12th NCERT INTEGRALS Application of integrals
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Exercise 8.2 Questions 7

Question (1)

Find the area of the region bounded by the curve y2 = x and the line x=1, x=4 and the x-axis.

Solution

x2 + y2 = 9/4 is equation of circle with centre C at (0, 0) and radius r=3/2
x2 = 4y is equation of parabla along y-axis.
First find point of intersection of circle and parabola for that replace x2 = 4y in equatiion of circle
4x2 + 4y2 =9
4(4y) + 16y - 9 = 0
4y2 +18y - 2y - 9 = 0
2y(2y+9) - (2y +9) = 0
(2y+9) (2y-1) = 0 2y+9 = 0 and 2y - 1 = 0
y=-9/2 and y = 1/2
if y = -9/2, then x2 = 4 × (-9/2) = -18 which is not possible
∴ y = 1/2
So x2 = 4(1/2) = 2 or x= ±√2 So pint of intersection A(√2, 1/2) and B(-√2, 1/2)
AM ⊥ x-axis, M=(√2, 0)
Circle and parabola are symmetric about y-axis
[IMAGE 8.2.1] ∴ area of region enclosed by circle and parabola (A)
A = 2 × area of region (CDAFGC)
\[A = 2\left[ {\int\limits_{\scriptstyle0\atop\text{circle}}^{\sqrt 2 } {ydx - \int\limits_{\scriptstyle0\atop\text{parabola}}^{\sqrt 2 } {ydx} } } \right]\] \[{x^2} + {y^2} = \frac{9}{4}\] \[{y^2} = \frac{9}{4} - {x^2}\] \[y = \sqrt {\frac{9}{4} - {x^2}} \] and \[{x^2} = 4y\] \[y = \frac{{{x^2}}}{4}\] Area A \[A = 2\left[ {\int\limits_0^{\sqrt 2 } {\sqrt {\frac{9}{4} - {x^2}} dx - \int\limits_0^{\sqrt 2 } {\frac{{{x^2}}}{4}dx} } } \right]\] \[A = \require{cancel}\cancel{2}\left[ {\frac{x}{\cancel{2}}\sqrt {\frac{9}{4} - {x^2}} } \right]_0^{\sqrt 2 } + \cancel{2}\left[ {\frac{9}{\cancel{8}_4}{{\sin }^{ - 1}}\left( {\frac{{2x}}{2}} \right)} \right]_0^{\sqrt 2 } - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_0^{\sqrt 2 }\] \[A = \left( {\sqrt 2 \sqrt {\frac{9}{4} - 2} - 0} \right) + \frac{9}{4}\left[ {{{\sin }^{ - 1}}\left( {\frac{{2\sqrt 2 }}{3}} \right) - {{\sin }^{ - 1}}0} \right] - \frac{1}{\cancel{6}_3}\left[ {2\sqrt 2 } \right]\] \[A = \sqrt 2 \frac{1}{2} + \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3} - \frac{{\cancel{2}\sqrt 2 }}{\cancel{6}_3}\] \[A = \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{3} + \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\] \[A = \frac{{\sqrt 2 }}{6} + \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}\] \[A = \int\limits_0^{\frac{{4a}}{{{m^2}}}} {2\sqrt a \sqrt {x\,} dx} - \int\limits_0^{\frac{{4a}}{{{m^2}}}} {mx{\kern 1pt} dx} \]

Question (2)

Find the area of the region bounded by curve (x-1)2+y2 =1 and x2 + y2

Solution

S1: (x-1)2+y2 =1 is equation of circle with centre C1 abd radius r1
S2 : x1 + y2 = 1 is the equation of circle with centre C2 at (0, 0) and radius r=1
Let A and E are points of intersection of both circles to find its coordinate. Solve both equation
x1 + y2 = 1
y2 = 1 - x1
(x-1)2+y2 =1
replace the value of y2
(x-1)2 + 1 - x1 = 1
(x-1-x)(x-1+x) = 0
-1(2x-1) = 0
2x-1=0
x = 1/2
Now y2 = 1 - x2
y2 = 1 - (1/4) = 3/4
∴ y = ± √3/2
Co-ordinate of A is (1/2, √3/2) and E is (1/2, -√3/2)
Let AM ⊥ x-axis
So coordinate of M is (1/2, 0). circle are symmetric about x-axis
Area enclosed between two circle A
A= 2 [ Area of region (C2DABC1C2)]
A=2[Area of region (C2DAMC2) + area of regiion (C1BAMC1)]
\[A = 2\left[ {\int\limits_{\scriptstyle0\atop\scriptstyle(ofS_1)}^{\frac{1}{2}} {ydx + \int\limits_{\scriptstyle\frac{1}{2}\atop\scriptstyle(ofS_2)}^1 {ydx} } } \right]\] \[A = 2\left[ {\int\limits_1^{\frac{1}{2}} {\sqrt {1 - {{\left( {x - 1} \right)}^2}} } dx + \int\limits_{\frac{1}{2}}^1 {\sqrt {1 - {x^2}} } dx} \right]\]
\[A = 2\left[ {\left( {\frac{{x - 1}}{2}\sqrt {1 - {{\left( {x - 1} \right)}^2}} } \right) + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{x - 1}}{1}} \right)} \right]_0^{\frac{1}{2}} + \left[ {\left( {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}\frac{x}{1}} \right)} \right]_{\frac{1}{2}}^1\]
\[A = 2\left[ {\left( {\frac{{x - 1}}{2}\sqrt {1 - {{\left( {x - 1} \right)}^2}} } \right) + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{x - 1}}{1}} \right)} \right]_0^{\frac{1}{2}}\] \[\qquad \qquad + 2 \left[ {\left( {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}\frac{x}{1}} \right)} \right]_{\frac{1}{2}}^1\]
\[A = 2\left[ {\frac{{\frac{1}{2} - 1}}{2}\sqrt {1 - {{\left( {\frac{1}{2} - 1} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2} - 1} \right) - \left( {\frac{{0 - 1}}{2}} \right)\sqrt {1 - {{\left( {0 - 1} \right)}^2}} - \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{0 - 1}}{1}} \right)} \right]\] \[\begin{array}{l}A = 2\left[ {\frac{{\frac{1}{2} - 1}}{2}\sqrt {1 - {{\left( {\frac{1}{2} - 1} \right)}^2}} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{1}{2} - 1} \right) - \left( {\frac{{0 - 1}}{2}} \right)\sqrt {1 - {{\left( {0 - 1} \right)}^2}} - \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{0 - 1}}{1}} \right)} \right]\\ + 2\left[ {\frac{1}{2}\sqrt {1 - 1} + \frac{1}{2}{{\sin }^{ - 1}}1 - \frac{{\frac{1}{2}}}{2}\sqrt {1 - \frac{1}{4}} - \frac{1}{2}{{\sin }^{ - 1}}\frac{1}{2}} \right]\end{array}\]
\[A = 2\left[ {\frac{{ - 1}}{4}\frac{{\sqrt 3 }}{2} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) - \frac{1}{2}\left( 0 \right) - \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right] + 2\left[ {0 + \frac{1}{2}{{\sin }^{ - 1}}1 - \frac{1}{4}\frac{{\sqrt 3 }}{2} - \frac{1}{2}{{\sin }^{ - 1}}\frac{1}{2}} \right]\]
\[A = 2\left[ {\frac{{ - 1}}{4}\frac{{\sqrt 3 }}{2} + \frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{ - 1}}{2}} \right) - \frac{1}{2}\left( 0 \right) - \frac{1}{2}{{\sin }^{ - 1}}\left( { - 1} \right)} \right]\] \[\qquad + 2\left[ {0 + \frac{1}{2}{{\sin }^{ - 1}}1 - \frac{1}{4}\frac{{\sqrt 3 }}{2} - \frac{1}{2}{{\sin }^{ - 1}}\frac{1}{2}} \right]\]
\[A = \frac{{ - \sqrt 3 }}{4} - {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) + {\sin ^{ - 1}}\left( 1 \right) + {\sin ^{ - 1}}\left( 1 \right) - \frac{{\sqrt 3 }}{4} - {\sin ^{ - 1}}\frac{1}{2}\] \[A = \frac{{ - \sqrt 3 }}{4} - \frac{\pi }{6} + \frac{\pi }{2} + \frac{\pi }{2} - \frac{{\sqrt 3 }}{2} - \frac{\pi }{6}\] \[A = \pi - \frac{\pi }{3} - \frac{{\cancel{2}\sqrt 3 }}{\cancel{4}_2}\] \[A = \frac{{2\pi }}{3} - \frac{{\sqrt 3 }}{2}\]

Question (3)

Find the area of the region bounded by curve y=x2+2, y = x , x=0 and x=3

Solution

y = x2 + 2 x2 = y - 2
It is a equation of parabola along y-axis and whose vertex is at (0, 1)
y = x is equation of line passing through O(0, 0)
x= 0 and x = 3 are vertical line passing through (0, 0) and (3, 0)
First we will find point of intersection of parabola and line y=x. Replace x = y in parabola we get
x = x2 + 2
x2 - x + 2 = 0
as a=1, b = -1, c=2
D = b2 -4ac
D= 1 - 8
D= -7 <0 No real roots
So parabola and line do not intersect each other when x=0, y=0+2 = 2
x = 0, y= 32 + 2 = 9+2 = 11
So point of intersection of line x = 3 and curve is C(3, 11)
D is point of co-ordinate of D(3, 3)
Area enclosed by curve, line y=x, x=0, and x=3 is A
A= Area of region OABCDO
A = area of region of OABCEO - area of region (ODEO)
\[A = \int\limits_{\begin{array}{*{20}{c}}0\\{(curve)}\end{array}}^3 {ydx - \int\limits_{\begin{array}{*{20}{c}}0\\{(line)}\end{array}}^3 {ydx} } \] \[A = \int\limits_0^3 {\left( {{x^2} + 2} \right)dx - \int\limits_0^3 {xdx} } \] \[A = \frac{{\left[ {{x^3}} \right]_0^3}}{3} + 2\left[ x \right]_0^3 - \frac{{\left[ {{x^2}} \right]_0^3}}{2}\] \[A = \frac{1}{3}\left( {27 - 0} \right) + 2\left( {3 - 0} \right) - \frac{1}{2}\left( {9 - 0} \right)\] \[A = 9 + 6 - \frac{9}{2}\] \[A = 15 - \frac{9}{2}\] \[A = \frac{{21}}{2}\]

Question (4)

Using iintegration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3.2)

Solution

Plot the points A, B and C, MB ⊥ x-axis, so coordinate of M(1, 0) CN ⊥ x-axis coordinate of N(3, 0)
Area of ΔANC = Area (ABM) + Area (BMNC) - Area (ACN)
\[A = \int\limits_{\scriptstyle - 1\atop\scriptstyle\overleftrightarrow {AB} }^1 {ydx} + \int\limits_{\scriptstyle1\atop\scriptstyle \overleftrightarrow {BC} }^3 {ydx} - \int\limits_{\scriptstyle - 1\atop\scriptstyle\overleftrightarrow {AC} }^3 {ydx} \] \[\text{We will find the equation of}\overleftrightarrow {AB} ,\overrightarrow {BC} ,\overleftrightarrow {AC} \] \[\text{equation of} \overleftrightarrow {AB} \] is given by
\[\left| {\begin{array}{*{20}{c}}x&y&1\\{ - 1}&0&1\\1&3&1\end{array}} \right| = 0\] x(-3) - y(-2) + 1(-3) = 0
2y = 3x + 3
\[\overleftrightarrow {AB}\qquad y = \frac{{3x + 3}}{2}\] equation of vector BC
\[\left| {\begin{array}{*{20}{c}}x&y&1\\1&3&1\\3&2&1\end{array}} \right| = 0\] x(3-2) - y(1-3) +1(2-9) =0
x+2y-7 = 0
2y = 7 -x
\[\overleftrightarrow {BC}\qquad y = \frac{{7 - x}}{2}\] equation of line AC \[\left| {\begin{array}{*{20}{c}}x&y&1\\{ - 1}&0&1\\3&2&1\end{array}} \right| = 0\] x(-2) -y(-4)+(-2) = 0
-2x +4y - 2 = 0
2y = x+1
\[\overleftrightarrow {AC}\qquad y = \frac{{x + 1}}{2}\] Area of Δ ABC
\[A = \int\limits_{\scriptstyle - 1\atop\scriptstyle \overleftrightarrow {AB} }^1 {ydx} + \int\limits_{\scriptstyle 1\atop\scriptstyle \overleftrightarrow {BC} }^3 {ydx} - \int\limits_{\scriptstyle - 1\atop\scriptstyle\overleftrightarrow {AB} }^3 {ydx} \] \[ = \int\limits_{ - 1}^1 {\frac{{3x + 3}}{2}dx + \int\limits_1^3 {\frac{{7 - x}}{2}dx - \int\limits_{ - 1}^3 {\frac{{x + 1}}{2}} } } dx\] \[A = \frac{3}{2}\int\limits_{ - 1}^1 {xdx + \frac{3}{2}} \int\limits_{ - 1}^1 {dx + \frac{7}{2}\int\limits_1^3 {dx} } - \frac{1}{2}\int\limits_1^3 {xdx - \frac{1}{2}\int\limits_{ - 1}^3 {xdx - \frac{1}{2}\int\limits_{ - 1}^3 {dx} } } \] \[A = \frac{3}{2}\left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^1 + \frac{3}{2}\left[ x \right]_{ - 1}^1 + \frac{7}{2}\left[ x \right]_1^3 - \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_1^3 - \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^3 - \frac{1}{2}\left[ x \right]_{ - 1}^3\] \[A = \frac{3}{4}\left( {1 - 1} \right) + \frac{3}{2}\left( {1 + 1} \right) + \frac{7}{2}\left( {3 - 1} \right) - \frac{1}{4}\left( {9 - 1} \right) - \frac{1}{4}\left( {9 - 1} \right) - \frac{1}{2}\left( {3 - 1} \right)\] \[A = 0 + \frac{3}{ \cancel {2}} \cdot \cancel{ 2} + \frac{7}{ \cancel{2}} \cdot \cancel{2} - \frac{1}{ \cancel{4}} \cdot \cancel{8}^2 - \frac{1}{\cancel{4}} \cdot \cancel{8}^2 - \frac{1}{\cancel{2}} \cdot \cancel{4}^2 \] \[A = 3 + 7 - 2 - 2 - 2\] \[A = 4\]

Question (5)

Using integration find the area of region bounded by the triangular region whose sides have the equations y=2x+1, y=3x+1
and x=4

Solution

l1: y = 2x+1
l2: y=3x+1
l3: x=4
Let l1 ∩ l2 = {A} \[\begin{array}{l}y = \,\;\;\;\;2x + 1\\\underline { - y = - 3x - 1} \\0 = - x + 0\end{array}\] y=1, A(0, 1)
l2 ∩ l3 = {B}
x=4 and y=3x+1
y=3x+1
y=3(4)+1 = 13
B(4, 13)
l1 ∩ l3 = {C}
x= 4 and y=2x+1
y=2(4)+1 = 9
y=9
C(4, 9)
BM ⊥ x-axis
Area of Δ ABC
= Area of region OABMO - Area of OACMP
\[A = \int\limits_{\scriptstyle0\atop\scriptstyle\overleftrightarrow {AB}}^4 {ydx - \int\limits_{\scriptstyle0\atop\scriptstyle\overleftrightarrow {AC}}^4 {ydx} } \] \[A = \int\limits_0^4 {\left( {3x + 1} \right)dx} - \int\limits_0^4 {\left( {2x + 2} \right)dx} \] \[A = 3\int\limits_0^4 {xdx} + \int\limits_0^4 {dx} - 2\int\limits_0^4 {xdx} - \int\limits_0^4 {dx} \] \[A = \frac{3}{2}\left[ {{x^2}} \right]_0^4 + \left[ x \right]_0^4 - \frac{2}{2}\left[ x \right]_0^4 - \left[ x \right]_0^4\] \[A = \frac{3}{2}\left( {16 - 0} \right) + \left( {4 - 0} \right) - \left( {16 - 0} \right) - \left( {4 - 0} \right)\] \[A = 24 + 4 - 16 - 4\] \[A = 8\]

Choose the correct answer in the following exercises 6 and 7

Question (6)

Small area enclosed by the circle x2 + y2 = 4 and the line x+y=2 is...
(A) 2(π-2)     (B) π-2     (C) 2π - 1     (D) 2(π + 2)

Solution

S1 : x2 + y2 = 4 is aequation of circle
with centre (0, 0) and radius r=2, co-ordinate of A(2, 0) and B(0, 2)
x + y = 2 is equation of line passing through the points (0, 2) and (2, 0)
Area of region bounded by circle and line = Area of region (BCAB)
Area = Area of region OBCAO - area of region OBAO
\[A = \int\limits_{\begin{array}{*{20}{c}}0\\{circle}\end{array}}^2 {ydx} - \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^2 {ydx} \] Now x2 + y2 = 4
y2 = 4-x2
∴ y = √(4-x2)
x + y = 2 ∴ y = 2-x \[A = \int\limits_{\begin{array}{*{20}{c}}0\\{circle}\end{array}}^2 {ydx} - \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^2 {ydx} \]\[A = \int\limits_0^2 {\sqrt {4 - {x^2}} dx} - \int\limits_0^2 {\left( {2 - x} \right)} dx\] \[A = \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^2 - \left[ {2x - \frac{{{x^2}}}{2}} \right]_0^2\] \[A = \left[ {\left( {\frac{2}{2}\sqrt 0 + 2{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right)} \right) - \left( {\frac{0}{2}\sqrt 4 + 2{{\sin }^{ - 1}}0} \right)} \right] - \left[ {\left( {4 - \frac{4}{2}} \right) - \left( {0 - 0} \right)} \right]\] \[A = 2{\sin ^{ - 1}}1 - 2{\sin ^{ - 1}}0 - 2\] \[A = 2 \times \frac{\pi }{2} - 0 - 2\] \[A = \pi - 2\] So correct option is (B)

Question (7)

Area lying between the curves y2 = 4x and y=2x is (A) 2/3     (B) 1/3     (C) 1/4     (D) 3/4

Solution

y2 = 4x is the equation of parabola along x-axis. Vertex is at origine (0 ,0), y=2x is equation of line passing the line (0, 0)
To get the point of intersectiion replace y=2x in parabola
(2x)2 = 4x
4x2 - 4x = 0
4x(x-1) = 0
x = 0 and x = 1
When x= 1, y=2
So B(1, 2) is other point of intersection of parabole and liine.
Let BM ⊥ x-axis
So co-ordinate of M(1, 0)
Area of region bounded by curve and liine = Area of region ACBA
A= Area of region ACBMA - Area of region ABMA
\[A = \int\limits_{\begin{array}{*{20}{c}}0\\{circle}\end{array}}^1 {ydx} - \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^1 {ydx} \]\[A = 2\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1 - 2\left[ {\frac{{{x^2}}}{2}} \right]_0^1\] \[A = \frac{4}{3}\left( {{1^{\frac{3}{2}}} - 0} \right) - \left( {{1^2} - {0^2}} \right)\] \[A = \frac{4}{3} - 1\] \[A = \frac{1}{3}\] So correct option is (B)
Exercise 8.1 ⇐
⇒ Miscellaneous Exercise