Ncert Solutions For Class 12 Maths 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

Related Links
Ncert Exemplar Solutions For Class 11 Maths Pdf	Ncert Books
Ncert Books For Class 6	Ncert Books For Class 7
Ncert Science Book Class 9 Pdf Download	10th Ncert Book
Ncert Books Pdf Class 11	Ncert Books Download For Class 12
Ncert Maths Book Class 11 Solutions Pdf	Class 12 Maths Ncert Book
Ncert Maths Book Class 7 Solutions Pdf	Ncert 10th Maths Book