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Area Under Curves

JEE Mathematics · 107 questions · Page 3 of 11 · Click an option or "Show Solution" to reveal answer

Q21
The area (in sq. units) of the region described by {(x,y):y2≤2x\left\{ {\left( {x,y} \right):{y^2} \le 2x} \right. and y≥4x−1}\left. {y \ge 4x - 1} \right\} is :
A 1564{{15} \over {64}}
B 932{{9} \over {32}}
C 732{{7} \over {32}}
D 564{{5} \over {64}}
Correct Answer
Option B
Solution

Required area == Area of

ABCDABCD

−-

arar
(ABOCD)(ABOCD)
=14[y22+y]−1/21  −12[y33]−11= {1 \over 4}\left[ {{{{y^2}} \over 2} + y} \right]_{ - 1/2}^1\,\, - {1 \over 2}\left[ {{{{y^3}} \over 3}} \right]_{ - 1}^1
=14[32+38]−948= {1 \over 4}\left[ {{3 \over 2} + {3 \over 8}} \right] - {9 \over {48}}
=1532−948=2796=932= {{15} \over {32}} - {9 \over {48}} = {{27} \over {96}} = {9 \over {32}}
Q22
The area (in sq. units) of the region described by A= {(x, y) ∣\left| {} \right.y≥ \ge x2 −- 5x + 4, x + y ≥ \ge 1, y ≤ \le 0} is :
A 72{7 \over 2}
B 196{{19} \over 6}
C 136{{13} \over 6}
D 176{{17} \over 6}
Correct Answer
Option B
Solution

Required Area = A1 + A2 =

∣∫13(1−x)dx∣+∣∫34(x2−5x+4)dx∣\left| {\int\limits_1^3 {\left( {1 - x} \right)} dx} \right| + \left| {\int\limits_3^4 {\left( {{x^2} - 5x + 4} \right)dx} } \right|

=

∣[x−x22]13∣+∣[x33−52x2+4x]34∣\left| {\left[ {x - {{{x^2}} \over 2}} \right]_1^3} \right| + \left| {\left[ {{{{x^3}} \over 3} - {5 \over 2}{x^2} + 4x} \right]_3^4} \right|

=

∣[(3−92)−(1−12)]∣+∣[(643−40+16)−(9−452+12)]∣\left| {\left[ {\left( {3 - {9 \over 2}} \right) - \left( {1 - {1 \over 2}} \right)} \right]} \right| + \left| {\left[ {\left( {{{64} \over 3} - 40 + 16} \right) - \left( {9 - {{45} \over 2} + 12} \right)} \right]} \right|

=

∣(2−4)∣+∣(−83+32)∣\left| {\left( {2 - 4} \right)} \right| + \left| {\left( {{{ - 8} \over 3} + {3 \over 2}} \right)} \right|

= 2 +

76{7 \over 6}

=

196{{19} \over 6}

sq. unit.

Q23
The area (in sq. units) of the region {(x,y):y2≥2x   and   x2+y2≤4x,x≥0,y≥0}\left\{ {\left( {x,y} \right):{y^2} \ge 2x\,\,\,and\,\,\,{x^2} + {y^2} \le 4x,x \ge 0,y \ge 0} \right\} is :
A π−423\pi - {{4\sqrt 2 } \over 3}
B π2−223{\pi \over 2} - {{2\sqrt 2 } \over 3}
C π−43\pi - {4 \over 3}
D π−83\pi - {8 \over 3}
Correct Answer
Option D
Solution

Points of intersection of the two curves are

(0,0),(2,2)\left( {0,0} \right),\left( {2,2} \right)

and

(2,−2)\left( {2, - 2} \right)

Area == Area

(OAB)−(OAB)-

area under parabola (

00

to

22

)

=π×(2)24−∫022x dx= {{\pi \times {{\left( 2 \right)}^2}} \over 4} - \int\limits_0^2 {\sqrt 2 \sqrt x } \,dx
=π−83= \pi - {8 \over 3}
Q24
The area (in sq. units) of the region {(x,y):x≥0,x+y≤3,x2≤4y and y≤1+x}\left\{ {\left( {x,y} \right):x \ge 0,x + y \le 3,{x^2} \le 4y\,and\,y \le 1 + \sqrt x } \right\} is
A 32{3 \over 2}
B 73{7 \over 3}
C 52{5 \over 2}
D 5912{59 \over 12}
Correct Answer
Option C
Solution

Area of shaded region =

∫01(1+x)dx+∫12(3−x)dx−∫02x24dx\int\limits_0^1 {\left( {1 + \sqrt x } \right)dx} + \int\limits_1^2 {\left( {3 - x} \right)dx} - \int\limits_0^2 {{{{x^2}} \over 4}dx}

=

[x]01+[x3232]01\left[ x \right]_0^1 + \left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^1

+

3[x]12−[x22]12−[x312]023\left[ x \right]_1^2 - \left[ {{{{x^2}} \over 2}} \right]_1^2 - \left[ {{{{x^3}} \over {12}}} \right]_0^2

=

52{5 \over 2}

sq. unit

Q25
The area (in sq. units) of the smaller portion enclosed between the curves, x2 + y2 = 4 and y2 = 3x, is :
A 123+π3{1 \over {2\sqrt 3 }} + {\pi \over 3}
B 13+2π3{1 \over {\sqrt 3 }} + {{2\pi } \over 3}
C 123+2π3{1 \over {2\sqrt 3 }} + {{2\pi } \over 3}
D 13+4π3{1 \over {\sqrt 3 }} + {{4\pi } \over 3}
Correct Answer
Option D
Solution

The given equation

x2+y2=4{x^2} + {y^2} = 4

is equation of circle of radius 2 centred at origin and equation

y2=3x{y^2} = 3x

is the equation of parabola.

x2+y2=4{x^2} + {y^2} = 4

..... (1)

y2=3x{y^2} = 3x

..... (2) Substituting Eq. (2) in Eq. (1), we get

x2+3x−4=0{x^2} + 3x - 4 = 0
⇒x2+4x−x−4=0\Rightarrow {x^2} + 4x - x - 4 = 0
⇒x(x+4)−1(x+4)=0\Rightarrow x(x + 4) - 1(x + 4) = 0
⇒(x−1)(x+4)=0\Rightarrow (x - 1)(x + 4) = 0
⇒(x−1)=0\Rightarrow (x - 1) = 0

and

(x+4)=0(x + 4) = 0

Therefore, x = 1, −-4. Considering x = 1, then from Eq. (2), we get

y=3,−3y = \sqrt 3 , - \sqrt 3

. Therefore

(1,3)(1,\sqrt 3 )

and

(1,−3)(1, - \sqrt 3 )

are the points of intersection of parabola and circle.

The required area (A) is the area of the shaded region shown in the figure.

Therefore,

A=2[∫01y2dx+∫12y1dx]A = 2\left[ {\int\limits_0^1 {{y_2}dx + \int\limits_1^2 {{y_1}dx} } } \right]

From Eq. (1), we get

y1=4−x2{y_1} = \sqrt {4 - {x^2}}

From Eq. (2), we get

y2=3x{y_2} = \sqrt {3x}

Therefore,

A=2[∫013xdx+∫124−x2dx]A = 2\left[ {\int\limits_0^1 {\sqrt {3x} dx + \int\limits_1^2 {\sqrt {4 - {x^2}} dx} } } \right]
=2[∫013x1/2dx+∫1222−x2dx]= 2\left[ {\int\limits_0^1 {\sqrt 3 {x^{1/2}}dx + \int\limits_1^2 {\sqrt {{2^2} - {x^2}} dx} } } \right]

Using standard integral :

∫xndx=xn−1n+1\int {{x^n}dx = {{{x^{n - 1}}} \over {n + 1}}}

, we have

∫a2−x2dx=x2a2−x2+a22tan⁡−1xa2−x2\int {\sqrt {{a^2} - {x^2}} dx = {x \over 2}\sqrt {{a^2} - {x^2}} + {{{a^2}} \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {{a^2} - {x^2}} }}}

Therefore,

A=2[(3x3/23/2)∣01+(x24−x2+42tan⁡−1x4−x2)∣12]A = 2\left[ {\left. {\left( {\sqrt 3 {{{x^{3/2}}} \over {3/2}}} \right)} \right|_0^1 + \left. {\left( {{x \over 2}\sqrt {4 - {x^2}} + {4 \over 2}{{\tan }^{ - 1}}{x \over {\sqrt {4 - {x^2}} }}} \right)} \right|_1^2} \right]
=2[3.13/2−0+224−4+2tan⁡−124−4−124−1−2tan⁡−114−1]= 2\left[ {\sqrt 3 .{1 \over {3/2}} - 0 + {2 \over 2}\sqrt {4 - 4} + 2{{\tan }^{ - 1}}{2 \over {\sqrt {4 - 4} }} - {1 \over 2}\sqrt {4 - 1} - 2{{\tan }^{ - 1}}{1 \over {\sqrt {4 - 1} }}} \right]
=2[233+tan⁡−1(∞)−32−2tan⁡−1(13)]= 2\left[ {{{2\sqrt 3 } \over 3} + {{\tan }^{ - 1}}(\infty ) - {{\sqrt 3 } \over 2} - 2{{\tan }^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)} \right]

Now,

tan⁡π2=∞\tan {\pi \over 2} = \infty

and

tan⁡π6=13\tan {\pi \over 6} = {1 \over {\sqrt 3 }}

. Therefore, the area of the smaller portion enclosed between the two curves is obtained as follows:

A=2[233−32+2tan⁡−1(tan⁡π2)−2tan⁡−1(tan⁡π6)]A = 2\left[ {{{2\sqrt 3 } \over 3} - {{\sqrt 3 } \over 2} + 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 2}} \right) - 2{{\tan }^{ - 1}}\left( {\tan {\pi \over 6}} \right)} \right]
=2[316+2π2−2π6]= 2\left[ {\sqrt 3 {1 \over 6} + 2{\pi \over 2} - 2{\pi \over 6}} \right]
=2[3+16+22π6]=2[36+4π6]= 2\left[ {\sqrt 3 + {1 \over 6} + 2{{2\pi } \over 6}} \right] = 2\left[ {{{\sqrt 3 } \over 6} + {{4\pi } \over 6}} \right]
=33+4π3=(13+4π3)= {{\sqrt 3 } \over 3} + {{4\pi } \over 3} = \left( {{1 \over {\sqrt 3 }} + {{4\pi } \over 3}} \right)

sq. units

Q26
Let g(x) = cosx2, f(x) = x\sqrt x and α,β(α<β)\alpha ,\beta \left( {\alpha < \beta } \right) be the roots of the quadratic equation 18x2 - 9π\pi x + π2{\pi ^2} = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x=αx = \alpha , x=βx = \beta and y = 0 is :
A 12(2−1){1 \over 2}\left( {\sqrt 2 - 1} \right)
B 12(3−1){1 \over 2}\left( {\sqrt 3 - 1} \right)
C 12(3+1){1 \over 2}\left( {\sqrt 3 + 1} \right)
D 12(3−2){1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)
Correct Answer
Option B
Solution

Given quadratic equation,

18x2−9πx+π2=018{x^2} - 9\pi x + {\pi ^2} = 0
⇒   18x2−6πx−3πx+π2=0\Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0
⇒    6x(3x−π)−π(3x−π)=0\Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0
⇒    (3x−π)(6x−π)=0\Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0

∴\therefore

    x=π3,π6\,\,\,\,x = {\pi \over 3},{\pi \over 6}

as

    ∝<B\,\,\,\, \propto < B

∴\therefore

    ∝=π6\,\,\,\, \propto = {\pi \over 6}
    \,\,\,\,

and

    \,\,\,\,
β=π3\beta = {\pi \over 3}

Given,

g(x)=cos⁡x2g\left( x \right) = \cos {x^2}

and

+(x)=x+ \left( x \right) = \sqrt x
y=(gof)xy = \left( {gof} \right)x
=     g(f(x))= \,\,\,\,\,g\left( {f\left( x \right)} \right)
=    cos⁡(f(x)2)= \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)
=    cos⁡(x)2= \,\,\,\,\cos {\left( {\sqrt x} \right)^2}
=    cos⁡x= \,\,\,\,\cos x

So, the required area in the curve is Area

=∫π6π3cos⁡  dx= \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx}
=[sin⁡x]π6π3= \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}
=sin⁡π3−sin⁡π6= \sin {\pi \over 3} - \sin {\pi \over 6}
=32−12= {{\sqrt 3} \over 2} - {1 \over 2}
=3−12= {{\sqrt 3 - 1} \over 2}
Q27
If the area (in sq. units) bounded by the parabola y2 = 4λ\lambda x and the line y = λ\lambda x, λ\lambda > 0, is 19{1 \over 9} , then λ\lambda is equal to :
A 434\sqrt 3
B 26\sqrt 6
C 48
D 24
Correct Answer
Option D
Solution

y2 = 4λ\lambdax and y =

λx\lambda x

If λ\lambda > 0 then Hence

∫04/λ(2λx−λx)dx=19\int\limits_0^{4/\lambda } {(2\sqrt \lambda \sqrt x } - \lambda x)dx = {1 \over 9}

⇒\Rightarrow

(2λx3/23/2−λx22)4λ=19{\left( {{{2\sqrt \lambda {x^{3/2}}} \over {3/2}} - {{\lambda {x^2}} \over 2}} \right)^{{4 \over \lambda }}} = {1 \over 9}

⇒\Rightarrow

43λ8λ3/2−λ8λ2=19{4 \over 3}\sqrt \lambda {8 \over {{\lambda ^{3/2}}}} - \lambda {8 \over {{\lambda ^2}}} = {1 \over 9}

⇒\Rightarrow

323λ−8λ=19{{32} \over {3\lambda }} - {8 \over \lambda } = {1 \over 9}

⇒\Rightarrow

83λ=19{8 \over 3\lambda } = {1 \over 9}

⇒\Rightarrow λ\lambda = 24

Q28
If the area (in sq. units) of the region {(x, y) : y2 ≤ \le 4x, x + y ≤ \le 1, x ≥ \ge 0, y ≥ \ge 0} is a 2\sqrt 2 + b, then a – b is equal to :
A 83{8 \over 3}
B −23 - {2 \over 3}
C 6
D 103{{10} \over 3}
Correct Answer
Option C
Solution

Let P be the point common to x + y = 1 and y2 = 4x Then y2 = 4(1-y) ⇒\Rightarrow y2 - 4y - 4 = 0

⇒y=−4±16+162\Rightarrow y = {{ - 4 \pm \sqrt {16 + 16} } \over 2}
⇒y=−2+22\Rightarrow y = - 2 + 2\sqrt 2

Then P is (3 -

222\sqrt 2

, -2 +

222\sqrt 2

). Hence shaded area = Area of region (OPN) + area of (

ΔOPQ\Delta OPQ

)

⇒∫03−222xdx+12[1−(3−22)]2\Rightarrow {\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt {xdx} + {1 \over 2}\left[ {1 - (3 - 2\sqrt 2 )} \right]} ^2}
⇒232(2−1)(3−22)+12[2(2−1)]2\Rightarrow {2 \over 3}2\left( {\sqrt 2 - 1} \right)\left( {3 - 2\sqrt 2 } \right) + {1 \over 2}{\left[ {2(\sqrt 2 - 1)} \right]^2}
⇒43{−7+52}+2(3−22)\Rightarrow {4 \over 3}\left\{ { - 7 + 5\sqrt 2 } \right\} + 2\left( {3 - 2\sqrt 2 } \right)
⇒(203−4)2+6−283\Rightarrow \left( {{{20} \over 3} - 4} \right)\sqrt 2 + 6 - {{28} \over 3}
⇒832−103\Rightarrow {8 \over 3}\sqrt 2 - {{10} \over 3}

Then a =

83{8 \over 3}

and b =

−103{-10 \over 3}

, so a - b = 6

Q29
The area (in sq.units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is :
A 12{1 \over 2}
B 32{3 \over 2}
C 32−1log⁡e2{3 \over 2} - {1 \over {\log _e^2}}
D log⁡e2+32\log _e^2 + {3 \over 2}
Correct Answer
Option C
Solution

Required area =

∫01((x+1)−2x)dx\int\limits_0^1 {((x + 1) - {2^x})dx}
⇒(x22+x−2xlog⁡e2)01\Rightarrow \left( {{{{x^2}} \over 2} + x - {{{2^x}} \over {{{\log }_e}2}}} \right)_0^1
⇒32−1log⁡e2\Rightarrow {3 \over 2} - {1 \over {{{\log }_e}2}}
Q30
The area (in sq. units) of the region A = {(x, y) : y22{{y{}^2} \over 2} ≤ \le x ≤ \le y + 4} is :-
A 30
B 18
C 533{{53} \over 3}
D 16
Correct Answer
Option B
Solution

y2 = 2x ...........(1) and x = y + 4 .............(

2) Solving (1) and (2) (x - 4)2 = 2x ⇒\Rightarrow x2 - 10x + 16 = 0 ⇒\Rightarrow x = 8, 2 and y = 4, -2 Integrating in y direction from A to B Required area (A) =

∫−24(y+4−y22)dy\int\limits_{ - 2}^4 {\left( {y + 4 - {{{y^2}} \over 2}} \right)dy}

=

[y22+4y−y36]−24\left[ {{{{y^2}} \over 2} + 4y - {{{y^3}} \over 6}} \right]_{ - 2}^4

=

(162+16−646){\left( {{{16} \over 2} + 16 - {{64} \over 6}} \right)}

-

(42−8+86){\left( {{4 \over 2} - 8 + {8 \over 6}} \right)}

= 30 - 12 sq. unit = 18 sq. unit

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