12th NCERT INTEGRALS Application of integrals Exercise 8.1 Questions 13
Hi

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

y2 = x is equation of parabola along x-axis. x=1 and x=4 equation of vertical lines.
Area of region bound between curve line and x-axis is as shown in figure
Here perpendicular drawn on x-axis and we have it as (1, 0) and (4, 0) so we have \[A = I = \int {ydx} \]
Area of shaded region = Area of region BCDEGB
\[{y^2} = x\] \[y = \sqrt x \] \[A = \int\limits_1^4 {ydx} \] \[A = \int\limits_1^4 {\sqrt x } dx\] \[A = \frac{{\left[ {{x^{\frac{3}{2}}}} \right]_1^4}}{{\frac{3}{2}}}\] \[A = \frac{2}{3}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 1 \right)}^{\frac{3}{2}}}} \right]\] \[A = \frac{2}{3}\left[ {{{\left( {{2^2}} \right)}^{\frac{3}{2}}} - {{\left( 1 \right)}^{\frac{3}{2}}}} \right]\] \[A = \frac{2}{3}\left[ {8 - 1} \right]\] \[A = \frac{2}{3} \times 7\] \[A = \frac{{14}}{3}\]

Question (2)

Find the area of the region bounded by y2 = 9x, x=2, x=4 and the x-axis in the first quadrant

Solution

y2 = 9x is equation of parabola along x-axis. x=2 and x=4 are vertical lines intersect x-axis at point B(2, 0) and G(4, 0)
Area of region bounded by curve, line and x-axs is haded region in diagram

y2 = 9x
y=3√x
Area of shaded region = area of region (BCDEG)
\[A = \int\limits_2^4 {ydx} \] \[A = \int\limits_2^4 {3\sqrt x dx} \] \[A = 3\frac{{\left[ {{x^{\frac{3}{2}}}} \right]_2^4}}{{\frac{3}{2}}}\] \[A = \require{cancel}\cancel{3} \times \frac{2}{\cancel{3}}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 2 \right)}^{\frac{3}{2}}}} \right]\] \[A = 2\left[ {8 - 2\sqrt 2 } \right]\] \[A = 16 - 4\sqrt 2 \]

Question (3)

Find the area of the region bounded by x2 = 4y, y=2, y=4 and the y-axis in the first quadrant

Solution

x2 = 4y is equation of parabola along y-axis, y=2 and y=4 are horizontal line drawn from (0, 2) and (0, 4) respectively.
Area of region bounded by curve two lines and y-axis in 1st quadrants is shown by shading in diagram
As 1n diagram perpendiculars are drown on y-axis at (0, 2) and (4, 0). So have two value of y as y=2 and y=4 so
\[I = \int {xdy} \]
Area bounded = Area of region by BCDEG
\[I = \int\limits_2^4 {xdy} \] x2 = 4y
x= 2√y
\[I = \int\limits_2^4 {2\sqrt y dy} \] \[I = 2\frac{{\left[ {{y^{\frac{3}{2}}}} \right]_2^4}}{{\frac{3}{2}}}\] \[I = \frac{4}{3}\left[ {{{\left( 4 \right)}^{\frac{3}{2}}} - {{\left( 2 \right)}^{\frac{3}{2}}}} \right]\] \[I = \frac{4}{3}\left[ {8 - 2\sqrt 2 } \right]\] \[I = \frac{{32 - 8\sqrt 2 }}{3}\]

Question (4)

Find the area of the region bounded by the ellipse
\[\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\]

Solution

\[\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\] is equation of ellipse.
Comparing to standard form of ellipse we get
a2 = 16 ⇒ a = 4
b2 = 9 ⇒ b = 3
as a > b, ellipse is along x-axis co-ordinate of vertex A(4, 0), B(0, 3), C(0, 0) is centre of ellipse.
Ellipse is always symmetric about x-axis of y-axis.
So area of region bounded by ellipse= 4 × area of region bounded by it in 1st quadrant
region bounded by ellipse= 4 × area of region (CADBC)
region bounded by ellipse= 4 I, where
\[I = \int\limits_0^3 {xdy} \]
Here we take y-cordinate of A to y-cordinate of B i.e. y=0 to y=3, we will find it by ∫xdy
\[\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\] \[\frac{{{x^2}}}{{16}} = 1 - \frac{{{y^2}}}{9}\] \[\frac{{{x^2}}}{{16}} = \frac{{9 - {y^2}}}{9}\] \[{x^2} = \frac{{16}}{9}\left( {9 - {y^2}} \right)\] \[x = \frac{4}{3}\sqrt {9 - {y^2}} \] \[\text{now}\quad \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\] \[I = \int\limits_0^3 {\frac{4}{3}\sqrt {9 - {y^2}} dy} \] \[I = \frac{4}{3}\int\limits_0^3 {\sqrt {9 - {y^2}} } dy\]
\[\int {\sqrt {{a^2} - {x^2}} } dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right)\]
\[I = \frac{4}{3}\left[ {\frac{y}{2}\sqrt {9 - {y^2}} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{y}{3}} \right)} \right]_0^3\] \[I = \frac{4}{3}\left[ {\left( {\frac{3}{2}\sqrt {9 - 9} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{3}{3}} \right)} \right) - \left( {\frac{0}{2}\sqrt {9 - 0} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{0}{3}} \right)} \right)} \right]\] \[I = \frac{4}{3}\left[ {0 + \frac{9}{2}{{\sin }^{ - 1}} - 0 - \frac{9}{2}{{\sin }^{ - 1}}0} \right]\]
\[\text{since} \quad \begin{array}{l}{\sin ^{ - 1}}1 = \frac{\pi }{2}\\{\sin ^{ - 1}}0 = 0\end{array}\]
\[A = \frac{4}{3}\left[ {\frac{9}{2} \times \frac{\pi }{2} - 0} \right]\] \[A = \frac{\cancel{4}}{\cancel{3}}\left[ {\frac{{\cancel{9}^3\pi }}{\cancel{4}}} \right] =3\pi\] Now area enclosed by the ellipse
I = 4A =4(3π)
I= 12π

Question (5)

Find the area of the region bounded by the ellipse
\[\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\]

Solution

\[\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\] is equation of ellipse.
Comparing to standard form of ellipse we get
a2 = 4 ⇒ a = 2
b2 = 9 ⇒ b = 3
as a < b, ellipse is along y-axis co-ordinate of vertex A(2, 0), B(0, 3), C(0, 0) is centre of ellipse.
Ellipse is always symmetric about x-axis of y-axis.
So area of region bounded by ellipse= 4 × area of region bounded by it in 1st quadrant
region bounded by ellipse= 4 × area of region (CADB)
region bounded by ellipse= 4 I, where
If take y-cordinate of A i.e 0 to y-cordinate of B i.e 3
\[I = \int\limits_0^3 {xdy} \] If we take x-cordinate of A , x=2 to x-cordinate of B i.e 3 then
\[I = \int\limits_2^0 {ydx} \] In both cases answer will be same, We will to solved by
\[I = \int\limits_0^3 {xdy} \] \[\text{Now}\quad \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\] \[\frac{{{x^2}}}{4} = 1 - \frac{{{y^2}}}{9}\] \[\frac{{{x^2}}}{4} = \frac{{9 - {y^2}}}{9}\] \[{x^2} = \frac{4}{9}\left( {9 - {y^2}} \right)\] \[x = \frac{2}{3}\sqrt {9 - {y^2}} \] \[I = \int\limits_0^3 {xdy} \] \[I = \int\limits_0^3 {\frac{2}{3}} \sqrt {9 - {y^2}} dy\] \[I = \frac{2}{3}\int\limits_0^3 {\sqrt {9 - {y^2}} dy} \]
\[\int {\sqrt {{a^2} - {x^2}} } dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right)\]
\[I = \frac{2}{3}\left[ {\frac{y}{2}\sqrt {9 - {y^2}} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{y}{3}} \right)} \right]_0^3\] \[I = \frac{2}{3}\left[ {\left( {\frac{3}{2}\sqrt {9 - 9} + \frac{9}{2}{{\sin }^{ - 1}}\frac{3}{3}} \right) - \left( {\frac{0}{2}\sqrt {9 - 0} + \frac{9}{2}\sin \frac{0}{3}} \right)} \right]\] \[I = \frac{2}{3}\left[ {0 + \frac{9}{2}{{\sin }^{ - 1}}1 - 0 - \frac{9}{2}{{\sin }^{ - 1}}0} \right]\] \[I = \frac{2}{3} \times \left( {\frac{9}{2} \times \frac{\pi }{2} - 0} \right)\] \[I = \frac{\cancel{2}}{\cancel{3}} \times \frac{{\cancel{9}^3\pi }}{\cancel{4}_2}\] \[I = \frac{{3\pi }}{2}\] Area enclosed by ellipse A = 4I
\[A = 4 \times \frac{{3\pi }}{2} = 6\pi \]

Question (6)

Find the area of the region in the first quadrant enclosed by x-axis, line x =√3y and circle x2 + y2 =4

Solution

x2 + y2 =4 is equation of circle whose centre is (0, 0) and radius r=2
x=√y is the equation of line passes through (0, 0)
To get point of integration of line and circle replace x=√3y in circle we get
(√3 y)2 + y2 = 4
3y2 + y2 = 4
4y2 = 4
y= ± 1
Now x= √3 y x = √3(± 1) =± √3
Area of region of first quadrant is required. point of inter section of line and circle is B(√3, 1)
Let BM ⊥ x-axis
∴ co-ordinate of M = (√3, 0)
Area of region bound by circle, line in 1st quadrant
= Area of region (CBMC) + area of region (MBDAM)
\[A = \int\limits_0^{\sqrt 3 } {ydx} + \int\limits_{\sqrt 3 }^2 {ydx} \] Now x2 + y2 and x = √y
∴ y= √(4-x2) and y = x/√3 \[A = \int\limits_0^{\sqrt 3 } {\frac{x}{{\sqrt 3 }}dx} + \int\limits_{\sqrt 3 }^2 {\sqrt {4 - {x^2}} dx} \] \[A = \frac{1}{{\sqrt 3 }}\frac{{\left[ {{x^2}} \right]_0^{\sqrt 3 }}}{2} + \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_{\sqrt 3 }^2\]
\[A = \frac{1}{{2\sqrt 3 }}\left( {3 - 0} \right) + \left[ {\left( {\frac{2}{3}\sqrt {4 - 4} + 2{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right)} \right) - \left( {\frac{{\sqrt 3 }}{2}\sqrt {4 - 3} + 2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right)} \right]\]
\[A = \frac{1}{{2\sqrt 3 }}\left( {3 - 0} \right) + \] \[\left[ {\left( {\frac{2}{3}\sqrt {4 - 4} + 2{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right)} \right) - \left( {\frac{{\sqrt 3 }}{2}\sqrt {4 - 3} + 2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right)} \right]\]
\[A = \frac{2}{{2\sqrt 3 }} + \left[ {0 + 2{{\sin }^{ - 1}}1 - \frac{{\sqrt 3 }}{2} - 2{{\sin }^{ - 1}}\frac{{\sqrt 3 }}{2}} \right]\]
\[\text{since} \quad \begin{array}{l}{\sin ^{ - 1}}1 = \frac{\pi }{2}\\{\sin ^{ - 1}}\frac{{\sqrt 3 }}{2} = \frac{\pi }{3}\end{array}\]
\[A = \cancel{\frac{{\sqrt 3 }}{2}} + \cancel{2}\frac{\pi }{\cancel{2}} - \cancel{\frac{{\sqrt 3 }}{2}} - 2\frac{\pi }{3}\] \[A = \pi - \frac{{2\pi }}{3} = \frac{\pi }{3}\]

Question (7)

Find the area of the smaller part of the circle x2 + y2 =a2 cut off by the line x= a/√2

Solution

x2 + y2 =a2 is equation of circle. Whose centre C iis at (0, 0) and radius "a", x=a/√2 is equation of vertical line parallel y-axis
Area enclosed = Area of shaded region
Circle is symmetric about x-axis
∴ Area enclosed between circle and line, A = 2× Area of region (MAEDM)
\[A = \int\limits_{\frac{a}{{\sqrt 2 }}}^a {ydx} \] Now x2 + y2 =a2
y2 =a2 -x2
\[y = \sqrt {{a^2} - {y^2}} \] \[A = 2\int\limits_{\frac{a}{{\sqrt 2 }}}^a {\sqrt {{a^2} - {y^2}} } dx\] \[A = 2\left[ {\frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{a}} \right)} \right]_{\frac{a}{{\sqrt 2 }}}^a\]
\[A = 2\left[ {\left( {\frac{a}{2}\sqrt {{a^2} - {a^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{a}{a}} \right)} \right) - \left( {\frac{a}{{a\sqrt 2 }}\sqrt {a - \frac{{{a^2}}}{2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{a}{{\sqrt 2 a}}} \right)} \right)} \right]\]
\[A = 2\left[ {\left( {\frac{a}{2}\sqrt {{a^2} - {a^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{a}{a}} \right)} \right)} \right]\] \[-2\left[ {\left( {\frac{a}{{a\sqrt 2 }}\sqrt {a - \frac{{{a^2}}}{2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{a}{{\sqrt 2 a}}} \right)} \right)} \right]\]
\[A = 2\left[ {\left( {0 + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}1} \right) - \left( {\frac{a}{{2\sqrt 2 }} \cdot \frac{a}{{\sqrt 2 }}} \right) - \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)} \right]\] \[A = 2\left[ {\frac{{{a^2}}}{2} \cdot \frac{\pi }{2} - \frac{{{a^2}}}{4} - \frac{{{a^2}}}{2} \cdot \frac{\pi }{4}} \right]\] \[A = 2\left[ {\frac{{\pi {a^2}}}{4} - \frac{{{a^2}}}{4} - \frac{{\pi {a^2}}}{8}} \right]\] \[A = 2\left[ {\frac{{2\pi {a^2} - \pi {a^2}}}{8} - \frac{{{a^2}}}{4}} \right]\] \[A = 2\left[ {\frac{{\pi {a^2}}}{8} - \frac{{{a^2}}}{4}} \right]\] \[A = \frac{{\cancel{2}{a^2}}}{\cancel{4}_2}\left( {\frac{\pi }{2} - 1} \right)\] \[A = \frac{{{a^2}}}{2}\left( {\frac{\pi }{2} - 1} \right)\]

Question (8)

The area between x=y2 and x=4 is divided into two equal parts by the line x=a, finf the value of a

Solution

y2 = x is equation of parabola along x-axis
x=4 and x=4 are equations of vertical lines parallel to y-axis. Line x=a divide the area under the curve and line x=4 in two equal part, co-ordinate E(a, 0) and D(4, 0)
∴ Area of region AEBFA = Area (EBGCDE)
\[\int\limits_0^a {ydx = \int\limits_a^4 {ydx} } \] \[\int\limits_0^a {\sqrt x } dx = \int\limits_a^4 {\sqrt x } dx\] \[\frac{{\left[ {{x^{\frac{3}{2}}}} \right]_0^a}}{{\cancel{\frac{3}{2}}}} = \frac{{\left[ {{x^{\frac{3}{2}}}} \right]_a^4}}{{\cancel{\frac{3}{2}}}}\] \[{a^{\frac{3}{2}}} - {0^{\frac{3}{2}}} = {\left( 4 \right)^{\frac{3}{2}}} - {a^{\frac{3}{2}}}\] \[{a^{\frac{3}{2}}} = 8 - {a^{\frac{3}{2}}}\] \[2{a^{\frac{3}{2}}} = 8\] \[{a^{\frac{3}{2}}} = 4\] \[a = {4^{\frac{2}{3}}}\]

Question (9)

Find the area bounded by the parabola y=x2and y=|x|

Solution

x2= y is the equation of parabola along y-axis,
y=|x| ∴ y = ± x
y =x and y=-x are two lines passing through (0, 0)
We will find points of intersection of y=x2 and y=x for replace y=x in parabola we get
x= x2 ∴ x2 - x = 0
x(x-1) = 0
x=0 or x = 1
∴ point of intersection is (0, 0) and (1, 1)
Area of region bound by y=x2 and y=|x|
A = 2 × Area of region bound by y=x2 and y=x ( as it is symmetric about y-axis)
A= 2[ (area of OABO) - (area of region OCABO)
\[A = 2\left[ {\int\limits_{\scriptstyle0\atop \text{Line}}^1 {ydx - \int\limits_{\scriptstyle0\atop \text{Curve}}^1 {ydx} } } \right]\] \[A = 2\left[ {\int\limits_0^1 {xdx - \int\limits_0^1 {{x^2}dx} } } \right]\] \[A = 2\left[ {\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \left[ {\frac{{{x^3}}}{3}} \right]_0^1} \right]\] \[A = 2\left[ {\frac{1}{2}\left( {1 - 0} \right) - \frac{1}{3}\left( {1 - 0} \right)} \right]\] \[A = 2\left[ {\frac{1}{2} - \frac{1}{3}} \right]\]\[A = 2\left( {\frac{{3 - 2}}{6}} \right)\] \[A = \cancel{2}\left( {\frac{{3 - 2}}{\cancel{6}_3}} \right)\] \[A = \frac{1}{3}\]

Question (10)

Find the area bounded by the curve x2 = 4y and the line x=4y-2

Solution

x2 = 4y is equation of parabola along y-axis, x=4y-2 is equation of line.
To find the point of interaction replace x+2 = 4y n parabola equation
x2 = 4y
x2 = x + 2
x2 - x - 2 = 0
(x-2)(x+1) = 0
x=2, x=1
If x=-1, 4y =3 ⇒ y=3/4
f x = 2, 4y = 4 ⇒ y=1
Co-ordinate of A=(-1, 3/4)
B(2, 1)
AM ⊥ on x-axis, M= (-1, 0) BN ⊥ x-axis N= (2, 0)
Area bounded by x2 = 4y and x=4y-2
A=Area of shaded region
\[A = \int\limits_{\scriptstyle - 1\atop\text {line}}^2 {ydx} - \int\limits_{\scriptstyle - 1\atop\text{curve}}^2 {ydx} \] Now x = 4y-2 and x2 = 4y
\[\therefore \quad y = \frac{{x + 2}}{4},y = \frac{{{x^2}}}{4}\] \[A = \int\limits_{ - 1}^2 {\frac{{x + 2}}{4}} dx - \int\limits_{ - 1}^2 {\frac{{{x^2}}}{4}dx} \] \[A = \int\limits_{ - 1}^2 {\frac{x}{4}} dx + \int\limits_{ - 1}^2 {\frac{2}{4}dx} - \int\limits_{ - 1}^2 {\frac{{{x^2}}}{4}dx} \] \[A = \frac{1}{4}\left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^2 + \frac{1}{2}\left[ x \right]_{ - 1}^2 - \frac{1}{4}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^2\] \[A = \frac{1}{8}\left( {4 - 1} \right) + \frac{1}{2}\left[ {2 - \left( { - 1} \right)} \right] - \frac{1}{{12}}\left[ {8 - \left( { - 1} \right)} \right]\] \[A = \frac{3}{8} + \frac{3}{2} - \frac{\cancel{9}^3}{{\cancel{12}_4}}\] \[A = \frac{{3 + 12 - 6}}{8} = \frac{9}{8}\]

Question (11)

Find the area bounded by the curve y2 = 4x and the line x=3

Solution

y2 = 4x is equation of parabola along x-axis, x=3 is equation of lne parallel to y-axis through (3, 0)
As parabola is symmetric about x-axis
Area enclosed by y2 = 4x and x=3
A = 2 × area of region ABCDA
\[A = 2\left[ {\int\limits_0^3 {ydx} } \right]\] \[A = 2\left[ {\int\limits_0^3 {2\sqrt x dx} } \right]\] \[A = 4\int\limits_0^3 {\sqrt x dx} \] \[A = 4\frac{{\left[ {{x^{\frac{3}{2}}}} \right]_0^3}}{{\frac{3}{2}}}\] \[A = \frac{8}{3}\left[ {{{\left( 3 \right)}^{\frac{3}{2}}} - 0} \right]\] \[A = \frac{8}{\cancel{3}} \times \cancel{3}\sqrt 3 \] \[A = 8\sqrt 3 \]

Choose the correct answer in the following Exercise 12 and 13

Question (12)

Area lying in the first quadrant and bounded by the circle x2+y2=4 and the lines x=0 and x=2 is
(A) π  (B) π/2   (C) π/3   (D) π/4

Solution


x2+y2=4 is equation of circle with centre (0, 0) and radius r=2
Now y = √(4-x2) Co-ordinate of A=(2, 0) area bound A
\[A = \int\limits_0^2 {ydx} \] \[A = \int\limits_0^2 {\sqrt {4 - {x^2}} dx} \] \[A = \left[ {\frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{{\sin }^{ - 1}}\frac{x}{2}} \right]_0^2\] \[A = \left( {\frac{2}{2}\sqrt {4 - 4} + 2{{\sin }^{ - 1}}\frac{2}{2}} \right) - \left( {\frac{0}{2}\sqrt {4 - 0} + 2{{\sin }^{ - 1}}0} \right)\] \[A = 2\left[ {\frac{\pi }{2} - 0} \right] = \pi \] correct option is (A)

Question (13)

Area of the region bounded by the curve y2 = 4x, y-axis and the line y=3 is
(A) 2   (B) 9/4   (C) 9/3   (D) 9/2

Solution

y2 = 4x is equation of parabola along x-axis.
y=3 iis equation of line parallel to x-axis
Point B of intersection of parabola and the co-ordinate of B(9/4, 3)
BM ⊥ y-axis so coordinate of M=(3, 0)
Area enclosed by curve and line and y-axis
A= area of region
A= Area of region AMBCA
\[A = \int\limits_0^3 {xdy} \] Now y2 4x
x= y2/4
\[A = \int\limits_0^3 {\frac{{{y^3}}}{4}dy} \] \[A = \frac{1}{4}\left[ {\frac{{{y^3}}}{3}} \right]_0^3\] \[A = \frac{1}{{12}}\left( {27 - 0} \right)\] \[A = \frac{{\cancel{27}9}}{{\cancel{12}_4}} = \frac{9}{4}\]
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⇒ Exercise 8.2