The Area enclosed by the curve ACOA [ 2x2 = y ]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} 2x^{2}\;dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | \frac{2\;x^{3}}{3}\right |_{-2}^{0}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left |(0-\frac{-16}{3}) \right |=\frac{16}{3}}$$ unit2

The Area enclosed by the curve ABC [ y = 2x + 12 ] :

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{Area\;of\;\Delta ABC=\frac{1}{2}\times Base\times Altitude}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times |BC|\times |AC|=\frac{1}{2}\times \left | 4 \right |\times |8|=16}$$ unit2

The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]

$$\Rightarrow \frac{16}{3}+16=\frac{64}{3}$$ unit2

Therefore, the Area of shaded region ABCOA $$=\frac{64}{3}$$unit2

Q.11: Plot the curve y = |x + 4| and hence evaluate $$\int_{-9}^{0}|x+4|\;dx$$

Sol:

From the given equation the corresponding values of x and y are given in the following table.

 X -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 Y 5 4 3 2 1 0 1 2 3 4 5 6 7

Now, on using these values of x and y, we will plot the graph of y = |x + 4|

From the above graph, the required Area = the Area enclosed by the curve ABCA + the Area enclosed by the curve CDOC

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{-9}^{0}|x+4|\;dx+\int_{-9}^{-4}(x+4)\;dx+\int_{-4}^{0}(x+4)\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-9}^{-4}+\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-4}^{0}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left | 8-16-\frac{81}{2}+36 \right |+\left | 0-(8-16) \right |=\left | \frac{-25}{2} \right |+8}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{41}{2}}$$ unit2

Therefore, the area of shaded region = $$\frac{41}{2}$$ unit2

Q.12: Find the area enclosed by the curve y = sin x between 0 x ≤ 2π

Sol:

From the above figure, the required Area is represented by the curve OABCD.

Now, the Area bounded by the curve OABCD = the Area bounded by the curve OABO + the Area bounded by the curve BCDB

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\pi } sin(x)\;dx+\left | \int_{\pi }^{2\pi } sin(x)\;dx\right |}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos(x)\right ]_{0}^{\pi }+\left | \left [ -cos (x) \right ] _{\pi }^{2\pi }\right |}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{[-cos(\pi )+cos(0)]+\left | [-cos(2\pi )+cos(\pi )] \right |}\\$$

$$\boldsymbol{\Rightarrow }$$ [1+1] + [ |– 1 – 1| ] unit2

Therefore, the area of shaded region = 4 unit2

Q.13: Find the area of smaller region enclosed by the curve $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ and the line $$\frac{x}{2}+\frac{y}{3}=1$$

Sol:

The Equation $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ represents an ellipse.

The Equation $$\frac{x}{2}+\frac{y}{3}=1$$ represents a line with x and y intercepts as 2 and 3 respectively.

Since, $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{y^{2}}{9}=1-\frac{x^{2}}{4}$$

$$\\\boldsymbol{\Rightarrow }$$ $$y=\frac{3}{2}\sqrt{4-x^{2}}$$

Therefore, the Area of smaller region enclosed by the Ellipse $$\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1$$ and the line $$\frac{x}{2}+\frac{y}{3}=1$$ is represented by curve ACBA

Now, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – Area enclosed by the curve ABOA

Now, the Area enclosed by the curve ACBOA:

$$\boldsymbol{\int_{0}^{2} y\;dx\Rightarrow\frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\\$$

Since, $$\ \int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}$$ unit2

Therefore, the Area enclosed by the curve ACBOA = $$\boldsymbol{\frac{3\pi}{2}}$$unit2

Now, the Area enclosed by the curve ABOA:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{Area \;of\;\Delta ABO = \frac{1}{2}\times AO\times BO}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{1}{2}\times 2\times 3=3}$$ unit2

Therefore, the Area enclosed by the curve ABOA = 3 unit2

Since, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – the Area enclosed by the curve ABOA

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)$$unit2

Therefore, the Area of shaded region $$=\frac{3}{2}(\pi -2)$$ unit2

Q.14: Find the area enclosed by the curve |x| + |y| = 2, by using the method of integration.

Sol:

Equation |x| + |y| = 2 represent a region bounded by the lines:

x + y = 2 . . . . . . (1)

x – y = 2 . . . . . . (2)

-x + y = 2 . . . . . . (3)

-x – y = 2 . . . . . . (4)

From equations (1), (2), (3) and (4) we conclude that the curve intersects x-axis and y-axis axis at points A (0, 2), B (2, 0), C (0, -2) and D (-2, 0) respectively.

From the above figure:

Since, the curve is symmetrical to x-axis and y-axis. Therefore, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA

Now, the Area of region bounded by the curve ABOA:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=4\int_{0}^{2}(2-x)dx}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 2x-\frac{x^{2}}{2} \right ]_{0}^{2}=(4-2)=2}\\$$ unit2

Therefore, the Area of region bounded by the curve ABOA = 2unit2

Since, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA

Therefore, the Area of region bounded by the curve ABCDA = (2 × 4) = 8 unit2

Therefore, the Area of shaded region = 8 unit2

Q.15: Find the area which is exterior to curve x2 = 2y and interior to curve x2 + y2 = 15.

Sol:

The Equation x2 = 2y represents a parabola symmetrical about y-axis.

The Equation x2 + y2 = 15 represents a circle with centre (0, 0) and radius units.

Now, on substituting the equation of parabola in the equation of circle we will get:

(2y) + y2 = 15 i.e. y2 + 2y – 15 = 0

Now, by splitting the middle term method solutions of this quadratic equation are:

y2 + (5 – 3)y – 15 = 0 $$\Rightarrow$$ y(y + 5) – 3(y +5) = 0

$$\Rightarrow$$ (y – 3) (y + 5) = 0

Neglecting y = -5 [gives absurd values of x]

Therefore, y = 3 which gives x = $$\pm \sqrt{6}$$

Hence, the coordinates of point B and point D are ($$\sqrt{6}$$, 3) ($$-\sqrt{6}$$, 3) respectively.

Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB+ Area of region bounded by the curve OAC’O)

Area of region bounded by the curve BAMB [ x 2 + y2 = 15 ]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{\sqrt{6}}^{\sqrt{15}}y\;dx=\int_{\sqrt{6}}^{\sqrt{15}}\sqrt{\left ( \sqrt{15} \right )^{2}-x^{2}}\;\;dx}\\$$

Since, $$\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{15- x^{2}}+\frac{15}{2}\sin^{-1}\frac{x}{\sqrt{15}} \right ]_{\sqrt{6}}^{\sqrt{15}}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac {\sqrt{15}}{2}\times \sqrt{{15-15}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{15}}{\sqrt{15}} \right ]-}\\$$ $$\\\boldsymbol{\left [\frac {\sqrt{6}}{2}\times \sqrt{{15-6}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{6}}{\sqrt{15}}\right]}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 0+\frac{15\pi }{4} \right ]-\left [\frac{3\sqrt{6}}{2}+\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}\right]}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{15\pi }{4}=\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5} \right ]}$$unit2

Area of region bounded by the curve OBMO [ x2 = 2y ]:

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\sqrt{6}}y\;dx=\int_{0}^{\sqrt{6}}\frac{x^{2}}{2}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^3}{6} \right ]_{0}^{\sqrt{6}}=\frac{6\sqrt{6}}{6}-0}\\$$ = $$\boldsymbol{\sqrt{6}}$$ unit2

Area of region bounded by the curve OAC’O:

$$\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( \sqrt{15} \right )^{2} }{4}}\\$$ = $$\boldsymbol{\frac{15\pi }{4}}$$unit2

Now, the Area of region bounded by the curve BAC’A’DOB = 2 × (Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB + Area of region bounded by the curve OAC’O)

Therefore, the Area of region bounded by the curve BAC’A’DOB:

$$\boldsymbol{\Rightarrow }$$ $$2\left [ \frac{15\pi }{4}-\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}+\sqrt{6} +\frac{15\pi }{4}\right ]\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\\$$unit2

Therefore, the Area of shaded region: $$\boldsymbol{ \left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}$$unit2

Q.16: Find the area enclosed by the curve y = x3, x-axis and the lines x = -2 and x = 2.

Sol.

The Equation y = x3 represents a cubic parabola which intersects the line x = 2 and x = -2 at points A and D respectively

From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}x^{3}\;dx+\left | \int_{-2}^{0}x^{3}\;dx \right |}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{4}}{4} \right ]_{0}^{2}+\left | \left [ \frac{x^{4}}{4} \right ]_{-2}^{0} \right |}$$

$$\\\boldsymbol{\Rightarrow }$$ 4 – 0 + 0 + 4 =16 unit2

Therefore, the Area of shaded region = 16 unit2

Q.17: Find the area enclosed by the curve y = x|x|, y – axis and the lines y = -1 and y = 3.

Sol:

Now, y = x|x| is equal to [y = x2] when x > o

And y = x|x| is equal to [y = -x2] when x < o

From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{3}x^{2}\;dx+\left | \int_{-1}^{0}-x^{2}\;dx \right |}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x^{3}}{3}\right ]_{0}^{3}+\left | \left [ \frac{x^{3}}{3} \right ]_{-1}^{0} \right |}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{9+\frac{1}{3}=\frac{28}{3}}$$unit2

Therefore, the area of shaded region $$=\frac{28}{3}$$unit2

Q.18: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and y – axis, when [0 ≤ x ≤ $$\frac{\pi }{2}$$].

Sol:

y = Cos(x) . . . . . . . . (1)

y = Sin(x) . . . . . . . . . (2)

Now, from equation (1) and equation (2):

Cos (x) = Sin (x) $$\Rightarrow$$ Cos (x) = Cos $$\left [ \frac{\pi }{2}-x \right ]$$

$$\Rightarrow$$ x = $$\left [ \frac{\pi }{2}-x \right ]$$ $$\Rightarrow$$ x = $$\frac{\pi }{4}$$

Therefore, the coordinates of point A are: $$\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )$$

Now, from the above figure:

The required Area = Area enclosed by the curve ADMA + Area enclosed by the curve AMOA

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\\$$

Since, $$\int \sin^{-1}y\;dy= y\sin^{-1}y+\sqrt{1-y^{2}}\\$$

And, $$\\\int \cos^{-1}y\;dy= y \cos^{-1}y-\sqrt{1-y^{2}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ y\sin^{-1}y+\sqrt{1-y^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}+\left [ y\cos^{-1}y-\sqrt{1-y^{2}} \right ]_{\frac{1}{\sqrt{2}}}^{1}}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{1}{\sqrt{2}}\times \frac{\pi }{4}+\frac{1}{\sqrt{2}}-(0+1) \right ]+\left [ 0-\left ( \frac{1}{\sqrt{2}}\times \frac{\pi }{4}-\frac{1}{\sqrt{2}} \right ) \right ]}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [\sqrt{2}-1 \right ]}$$unit2

Therefore, the Area of shaded region = $$\left [\sqrt{2}-1 \right ]$$unit2

Q.20: Find the area which is exterior to curve y2 = 6x and interior to curve x2 + y2 = 16.

Sol:

The Equation y2 = 6x represents a parabola symmetrical about x – axis.

The Equation x2 + y2 = 16 represents a circle with centre (0, 0) and radius 4 units.

Now, on substituting the equation of parabola in the equation of circle we will get:

x2 + (6x) = 16 i.e. x2 + 6x – 16 = 0

Now, by splitting of middle term method solutions of this quadratic equation are:

x2 + (8 – 2)x – 16 = 0 $$\Rightarrow$$ x(x + 8) – 2(x + 8) = 0

$$\Rightarrow$$ (x – 2) (x + 8) = 0

Neglecting x = -8 [gives absurd values of y]

Therefore, x = 2 which gives y = $$\pm 2\sqrt{3}$$

Hence, the coordinates of point B and point D are (2, $$2\sqrt{3}$$) (2, $$-2\sqrt{3}$$) respectively.

Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)

Area of region bounded by the curve BPMOB [x 2 + y2 = 16]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{\left ( 4\right )^{2}-x^{2}}\;\;dx}\\$$

Since, $$\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{x}{2}\sqrt{16- x^{2}}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]_{0}^{2}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{2}{2}\times \sqrt{{16-4}}+8\times \sin^{-1}\frac{1}{2} \right ]-0}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ 2\sqrt{3}+\frac{8\times \pi }{6} \right ]=\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]}\\$$unit2

Therefore, Area of region bounded by the curve BPMOB = $$\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]$$unit2

Area of region bounded by the curve POMP [y2 = 6x]:

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{6x}\;dx}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\left [ \frac{\sqrt{6}\times 2}{3}\times x^\frac{3}{2} \right ]_{0}^{2}=\frac{2\sqrt{6}}{3}\times 2\sqrt{2}-0\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\frac{8\sqrt{3}}{3}$$unit2

Therefore, Area of region bounded by the curve POMP : $$\frac{8\sqrt{3}}{3}$$unit2

Area of region bounded by the curve BOA’B:

$$\\\boldsymbol{\Rightarrow \frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( 4 \right )^{2} }{4}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ 4π unit2

Therefore, Area of region bounded by the curve BOA’B = 4π unit2

Now, the Area of region bounded by the curve PBA’B’NOP = 2 × (Area of region bounded by the curve BPMOB – Area of region bounded by the curve POMP + Area of region bounded by the curve BOA’B)

Therefore, the Area of region bounded by the curve PBA’B’NOP:

$$\\\boldsymbol{\Rightarrow }$$ $$2\left [2\sqrt{3}+\frac{4\pi }{3}-\frac{8\sqrt{3}}{3}+4\pi \right ]\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{32\pi }{3}-\frac{4\sqrt{3}}{3} \right ]=\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]}\\$$unit2

Therefore, the Area of shaded region = $$\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]$$unit2

Q.21: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and x – axis, when [0 ≤ x ≤ ].

Sol:

y = Cos(x) . . . . . . . . (1)

y = Sin(x) . . . . . . . . . (2)

Now, from equation (1) and equation (2):

Cos (x) = Sin (x) $$\Rightarrow$$ Cos (x) = Cos $$\left [ \frac{\pi }{2}-x \right ]\\$$

$$\\\Rightarrow$$ x = $$\left [ \frac{\pi }{2}-x \right ]\\$$ $$\\\Rightarrow$$ x = $$\frac{\pi }{4}$$

Therefore, the coordinates of point A are: $$\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )$$

Now, from the above figure:

The required Area = Area enclosed by the curve AONA + Area enclosed by the curve ANBA

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos(x) \right ]_{0}^{\frac{\pi }{4}}+[sin (x)]_{\frac{\pi }{4}}^{\frac{\pi }{2}}}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ -cos\frac{\pi }{4}+cos(0)+sin\frac{\pi }{2}-sin\frac{\pi }{4}\right ]}\\$$

$$\\\boldsymbol{\Rightarrow }$$ $$\boldsymbol{\left [ \frac{-1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}} \right ]=[2-\sqrt{2}\;]}$$unit2

Therefore, the Area of shaded region $$= [2-\sqrt{2}\;]$$unit2