12th NCERT INTEGRALS Application of integrals Miscellaneous Exercise
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## Miscellaneous Questions

Question (1)

Find the (i) y = x2, x = 1, x = 2 and x-axis

Solution

y=x2 is equation of parabola along y-axis and vector A(0, 0)
x=1 and x=2 are two vertical line passing through F(1,0) and E(2, 0)
Coordinate B(1, 1)n D(2, 4)
Area of region bounded by curve and two lines Area of region FBCDEF A = $A = \int\limits_{\begin{array}{*{20}{c}}1\\{curve}\end{array}}^2 {y{\mkern 1mu} dx}$ $A = \int\limits_1^2 {{x^2}dx}$ $A = \left[ {\frac{{{x^3}}}{3}} \right]_1^2$ $A = \frac{1}{3}\left( {8 - 1} \right) = \frac{7}{3}$ (ii) y=x4, x = 1, x = 5
y=x4 is equation of parabolic curve x=1, x = 5 are two vertical line passing through M(1, 0) and N(5,0)
A and B are point of intersection of curve and line.
So coordinate of A(1, 1) and B(5, 625) Area of region bounded by curve and line = Area of region MACBNM A
$A = \int\limits_1^5 {ydx}$ $A = \int\limits_1^5 {{x^4}dx}$ $A = \left[ {\frac{{{x^5}}}{5}} \right]_1^5$ $A = \frac{1}{5}\left[ {{5^5} - 1} \right]$ $A = \frac{1}{5}\left[ {3125 - 1} \right]$ $A = \frac{{3124}}{5} = 624.8$

Question (2)

Find the area between the curves y=x and y=x2

Solution

x2 = y is the equation of parabola along y-axis , y=x is equation of line passing the origin O(0, 0)
To get the point of intersection of line and parabola, replace y=x in parabola
y=x2
x=x2
x2 -x = 0
x(x - 1) = 0
x = 0 an x = 1
If x = 0, y = 0 O(0, 0)
If x = 1, y = 1 A(1, 1)
Let AM ⊥ x-axis
So coordinates of M is (1, 0)
Area of region bounded by curve and line = Area of region OABO
Area (A) = Area of region OAMO - area of region OBAMO
$A = \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^1 {y\;dx} - \int\limits_{\begin{array}{*{20}{c}}0\\{curve}\end{array}}^1 {y\;dx}$ $A = \int\limits_0^1 {xdx - \int\limits_0^1 {{x^2}dx} }$ $A = \left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \frac{1}{3}\left[ {{x^3}} \right]_0^1$ $A = \frac{1}{2}\left( {1 - 0} \right) - \frac{1}{3}\left( {1 - 0} \right)$ $A = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$

Question (3)

Find the area of the region lying in the first quadrant and bounded by y=4x2, x=0, y=1, and y=4

Solution

y=4x2 is equation of parabola along y-axis, y = 1 and y=4 are parallel lines parallel to x-axis,
Point A and B are y-axis so co-ordinates are A(0, 1) and b(0, 4), C and E are on parabola of lines. So co-ordinate of C(1, 4), E(½ , 1)
Area of region bounded by parabola and line in first quadrant (A) = Area of region ABCDEA
$A = \int\limits_1^4 {xdy}$ Substituting value of x = √y /2
$A = \int\limits_1^4 {\frac{{\sqrt y }}{2}} dy$ $A = \frac{1}{2}\left[ {\frac{{{y^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^4$ $A = \frac{1}{\require{cancel}\cancel{2}} \times \frac{\cancel{2}}{3}\left[ {{4^{\frac{3}{2}}} - {1^{\frac{3}{2}}}} \right]$ $A = \frac{1}{3}\left[ {8 - 1} \right] = \frac{7}{3}$

Question (4)

Sketch the graph of y =|x+3| and evaluate $\int\limits_{ - 6}^0 {\left| {x + 3} \right|{\kern 1pt} } dx$

Solution

y=|x+3|
y = (x+3)
y= -(x+3)
for y = x+3
 x -2 -1 0 1 y 1 2 3 4
for y = -(x+3)
 x -4 -5 -6 y 1 2 3
$A = \int\limits_{ - 6}^0 {\left| {x + 3} \right|} dx$ at x+3 = 0 ⇒ x = -3 function breaks $A = \int\limits_{ - 6}^{ - 3} { - \left( {x + 3} \right)dx} + \int\limits_{ - 3}^0 {\left( {x + 3} \right)dx}$ $A = - \int\limits_{ - 6}^{ - 3} {xdx} - \int\limits_{ - 6}^{ - 3} {3dx} + \int\limits_{ - 3}^0 {xdx} + \int\limits_{ - 3}^0 {3dx}$ $A = - \frac{{\left[ {{x^2}} \right]_{ - 6}^{ - 3}}}{2} - 3\left[ x \right]_{ - 6}^{ - 3} + \frac{{\left[ {{x^2}} \right]_{ - 3}^{0.}}}{2} + 3\left[ x \right]_{ - 3}^0$ $A = - \frac{1}{2}\left( {9 - 36} \right) - 3\left( { - 3 + 6} \right) + \frac{1}{2}\left( {0 - 9} \right) + \left( {0 + 3} \right)$ $A = \frac{{27}}{2} - 9 - \frac{9}{2} + 9$ $A = \frac{{18}}{2} = 9$

Question (5)

Find the area bounded by the curve y=sinx between x=0 and x=2π

Solution

y= sinx, x = 0, x = 2π
y=0 when x=0, π, 2π
y=1 when x = π/2
y=-1 at x = 3π/2
Co-ordinates of B(π/2, 1)
BM ⊥ X-axis M(π/2, 0)
Area under the curve, y=sinx A
A= 4[Area of region OABMO]
$A = 4\left[ {\int\limits_0^{\frac{\pi }{2}} {y\;dx} } \right]$ $A = 4\left[ {\int\limits_0^{\frac{\pi }{2}} {\sin x\;dx} } \right]$ $A = 4\left[ { - \cos x} \right]_0^{\frac{\pi }{2}}$ $A = - 4\left( {\cos \frac{\pi }{2} - \cos 0} \right)$ $A = - 4\left( {0 - 1} \right)$ A=+4

Question (6)

Find the area enclosed between the parabola y2 = 4ax and the line y=mx

Solution

y2 = 4ax, y = mx
y2 = 4ax is equation of parabola along x-axis
y= mx is line through (0, 0)
To get point of intersection replace y = mx, in parabola
y2 = 4ax
or y = √(4ax)
m2x2 - 4ax = 0
x(m2x - 4a) = 0
x = 0, x = 4a/m2
If x = 0, y = 0, O(0, 0)
x = 4a/m2 then y= 4a/m
Coordinate of A ( 4a/m2, 4a/m)
Let AM ⊥ x-axis, coordinate of M (4a/m2, 0)
Area of region bounded by curve and line (A) = Area of region OCAO
A = area of region OCAMO - area of region OAMO
$A = \int\limits_{\begin{array}{*{20}{c}}0\\{curve}\end{array}}^{\frac{{4a}}{{{m^2}}}} {y\;dx} - \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^{\frac{{4a}}{{{m^2}}}} {y\;dx}$ $A = \int\limits_0^{\frac{{4a}}{{{m^2}}}} {2\sqrt a \sqrt {x\,} dx} - \int\limits_0^{\frac{{4a}}{{{m^2}}}} {mx{\kern 1pt} dx}$ $A = 2\sqrt a \frac{{\left[ {{x^{\frac{3}{2}}}} \right]_0^{\frac{{4a}}{{{m^2}}}}}}{{\frac{3}{2}}} - \frac{m}{2}\left[ {{x^2}} \right]_0^{\frac{{4a}}{{{m^2}}}}$ $A = \frac{4}{3}\sqrt a \left[ {{{\left( {\frac{{4a}}{{{m^2}}}} \right)}^{\frac{3}{2}}} - 0} \right] - \frac{m}{2}\left[ {{{\left( {\frac{{4a}}{{{m^2}}}} \right)}^2} - 0} \right]$ $A = \frac{4}{3}\sqrt a \left[ {\frac{{8{a^{\frac{3}{2}}}}}{{{m^3}}}} \right] - \frac{m}{2}\left( {\frac{{16{a^2}}}{{{m^4}}}} \right)$ $A = \frac{{32{a^2}}}{{3{m^3}}} - \frac{{\cancel{16}^8{a^2}}}{{\cancel{2}_1{m^3}}}$ $A = \frac{{32{a^2} - 24{a^2}}}{{3{m^3}}}$ $A = \frac{{8{a^2}}}{{3{m^3}}}$

Question (7)

Find the area enclosed between by the parabola 4y=3x2 and the line 2y=3x+12

Solution

4y = 3x2, 2y = 3x +12
4y = 3x2 ⇒ x2 = 4y/3 is equation of parabola along y-axis and 2y=3x+12 is equation of line To get point of intersection replace 2y=3x+12 in parabola
2(3x+12) = 3x2
6x+24 = 3x2
3x2 - 6x - 24 = 0
x2 - 2x - 8 = 0
(x-4) (x+2) = 0
x = 4 or x = -2
If x=-2, y = 2
If x =4, y = 12
Co-ordinate of A = (-2, 2), B = (4, 12)
Let AM ⊥ x-axis coordinate of M (-2, 0)
BN ⊥ x-axis coordinate of N (4, 0)
Let C and D be point on parabola
Area under the curve and line = area of region ABCODA
A= area of region ABCODA
A = area of region ABNMA - area of region ADCBNMA
$A = \int\limits_{\begin{array}{*{20}{c}}{ - 2}\\{line}\end{array}}^4 {y\;dx} - \int\limits_{\begin{array}{*{20}{c}}{ - 2}\\{curve}\end{array}}^4 {y\;dx}$ $A = \int\limits_{ - 2}^4 {\frac{{3x + 12}}{2}\,dx - \int\limits_{ - 2}^4 {\frac{3}{4}{x^2}\,dx} }$ $A = \frac{3}{2}\left[ {\frac{{{x^2}}}{2}} \right]_{ - 2}^4 + 6\left[ x \right]_{ - 2}^4 - \frac{3}{4}\left[ {\frac{{{x^3}}}{3}} \right]_{ - 2}^4$ $A = \frac{3}{4}\left( {16 - 4} \right) + 6\left( {4 + 2} \right) - \frac{1}{4}\left[ {64 - \left( { - 8} \right)} \right]$ $A = \frac{3}{\cancel{4}} \times \cancel{12}^3 + 36 - \frac{1}{\cancel{4}} \times \cancel{72}^{18}$ $A = 9 + 36 - 18 = 27$

Question (8)

Find the area of the smaller region bounded by the ellipse and line $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ and $\frac{x}{3} + \frac{y}{2} = 1$

Solution

$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\text{ is equation of ellipse}$ a2 = 9, b2 = 4
a = 3 and b =2, a > b
ellipse along x-axis $\frac{x}{3} + \frac{{{y}}}{2} = 1\text{ is equation of line}$ It passes through A(3, 0) and B(0, 2) Area of small er region bounded by ellipse and line (A) = Area of region BCAB
A= Area of region OBCAO - area OBAO $A = \int\limits_{\begin{array}{*{20}{c}}0\\{ellipse}\end{array}}^3 {ydx} - \int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^3 {ydx}$ $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ $\frac{{{y^2}}}{4} = 1 - \frac{{{x^2}}}{9}$ $\frac{{{y^2}}}{4} = \frac{{9 - {x^2}}}{9}$ ${y^2} = \frac{4}{9}\left( {9 - {x^2}} \right)$ $\Rightarrow y = \frac{2}{3}\sqrt {9 - {x^2}}$ $\frac{x}{3} + \frac{y}{2} = 1$ $\frac{y}{2} = 1 - \frac{x}{3}$ $\frac{y}{2} = \frac{{3 - x}}{3}$ $y = \frac{{2\left( {3 - x} \right)}}{3}$ $A = \int\limits_0^3 {\frac{2}{3}\sqrt {9 - {x^2}} dx - \int\limits_0^3 {\frac{2}{3}\left( {3 - x} \right)dx} }$ $A = \frac{2}{3}\left[ {\frac{x}{2}\sqrt {9 - {x^2}} + \frac{9}{2}{{\sin }^{ - 1}}\frac{x}{3}} \right]_0^3 - 2\left[ x \right]_0^3 + \frac{2}{3}\left[ {\frac{{{x^2}}}{2}} \right]_0^3$ $A = \frac{2}{3}\left[ {\left( {\frac{3}{2}\left( 0 \right) + \frac{9}{2}{{\sin }^{ - 1}}1} \right) - \left( {0 + \frac{9}{2}{{\sin }^{ - 1}}1} \right)} \right] - 2\left[ {\left( {3 - 0} \right) + \frac{1}{3}\left( {9 - 0} \right)} \right]$ $A = \frac{2}{3}\left[ {\frac{9}{2} \cdot \frac{\pi }{2}} \right] - 6 + 3$ $A = \frac{{3\pi }}{2} - 3$ $A = 3\left( {\frac{\pi }{2} - 1} \right)$ $A = \frac{3}{2}\left( {\pi - 2} \right)$

Question (9)

Find the area of the smaller region bounded by the ellipse and line $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{x}{a} + \frac{y}{b} = 1$

Solution

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\text{is equation of ellipse alongx-axis}$ Co-ordinate of A(a,0) and B(0, b)
x/a + y/b = 1 is equation of line passes through A(a, 0) and B(0, b) $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ $\frac{{{x^2}}}{{{a^2}}} = 1 - \frac{{{y^2}}}{{{b^2}}}$ ${x^2} = \frac{{{a^2}}}{{{b^2}}}\left( {{b^2} - {y^2}} \right)$ $x = \frac{a}{b}\sqrt {{b^2} - {y^2}}$ For line $\frac{x}{a} + \frac{y}{b} = 1$ $\frac{x}{a} = 1 - \frac{y}{b}$ $y = \frac{{a\left( { b - y} \right)}}{b}$ Area of region bounded by ellipse and line = area of region ACBA
A= area of region OBCAO - area of region OBAO
$= \int\limits_0^b {xdy} ofellipse - \int\limits_0^b {xdyofline}$ $= \int\limits_0^b {\frac{a}{b}\sqrt {{b^2} - {y^2}} dy} - \int\limits_0^b {\frac{a}{b}(b - y)dy}$ $= \frac{a}{b}\left[ {\frac{y}{2}\sqrt {{b^2} - {y^2}} + \frac{{{b^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{y}{b}} \right)} \right]_0^b - a\left[ y \right]_0^b + \frac{a}{b}\left[ {\frac{{{y^2}}}{2}} \right]_0^b$ $= \frac{a}{b}\left( {\frac{{{b^2}}}{2} \times \frac{\pi }{2}} \right) - ab + \frac{{ab}}{2}$ $= \frac{{\pi ab}}{4} - \frac{{ab}}{2}$ $= \frac{{ab}}{4}\left( {\pi - 2} \right)$

Question (10)

Find the area of the region enclosed by the parabola x2 = y, the line y=x+2 and the x-axis

Solution

x2 = y is equation of parabola along y-axis
y = x+2 is equation of line
To get point of intersection replace y = x + 2 in parabola x2 = x+2
x2 - x - 2 = 0
(x-2) (x+1) = 0If x = -1, y = 1 A(-1, 1)
If x =2, y = 4 B(2, 4)
Line passes through A and B
AM ⊥ x-axis, so coordinates of M = (-1, 0)
BN ⊥ x-axis, so coordinates of N = (2, 0)
C and D are on parabola Area of region bounded by curve and line = area of region ABCDA
A = area of ABNMA - area of region ADCBNMA $A = \int\limits_{\begin{array}{*{20}{c}}{ - 1}\\{line}\end{array}}^2 {ydx} - \int\limits_{\begin{array}{*{20}{c}}{ - 1}\\{curve}\end{array}}^2 {ydx}$ $A = \int\limits_{ - 1}^2 {\left( {x + 2} \right)dx} - \int\limits_{ - 1}^2 {{x^2}} dx$ $A = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^2 + 2\left[ x \right]_{ - 1}^2 - \left[ {\frac{{{x^3}}}{3}} \right]_{ - 1}^2$ $A = \frac{1}{2}\left( {4 - 1} \right) + 2\left[ {2 - \left( { - 1} \right)} \right] - \frac{1}{3}\left[ {8 - \left( { - 1} \right)} \right]$ $A = \frac{3}{2} + 6 - \frac{1}{\cancel{3}} \times \cancel{9}^3$ $A = \frac{3}{2} + 3 = \frac{9}{2}$

Question (11)

Using the method of integration find the area bounded by the curve |x|+|y|=1
[Hint: The require region is bounded by lines x+y=1, x-y=1, -x+y=1, and -x-y=1]

Solution

|x|+|y|=1 so equations for lines are l1 : x+y=1, l2 :x-y=1, l3 :-x+y=1, and l4 :-x-y=1
Solving these equations we get coordinates of A(0, 1), B(1,0), C(0,-1) and D (-1,0)
Area under (enclosed by lines ) area of (ABCD)
A = 4[area of ABOA] $A = 4\int\limits_{\scriptstyle0\atop\scriptstyle{l_1}}^1 {ydx} \;$ $A = 4\int\limits_0^1 {\left( {1 - x} \right)dx}$ $A = 4\left[ {\left[ x \right]_0^1 - \left[ {\frac{{{x^2}}}{2}} \right]_0^1} \right]$ $A = 4\left[ {\left( {1 - 0} \right) - \frac{1}{2}\left( {1 - 0} \right)} \right]$ $A = 4\left[ {\left( {1 - 0} \right) - \frac{1}{2}\left( {1 - 0} \right)} \right]$ $A = 4\left( {1 - \frac{1}{2}} \right) = 2$

Question (12)

Find the area bounded by the curve {(x, y): y≥ x2 and y=|x|}

Solution

{(x, y): y≥ x2 and y=|x|}
y ≥ x2 is equation of parabola along y-axis
y = |x| that is y=x and y=-x are lines passes through origin
To get point of intersection replace y=x in parabola
y = x2
x=x2
x2 - x = 0
x(x-1) = 0
x = 0 or x=1
If x = 0, y = 0 , O (0, 0)
If x= 1, y=1, A(1, 1) B(1, -1)
AM ⊥ x-axis M(1, 0)
Area of region between curve and lines = 2[area of OACO]
A=2[ area of OAMO - area of OCAMO]
$A = 2\left[ {\int\limits_{\begin{array}{*{20}{c}}0\\{line}\end{array}}^1 {ydx} - \int\limits_{\begin{array}{*{20}{c}}0\\{curve}\end{array}}^1 {ydx} } \right]$ $A = 2\left[ {\int\limits_0^1 {xdx - \int\limits_0^1 {{x^2}dx} } } \right]$ $A = 2\left[ {\left[ {\frac{{{x^2}}}{2}} \right]_0^1 - \frac{1}{3}\left[ {{x^3}} \right]_0^1} \right]$ $A = 2\left[ {\frac{1}{2}\left( {1 - 0} \right) - \frac{1}{3}\left( {1 - 0} \right)} \right]$ $A = 2\left[ {\frac{1}{2} - \frac{1}{3}} \right]$ $A = 2 \times \frac{1}{6} = \frac{1}{3}$

Question (13)

Using the method of integration find the area of the triangle ABC, coordination whose vertices are A(2, 0), B(4,5) and C(6, 3)

Solution

A(2, 0), B(4, 5), C(6, 3) Equation of AB is $\left| {\begin{array}{*{20}{c}}x&y&1\\2&0&1\\4&5&1\end{array}} \right| = 0$ -5x+2y+10= 0
$y = \frac{{5x}}{2} - 5$ Equation of BC is $\left| {\begin{array}{*{20}{c}}x&y&1\\4&5&1\\6&3&1\end{array}} \right| = 0$ 2x+2y-18 = 0
y=9-x
Equation of AC is $\left| {\begin{array}{*{20}{c}}x&y&1\\2&0&1\\6&3&1\end{array}} \right| = 0$ -3x+4y+6 = 0
$y = \frac{{3x}}{4} - \frac{3}{2}$ Area of ΔABC= A
$A = \int\limits_{\begin{array}{*{20}{c}}2\\\overleftrightarrow{AB}\end{array}}^4 {ydx + } \int\limits_{\begin{array}{*{20}{c}}4\\\overleftrightarrow{BC}\end{array}}^6 {ydx} - \int\limits_{\begin{array}{*{20}{c}}2\\\overleftrightarrow{AC}\end{array}}^6 {ydx}$ $A = \int\limits_2^4 {\left( {\frac{5}{2}x - 5} \right)dx} + \int\limits_4^6 {\left( {9 - x} \right)dx} - \int\limits_2^6 {\left( {\frac{{3x}}{4} - \frac{3}{2}} \right)dx}$ $= \frac{5}{2}\left[ {\frac{{{x^2}}}{2}} \right]_2^4 - 5\left[ x \right]_2^4 + 9\left[ x \right]_4^6 - \frac{1}{2}\left[ {{x^2}} \right]_4^6 - \frac{3}{4}\left[ {\frac{{{x^2}}}{2}} \right]_2^6 + \frac{3}{2}\left[ x \right]_2^6$ $= \frac{5}{4}\left( {16 - 4} \right) - 5\left( {4 - 2} \right) + \left( {6 - 4} \right) - \frac{1}{2}\left( {36 - 16} \right) - \frac{3}{8}\left( {36 - 4} \right) + \frac{3}{2}\left( {6 - 2} \right)$ $= \frac{5}{4} \times 12 - 5\left( 2 \right) + 9\left( 2 \right) - \frac{1}{2} \times 20 - \frac{3}{8} \times 32 + \frac{3}{2} \times 4$ $= 15 - 10 + 18 - 10 - 12 + 6$ $= 39 - 32 = 7$

Question (14)

Using the method of integration find the area of the region bounded by lines 2x+y=4, 3x-2y=6 and x-3y+5 = 0

Solution

l1: 2x+y = 4,
x = 0, y = 4 x= 2, y =0 l2: 3x-2y = 6,
x= 0, y = -3 l3: x-3y+5 = 0, let l1 ∩ l2 = {A}
$\begin{array}{l}2x + y = 4 \times 2\\3x - 2y = 6 \times 1\end{array}$ $\begin{array}{l}\quad 4x + 2y = 8\\ + \\\underline {\quad 3x - 2y = 6} \\\quad 7x + 0 = 14\\x = 2\end{array}$ x =2, y = 0
A=(2, 0)
l2 ∩ l3= {B}
$3x - 2y = 6 \times 3$ $x - 3y = - 5 \times 2$ $\begin{array}{l}\quad 9x - 6y = 18\\ - \\\underline {\quad 2x \mp 6y = \mp 10} \\\quad 7x + 0 = 28\\x = 4\end{array}$ When x = 4, y = 3 B(4, 3)
let l1 ∩ l3 ={C}
$2x + y = 4 \times 3$ $x - 3y = - 5$ $\begin{array}{l}\quad 6x + 3y = 12\\ + \\\underline {\quad x - 3y = - 5} \\\quad 7x + 0 = 7\\x = 1\end{array}$ When x = 1, y=2, C(1, 2)
Area of Δ ABC
$A = \int\limits_{\begin{array}{*{20}{c}}1\\\overleftrightarrow{BC}\end{array}}^4 {ydx - \int\limits_{\begin{array}{*{20}{c}}1\\\overleftrightarrow{AC}\end{array}}^2 {ydx - \int\limits_{\begin{array}{*{20}{c}}2\\\overleftrightarrow{AB}\end{array}}^4 {ydx} } }$ $A = \int\limits_1^4 {\frac{{x + 5}}{3}dx} - \int\limits_1^2 {\left( {4 - 2x} \right)dx} - \int\limits_2^4 {\left( {\frac{{3x - 6}}{2}} \right)dx}$ $= \frac{1}{3}\left[ {\frac{{{x^2}}}{2}} \right]_1^4 + \frac{5}{3}\left[ x \right]_1^4 - 4\left[ x \right]_1^2 + 2\left[ {\frac{{{x^2}}}{2}} \right]_1^2 - \frac{3}{2}\left[ {\frac{{{x^2}}}{2}} \right]_2^4 + 3\left[ x \right]_2^4$ $= \frac{1}{6}\left( {16 - 1} \right) + \frac{5}{3}\left( 3 \right) - 4\left( {2 - 1} \right) + \left( {4 - 1} \right) - \frac{3}{4}\left( {16 - 4} \right) + 3\left( {4 - 2} \right)$ $A = \frac{{\cancel{15}^5}}{\cancel{6}_2} + 5 - 4 + 3 - \frac{3}{\cancel{4}} \times \cancel{12}^3 + 6$ $A = \frac{5}{2} + 4 - 9 + 6$ $A = \frac{5}{2} + 1 = \frac{7}{2}$

Question (15)

Find the area of region {(x, y) : y2 ≤ 4x, 4x2+4y2 ≤ 9}

Solution

{(x, y): y2 ≤ 4x: 4x2 + 4y2 ≤ 9} y2 = 4x is equation of parabola
4x2 + 4y2 = 9 is equation of circle
⇒ C (0, 0) and r = 3/2
4x2 +4(4x) - 9 = 0
4x2 + 16x - 9 = 0
4x2 + 18x - 2x - 9 = 0
2x(2x+9) - (2x+9) = 0
(2x+9) (2x-1) = 0
x = -9/2, x = 1/2
if x= -9/2, y2 = 4(-9/2) = -18
which is not possible
x = ½, y2 = 4(½) = 2
⇒ y = √2
A( ½ , √2)
Area between circle and parabola = 2[area ODACBO]
A = [ area of ODAMO + area of region ACBMA] $A = 2\left[ {\int\limits_{\begin{array}{*{20}{c}}0\\{parabola}\end{array}}^{\frac{1}{2}} {ydx + \int\limits_{\begin{array}{*{20}{c}}{\frac{1}{2}}\\{circle}\end{array}}^{\frac{3}{2}} {ydx} } } \right]$ $A = 2\left[ {\int\limits_0^{\frac{1}{2}} {2\sqrt x dx} + \int\limits_{\frac{1}{2}}^{\frac{3}{2}} {\sqrt {\frac{9}{4} - {x^2}} } dx} \right]$ $A = 4\left[ {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right] + 2\left[ {\frac{x}{2}\sqrt {\frac{9}{4}{-x^2}} + \frac{9}{{4\left( 2 \right)}}{{\sin }^{ - 1}}\left( {\frac{x}{{\frac{3}{2}}}} \right)} \right]_{\frac{1}{2}}^{\frac{3}{2}}$ $= 2\left[ {\frac{4}{3}\left( {\frac{1}{{2\sqrt 2 }}} \right)} \right] + \left[ {\left( {0 + \frac{9}{8}{{\sin }^{ - 1}}1} \right) - \frac{1}{4}\sqrt {\frac{9}{4} - \frac{1}{4}} - \frac{9}{8}{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right]$ $= \frac{4}{{3\sqrt 2 }} + 2 \times \frac{9}{8} \times \frac{\pi }{2} - \frac{2}{4}\sqrt 2 - 2 \times \frac{9}{8}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$ $A = \frac{4}{{3\sqrt 2 }} + \frac{{9\pi }}{8} - \frac{{\sqrt 2 }}{2} - \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$ $A = \frac{{\cancel{4}^2\sqrt 2 }}{\cancel{6}_3} - \frac{{\sqrt 2 }}{2} + \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$ $A = \frac{{\sqrt 2 }}{6} + \frac{9}{4}\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right)$ $A = \frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\left( {\frac{1}{3}} \right) + \frac{1}{{3\sqrt 2 }}$
Choose the correct answer in the following Exercise from 16 to 20

Question (16)

Area bounded by the curve y=x3, the x-axis and the ordinates x= -2 and x =1 is $(A) - 9$ $(B)\frac{{ - 15}}{4}$ $(C)\frac{{15}}{4}$ $\left( D \right)\frac{{17}}{4}$

Solution

y=x3 , x = -2 and x=1
x3 = -x3, x < 0
x3 = x3, x > 0
$Area = \int\limits_{ - 2}^1 {ydx}$ $A = \int\limits_{ - 2}^0 { - {x^3}dx} + \int\limits_0^1 {{x^3}dx}$ $A = - \left[ {\frac{{{x^4}}}{4}} \right]_{ - 2}^0 + \left[ {\frac{{{x^4}}}{4}} \right]_0^1$ $A = - \frac{1}{4}\left( {0 - 16} \right) + \frac{1}{4}\left( {1 - 0} \right)$ $A = - \left( {\frac{{ - 16}}{4}} \right) + \frac{1}{4}$ $A = 4 + \frac{1}{4}$ $A = \frac{{17}}{4}$ D is correct option

Question (17)

The area bounded by the curve y=x|x|, x-axis and the ordinates x = -1 and x=1 is given by
$(A)\;0$ $(B)\;\frac{1}{3}$ $(B)\;\frac{1}{3}$ $(D)\;\frac{4}{3}$

Solution

y=x|x|, x = -1, x = 1
|x| = x , x > O |x| = -x , x < O ∴ y = -x2 x < 0
y= x2 x < 1
Area under curve (A)
$A = \int\limits_{ - 1}^1 {ydx}$ $A = 2\int\limits_0^1 {ydx}$ $A = 2\int\limits_0^1 {{x^2}dx}$ $A = 2\left[ {\frac{{{x^3}}}{3}} \right]_0^1$ $A = \frac{2}{3}\left[ {1 - 0} \right] = \frac{2}{3}$ Correct option is "C"

Question (18)

The area of the circle x2 + y2 = 16 exterior to the parabola y2 = 6x is $(A)\frac{4}{3}\left( {4\pi - \sqrt 3 } \right)$ $(B)\frac{4}{3}\left( {4\pi + \sqrt 3 } \right)$ $(C)\frac{4}{3}\left( {8\pi - \sqrt 3 } \right)$ $(D)\frac{4}{3}\left( {8\pi + \sqrt 3 } \right)$

Solution

x2 + y2 = 16 is circle whose4 centre is at (0, 0) and radius is 4
y2 = 6x equation of parabola along x-axis Area of region bounded inside circle and exterior of parabola is area of shaded region
To get point of intersection of circle and parabola we will solve as
x2 + 6x - 16 = 0
(x+8) (x-2) = 0
x = -8 and x=2
If x = -8 then y2= 6(-8) = -48 wich is not possible
x = 2, y2 = 2(6) = 12
⇒ y = ±√12 = ±2√3 ∴ B and C are point of intersection B(2, 2√3) and C(2, -2√3)
Let BM ⊥ x-axis so co-ordinate of M=(2, 0)
First we will find area enclosed between circle and parabola
area of region OEBDDCFO = circle and parabola are symmetric aboout x-axis
So area of region (A) = 2[area of region OEBDAO]
$A = 2\left[ {\int\limits_{\begin{array}{*{20}{c}}0\\{parabola}\end{array}}^2 {ydx} + \int\limits_{\begin{array}{*{20}{c}}2\\{circle}\end{array}}^4 {ydx} } \right]$ $A = 2\left[ {\int\limits_0^2 {\sqrt {6x} dx} + \int\limits_2^4 {\sqrt {16 - {x^2}} dx} } \right]$ $A = 2\left[ {\left[ {\sqrt 6 \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^2 + \left[ {\frac{x}{2}\sqrt {16 - {x^2}} + \frac{{16}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_2^4} \right]$ $A = 2\left[ {\frac{{2\sqrt 6 }}{3}\left( {{2^{\frac{3}{2}}} - 0} \right) + \left( {\frac{4}{2}\left( 0 \right) + 8{{\sin }^{ - 1}}1 - \sqrt {12} - 8{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]$ $2\left[ {\frac{{2\sqrt 6 }}{3} \times 2\sqrt 2 + \frac{{8\pi }}{2} - 2\sqrt 3 - \frac{{8\pi }}{6}} \right]$ $2\left[ {\frac{{4\sqrt {12} }}{3} + 4\pi - 2\sqrt 3 - \frac{{4\pi }}{3}} \right]$ $2\left[ {\frac{{8\sqrt 3 }}{3} + 4\pi - 2\sqrt 3 - \frac{{4\pi }}{3}} \right]$ $= 2\left[ {\frac{{2\sqrt 3 }}{3} + \frac{{8\pi }}{3}} \right]$ $= \frac{4}{3}\left[ {\sqrt 3 + 4\pi } \right]$ Area of the region inside the circle and exteriior of the parabola = area of the circla - area of regon bounded between circle and parabola. $= \pi {r^2} - \frac{4}{3}\left( {\sqrt 3 + 4\pi } \right)$ $= 16\pi - \frac{{4\sqrt 3 }}{3} - \frac{{16\pi }}{3}$ $= \frac{{32\pi }}{3} - \frac{{4\sqrt 3 }}{3}$ $= \frac{4}{3}\left( {8\pi - \sqrt 3 } \right)$

Question (19)

The area bounded by the y-axis, y=cosx and y=sinx when 0≤ x ≤ π/2 is $(A)\,2\left( {\sqrt {2 - 1} } \right)$ $(B)\,\sqrt 2 - 1$ $(C)\,\sqrt 2 + 1$ $(D)\,\sqrt 2$

Solution

y = cosx, y = sinx, 0 < x < π/2, y-axis
sin0 = 0, sin π/2 = 1
cos0 = 1, cosπ/2
To get point of intersection compare sinx = cosx, it is possible at x = π/4, y=1/√2
coordinate of A = (π/4, 1/√2)
AM ⊥ x-axis , coordinate of M =(π/4, 0) Area under curve= A $A = \int\limits_0^{\frac{\pi }{4}} {ydx} - \int\limits_0^{\frac{\pi }{4}} {ydx}$ $A = \int\limits_0^{\frac{\pi }{4}} {coxdx} - \int\limits_0^{\frac{\pi }{4}} {\sin xdx}$ $A = \left[ {\sin x} \right]_0^{\frac{\pi }{4}} - \left[ { - \cos x} \right]_0^{\frac{\pi }{4}}$ $A = \left( {\frac{1}{{\sqrt 2 }} - 0} \right) + \left( {\frac{1}{{\sqrt 2 }} - 1} \right)$ $A = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} - 1$ $A = \sqrt 2 - 1$ Option B is correct