Ncert Solutions For Class 12 Maths Ex 8.1

Ncert Solutions For Class 12 Maths Chapter 8 Ex 8.1

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

Sol:

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

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The Area of the region bounded by the curve y = x2, y = 2, and y = 4, is the Area enclosed by the curve AA’B’BA.

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

Since, x2 = y

Therefore, x = \(\sqrt{y}\)

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

\(\boldsymbol{\Rightarrow }\) \(\int_{2}^{4}x\;dy\) = \(\int_{2}^{4}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{2}^{4}=\frac{2}{3}\times [(4^{\frac{3}{2}})-(2)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times (8-2\sqrt{2}) = \frac{4}{3}(4-\sqrt{2})}\) unit2

Therefore, the Area of region bounded by the curve ABNMA = \(\boldsymbol{\frac{4}{3}(4-\sqrt{2})}\)unit2

Hence, the Area of region bounded by the curve AA’B’BA = 2(Area of region bounded by the curve ABNMA)= \(\boldsymbol{\frac{8}{3}(4-\sqrt{2})}\)unit2

 

Q.2: Find the area enclosed by the curve y2 = 4x and lines x = 1, x = 3 and the x- axis in the first quadrant.

Sol:

10

Equation y2 = 4x represents a parabola, symmetrical about the x-axis as shown in the above figure.

The area of the region bounded by the curve y2 = 4x, x = 1, x = 3 and the x-axis is the area enclosed by the curve ABCDA.

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

Since, y2 = 4x

Therefore, y = \(2\sqrt{x}\)

Hence, the area of region bounded by the curve ABCDA [y2 = 4x]:

\(\boldsymbol{\Rightarrow }\) \(\int_{1}^{3}y\;dx\) = \(\int_{1}^{3}2\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=\frac{4}{3}\times [(3^{\frac{3}{2}})-(1)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4}{3}\times (3\sqrt{3}-1) = \frac{-4\;+\;12\sqrt{3}}{3}}\) unit2

Therefore, the area of region bounded by the curve ABCDA \(\boldsymbol{= \frac{-4\;+\;12\sqrt{3}}{3}}\)unit2

Q.3: Find the value of k if the line x = k divides the area enclosed by the curve y2 = 9x and the line x = 4 in to two equal parts.

Sol:

11

Equation y2 = 9x represents a parabola, symmetrical about the x-axis as shown in the above figure.

Since, the line x = k divides the Area OCBB’O in to two equal halves and the curve is symmetrical to x-axis. Therefore, Area of the region OADO = Area of the region ABCDA.

The Area of the region bounded by the curve y2 = 9x and the line x = k is the Area of region enclosed by the curve OADO.

Since, y2 = 9x

Therefore, y = \(3\sqrt{x}\)

Hence, the Area of region bounded by the curve OADO:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{k}y\;dx\) = \(\int_{0}^{k}3\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=2\times [(k^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2k^{\frac{3}{2}}}\)unit2

Therefore, the Area of region bounded by the curve OADO = \(\boldsymbol{2k^{\frac{3}{2}}}\)unit2

The Area of the region bounded by the curve y2 = 9x and the lines x = k and x = 4 is the Area under the curve ABCDA

Now, the Area of region bounded by the curve ABCDA [y2 = 9x]:

\(\boldsymbol{\Rightarrow }\) \(\int_{k}^{4}y\;dx\) = \(\int_{k}^{4}3\sqrt{x}\;dx\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{k}^{4}=2\times [(4^{\frac{3}{2}})-(k)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})}\) unit2

Therefore, the Area of region bounded by the curve ABCDA = \(\boldsymbol{\Rightarrow (16-2k^{\frac{3}{2}})}\) unit2

Since, the Area of region OADO = Area of region ABCDA [GIVEN]

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})=2k^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k^{\frac{3}{2}}=4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k = 4^{\frac{2}{3}}}\)

Therefore, the value of k = \(\boldsymbol{4^{\frac{2}{3}}}\)

Q.4: Find the area enclosed by the curve y2 = 16x, y-axis and the line y = 2.

Sol:

12

Equation y2 = 16x represents a parabola, symmetrical about the x-axis as shown in the above figure.

The Area of the region bounded by the curve y2 = 16x, y = 2 and the y-axis is the Area of region enclosed by the curve OABO.

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

Since, y2 = 16x

Therefore, x = \(\frac{y^{2}}{16}\)

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

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{2}x\;dy\) = \(\int_{0}^{2}\frac{y^{2}}{16}\;dy\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{1}{16}\times \frac{y^{3}}{3}\right |_{0}^{2}=\frac{1}{16}\times [\frac{8}{3}-0}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}}\) unit2

Therefore, Area of region bounded by the curve OABO \(\boldsymbol{=\frac{1}{6}}\)unit2

Q.5: Find the area enclosed by the curve y2 = 25x and the line x = 3

Sol:

13

Equation y2 = 25x represents a parabola, symmetrical about x-axis as shown in the above figure.

The area of the region bounded by the curve y2 = 25x and x = 3 is the Area enclosed by the curve BOCAB.

Now, the Area of region BOCAB = 2(Area of region OABO)

Since, y2 = 25x

Therefore, y = \(5\sqrt{x}\)

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

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx\) = \(\int_{0}^{3}5\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 5\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{0}^{3}=\frac{10}{3}\times [(3^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{10}{3}\times (3\sqrt{3})=10\sqrt{3}}\)unit2

Therefore, the Area of region bounded by the curve OABO\(\boldsymbol{=10\sqrt{3}}\) unit2

Now, the Area of region BOCAB = 2 × (Area of region OABO) \(\boldsymbol{=20\sqrt{3}}\) unit2

Q.6: Find the area enclosed by the curve x2 = 8y, y = 1, y = 9 and the y-axis in the first quadrant.

Sol:

14

Equation x2 = 8y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The area of the region bounded by the curve x2 = 8y, y = 1, y = 9 and the first quadrant is the area enclosed by the curve ABDCA.

Now, the Area of the region ABDCA:

Since, x2 = 8y

Therefore, x = \(2\sqrt{2y}\)

Hence, the Area of region bounded by the curve ABDCA:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{1}^{9}x\;dy\) = \(\int_{1}^{9}2\sqrt{2y}\;dy\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\sqrt{2}\times \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{9}=\frac{4\sqrt{2}}{3}\times [(9^{\frac{3}{2}})-(1)}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4\sqrt{2}}{3}\times (27-1)=104\times \frac{\sqrt{2}}{3}}\) unit2

Therefore, Area of region bounded by the curve ABDCA \(\boldsymbol{=104\times \frac{\sqrt{2}}{3}}\)unit2

Q.7: Find the area bounded by the curve whose equation is x2 = 9y and the line x = 6y – 3.

Sol:

15

Equation x2 = 9y represents a parabola, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola x2 = 9y and the line x =6y – 3 is the Area enclosed under the curve ABC0A.

Since, the parabola x2 = 9y and the line x = 6y – 3 intersect each other at points A and C, hence the coordinates of points A and C are given by:

(6y-3)2 = 9y

\(\boldsymbol{\Rightarrow }\) 36y2 – 45y + 9 = 0

By Hit and Trial method solutions of this quadratic equation are:

y = 1 and y = \(\frac{1}{4}\)

Hence, the co-ordinates of point A and point C are (3,1) and \((\frac{-3}{2},\frac{1}{4})\) respectively

Since, x2 = 9y

Therefore, x = \(3\sqrt{y}\)

The Area of region bounded by the curve ABCOA = [Area of region OBCbOArea of region OCbO] + [Area of region ABOaAArea of region OAaO]

The Area enclosed by the curve OBCbO:

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3} y\;dx = \boldsymbol{\int_{0}^{3}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(\frac{9}{2}+9)=\frac{9}{4}}\) unit2

The Area enclosed by the curve ABOaA:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx} = \boldsymbol{\int_{\frac{-3}{2}}^{0}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(0-\frac{9}{8}+\frac{9}{2})=\frac{9}{16}}\) unit2

The Area enclosed by the curve OAaO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx\; =\;\int_{\frac{-3}{2}}^{0} \frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{27}(0-\frac{-27}{8})=\frac{1}{8}}\)unit2

The Area enclosed by the curve OCbO:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3} y\;dx\; = \;\int_{0}^{3}\frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}(\frac{27}{3}-0)=1}\) unit2

Since, the Area of region bounded by the curve ABCOA = [Area of region OBCbO – Area of region OCbO] + [Area of region ABOaA – Area of region OAaO]

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{9}{4}-1]+[\frac{9}{16}-\frac{1}{8}]=\frac{27}{16}}\)

Therefore, the Area of region bounded by the curve ABCOA \(=\frac{27}{16}\) unit2

 

 

Q.8: Find the area enclosed by the curve 4y = x2 and y = |x|

Sol:

16

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

The area of the region bounded by the curve x2 = 4y and y = |x| is 2(OAEO) i.e. (area OCFO+ area OAEO)

Now, Area of region OAEO = OABO – OEABO

Since, x2 = 4y

\(\boldsymbol{\Rightarrow }\) \(y=\frac{x^{2}}{4}\)

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

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

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{4}\left | \frac{x^{3}}{3} \right |_{0}^{4}=\frac{16}{3}}\) unit2

Therefore, the area of region bounded by the curve OEABO = \(=\frac{16}{3}\) unit2

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

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;\Rightarrow \;\int_{0}^{4}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{4}=8}\) unit2

Therefore, the Area of the region bounded by the curve OABO = 8 unit2

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

\(\boldsymbol{\Rightarrow }\) \(8-\frac{16}{3}\) = \(\frac{8}{3}\) unit2

Therefore, the total Area of shaded region = 2×\(\frac{8}{3}\) = \(\frac{16}{3}\)unit2

Q.9: Find the area enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)

Sol:

17

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

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 region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)

Now, the Area enclosed by the curve ABOA:

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

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 ABOA \(\boldsymbol{=\frac{3\pi}{2}}\) unit2

Hence, the total Area enclosed by the ellipse ABA’B’ = 4 × \(\boldsymbol{\frac{3\pi}{2}}\) unit2

= 6π unit2

Q.10: Find the area enclosed by the curve x2 + y2 = 9, line x = \(2\sqrt{2}y\) and the first quadrant.

Sol:

18

Equation x2 + y2 = 9 represents a circle with radius equal to 3units.

Since, x = \(2\sqrt{2}y\)

Therefore (\(2\sqrt{2}y\))2 + y2 = 9 (for points of intersection of both the curves)

Hence, the coordinates of point B = (\(2\sqrt{2}y\),1)

Now, the area of region bounded by the curve x2 + y2 = 9, x = \(2\sqrt{2}y\), and the first quadrant is the area enclosed by the curve OMABO.

i.e. Area of triangle OMB + Area under the curve MBAM.

Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;=\;\frac{1}{2}\times2\sqrt{2}y\times 1 =\sqrt{2}\;y\;\)unit2

Now, the Area under the curve MBAM:

\(\boldsymbol{\Rightarrow }\) \(\int_{2\sqrt{2}}^{3}\;y\;dx\)

Since, x2 +y2 =9

Therefore, y2 = 9 – x2

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{9-x^{2}}}\)

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

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2\sqrt{2}}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\;\sin^{-1}\frac{x}{3} \right |_{2\sqrt{2}}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\;\boldsymbol{[\frac{3}{2}\times\ \sqrt{9-9}]+[\frac{9}{2}\;\sin^{-1}(1)]-[\frac{2\sqrt{2}}{2}\times \sqrt{9-8}]-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\) unit2

Therefore, the Area under the curve MBAM:

= \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit2

Now, total Area under the shaded region = Area under the curve MBAM + Area of the triangle OMB

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]+\sqrt{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\) unit2

Hence, the required Area is given by the region OMABO:

\(\boldsymbol{=\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit2

Q.11: Find the area of larger part of circle x2 + y2 = 16 cut off by the line x = \(\frac{4}{\sqrt{2}}\) in the first quadrant.

Sol:

19

Equation x2 + y2 = 16 represents a circle with radius = 4 units.

For coordinates of point B:

\(\boldsymbol{\Rightarrow }\) \((\frac{4}{\sqrt{2}})^{2}+y^{2}=16\)

\(\boldsymbol{\Rightarrow }\) 8 + y2 = 16

\(\boldsymbol{\Rightarrow }\) y = \(2\sqrt{2}\)

Hence, the coordinates of point B are: \(\boldsymbol{(\frac{4}{\sqrt{2}},2\sqrt{2})}\)

The required Area is given by the curve OMBCO:

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2\sqrt{2}}y\;dx\;=\;\int_{0}^{2\sqrt{2}}\sqrt{16-x^{2}}\;dx}\)

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

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1} \frac{x}{4} \right |_{0}^{2\sqrt{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8\sin^{-1}\frac{1}{\sqrt{2}}]-0}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2}\times 2\sqrt{2}+8\times \frac{\pi }{4}=[4+2\pi] }\) unit2

Therefore, the Area of shaded region OMBCO = [4 + 2π]unit2

 

 

20

Let us assume two curves represented by the equation y = f(x) and y = g(x) in [a, b] as shown in the above figure. In this case the height of an elementary strip will be [f(x) – g(x)] and its width will be dx.

Now, Area of the elementary strip (dA) = [f(x) – g(x)] dx

Hence, the total Area of shaded region (A) = \(\int_{a}^{b} [f(x)-g(x)]\;dx\)

 

 

Example – 1: Find the area enclosed between two curves whose equations are: x2 = 4y and y2 = 4x.

Sol:

21

The Equation x2 = 4y represents a parabola symmetrical about y-axis and the equation y2 = 4x represents a parabola symmetrical about x-axis.

On solving both the equations:

\(\boldsymbol{\Rightarrow }\) \(\left ( \frac{y^{2}}{4} \right )^{2}=4y\)

\(\boldsymbol{\Rightarrow }\) y3 = 64 i.e. y = 4

Which gives x = 4. Hence, the coordinates of point N are (4, 4).

Now, The Area of region enclosed by the curve NAOBN = Area of region enclosed by the curve OANMO – Area of region enclosed by the curve OBNMO.

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}2\sqrt{x}\;dx-\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\times \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{3}}{12} \right ]_{0}^{4}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4}{3}\times (4)^{\frac{3}{2}}-0 \right ]-\left [ \frac{64}{12}-0 \right ]= \frac{16}{3}}\) unit2

Therefore, the Area of shaded region = \(\frac{16}{3}\) unit2

 

 

EXAMPLE – 2: Find the area enclosed by the sides of a triangle whose vertices have coordinates (1, 0) (3, 5) and (5, 4).

Sol:

23

Form the above figure:

Let, A (1, 0), B (3, 5) and C (5, 4) be the vertices of triangle ABC.

Now, the equation of line AB:

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

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{5\;-\;0}{3\;-\;1}\right]\)

\(\boldsymbol{\Rightarrow }\) 2y = 5x – 5

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{5x\;-\;5}{2}}\)

The Equation of line BC:

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

\(\boldsymbol{\Rightarrow }\) \((y-5)=(x-3)\times \left[\frac{4-5}{5-3}\right]\)

\(\boldsymbol{\Rightarrow }\) 2y – 10 = 3 – x

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{13\;-\;x}{2}}\)

The Equation of line AC:

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

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{4-0}{5-1}\right]\)

\(\boldsymbol{\Rightarrow }\) 4y = 4x – 4

\(\boldsymbol{\Rightarrow }\) y = x – 1

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 [2y = 5x – 5]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}y\;dx\;=\;\int_{1}^{3}\frac{5x-5}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{4}-\frac{5x}{2} \right ]_{1}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}=5}\) unit2

Therefore, Area under the curve ABMA = 5 unit2

The Area under the curve MBCN [2y = 13 – x]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}\frac{13-x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{13x}{2}-\frac{x^{2}}{4} \right ]_{3}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{65}{2}-\frac{25}{4}-\frac{39}{2}+\frac{9}{4}=9}\) unit2

Therefore, Area under curve MBCN = 9 unit2

The Area under the curve ACNA [y = x – 1]:

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{5}y\;dx\;=\;\int_{1}^{5}(x-1)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{2}-x \right ]_{1}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{2}-5-\frac{1}{2}+1=8}\) unit2

Therefore, Area under the curve ACNA = 8 unit2

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

Therefore, the Area of triangle ABC = 5 + 9 – 8 = 6 unit2