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