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Area Under Curves

JEE Mathematics · 107 questions · Page 4 of 11 · Click an option or "Show Solution" to reveal answer

Q31
The area (in sq. units) of the region A = {(x, y) : x2 ≤ \le y ≤ \le x + 2} is
A 316{{31 \over 6}}
B 103{{10 \over 3}}
C 136{{13 \over 6}}
D 92{{9 \over 2}}
Correct Answer
Option D
Solution

Parabola : x2 = y Straight line : y = x + 2 ∴\therefore x2 = x + 2 ⇒\Rightarrow x2 - x - 2 = 0 x = -1, 2 ∴\therefore y = 1, 4 Required area =

∫−12[(x+2)−x2]dx\int\limits_{ - 1}^2 {\left[ {\left( {x + 2} \right) - {x^2}} \right]} dx

=

[x22+2x−x33]−12\left[ {{{{x^2}} \over 2} + 2x - {{{x^3}} \over 3}} \right]_{ - 1}^2

=

(2+4−83)−(12−2+13)\left( {2 + 4 - {8 \over 3}} \right) - \left( {{1 \over 2} - 2 + {1 \over 3}} \right)

=

92{{9 \over 2}}
Q32
Let S(α\alpha ) = {(x, y) : y2 ≤ \le x, 0 ≤ \le x ≤ \le α\alpha } and A(α\alpha ) is area of the region S(α\alpha ). If for a λ\lambda , 0 < λ\lambda < 4, A(λ\lambda ) : A(4) = 2 : 5, then λ\lambda equals
A 2(425)132{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}
B 2(25)132{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}
C 4(425)134{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}
D 4(25)134{\left( {{2 \over {5}}} \right)^{{1 \over 3}}}
Correct Answer
Option C
Solution

A(λ\lambda) =

2∫0λxdx2\int\limits_0^\lambda {\sqrt x } dx

=

2[x3232]0λ2\left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^\lambda

=

43λ32{4 \over 3}{\lambda ^{{3 \over 2}}}

∴\therefore A(4) =

43(4)32{4 \over 3}{\left( 4 \right)^{{3 \over 2}}}

Given,

A(λ)A(4)=25{{A\left( \lambda \right)} \over {A\left( 4 \right)}} = {2 \over 5}

⇒\Rightarrow

43(λ)3243(4)32=25{{{4 \over 3}{{\left( \lambda \right)}^{{3 \over 2}}}} \over {{4 \over 3}{{\left( 4 \right)}^{{3 \over 2}}}}} = {2 \over 5}

⇒\Rightarrow

λ32=25×8{\lambda ^{{3 \over 2}}} = {2 \over 5} \times 8

⇒\Rightarrow λ\lambda =

(165)23{\left( {{{16} \over 5}} \right)^{{2 \over 3}}}

⇒\Rightarrow λ\lambda =

(25625)13{\left( {{{256} \over {25}}} \right)^{{1 \over 3}}}

⇒\Rightarrow λ\lambda =

(43×425)13{\left( {{{{4^3} \times 4} \over {25}}} \right)^{{1 \over 3}}}

=

4(425)134{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}
Q33
The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is
A 154{{15} \over 4}
B 152{{15} \over 2}
C 212{{21} \over 2}
D 174{{17} \over 4}
Correct Answer
Option B
Solution

Required area

=∫03(x2+2)dx−12.5.3=9+6−152= \int\limits_0^3 {\left( {{x^2} + 2} \right)dx - {1 \over 2}.5.3 = 9 + 6 - {{15} \over 2}} {}
=152= {{15} \over 2}
Q34
The area (in sq. units) in the first quadrant bounded by the parabola, y = x2 + 1, the tangent to it at the point (2, 5) and the coordinate axes is :
A 83{8 \over 3}
B 143{{14} \over 3}
C 18724{{187} \over {24}}
D 3724{{37} \over {24}}
Correct Answer
Option D
Solution

Area

=∫02(x2+1)dx−12(54)(5)=3724= \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}
Q35
The area (in sq. units) of the region enclosed by the curves y = x2 – 1 and y = 1 – x2 is equal to :
A 83{8 \over 3}
B 43{4 \over 3}
C 72{7 \over 2}
D 163{{16} \over 3}
Correct Answer
Option A
Solution

A =

∫−11((1−x2)−(x2−1))dx\int\limits_{ - 1}^1 {\left( {\left( {1 - {x^2}} \right) - \left( {{x^2} - 1} \right)} \right)dx}

=

∫−11(2−2x2)dx\int\limits_{ - 1}^1 {\left( {2 - 2{x^2}} \right)dx}

=

4∫01(1−x2)dx4\int\limits_0^1 {\left( {1 - {x^2}} \right)dx}

=

4(x−x33)014\left( {x - {{{x^3}} \over 3}} \right)_0^1

=

4(23)4\left( {{2 \over 3}} \right)

=

83{8 \over 3}
Q36
The area (in sq. units) of the region A = {(x, y) : |x| + |y| ≤ \le 1, 2y2 ≥ \ge |x|}
A 16{1 \over 6}
B 56{5 \over 6}
C 13{1 \over 3}
D 76{7 \over 6}
Correct Answer
Option B
Solution

For point of intersection x + y = 1 ⇒\Rightarrow x = 1 – y y2 =

x2{x \over 2}

⇒\Rightarrow 2y2 = x 2y2 = 1 – y ⇒\Rightarrow 2y2 + y – 1 = 0 ⇒\Rightarrow (2y – 1) (y + 1) = 0 ⇒\Rightarrow y =

12{1 \over 2}

or -1 Total area =

4∫012[(1−x)−(x2)]dx4\int\limits_0^{{1 \over 2}} {\left[ {\left( {1 - x} \right) - \left( {\sqrt {{x \over 2}} } \right)} \right]} dx

=

4[x−x22−12x3/23/2]0124\left[ {x - {{{x^2}} \over 2} - {1 \over {\sqrt 2 }}{{{x^{3/2}}} \over {3/2}}} \right]_0^{{1 \over 2}}

=

4[12−18−23(12)3/2]4\left[ {{1 \over 2} - {1 \over 8} - {{\sqrt 2 } \over 3}{{\left( {{1 \over 2}} \right)}^{3/2}}} \right]

= 4 ×\times

524{5 \over {24}}

=

56{5 \over 6}
Q37
The area (in sq. units) of the region A = {(x, y) : (x – 1)[x] ≤ \le y ≤ \le 2x\sqrt x , 0 ≤ \le x ≤ \le 2}, where [t] denotes the greatest integer function, is :
A 832−1{8 \over 3}\sqrt 2 - 1
B 432+1{4 \over 3}\sqrt 2 + 1
C 832−12{8 \over 3}\sqrt 2 - {1 \over 2}
D 432−12{4 \over 3}\sqrt 2 - {1 \over 2}
Correct Answer
Option C
Solution

y = (x – 1)[x] =

{0,0≤x<1x−1,1≤x<22,x=2\left\{ \begin{array}{ll}{0,} & {0 \le x < 1} \\ {x - 1,} & {1 \le x < 2} \\ {2,} & {x = 2} \end{array} \right.

A =

∫022xdx−12.1.1\int\limits_0^2 {2\sqrt x } dx - {1 \over 2}.1.1

= 2

[x3/232]02\left[ {{{{x^{3/2}}} \over {{3 \over 2}}}} \right]_0^2

-

12{1 \over 2}

=

823−12{{8\sqrt 2 } \over 3} - {1 \over 2}
Q38
The area (in sq. units) of the region { (x, y) : 0 ≤ \le y ≤ \le x2 + 1, 0 ≤ \le y ≤ \le x + 1, 12{1 \over 2} ≤ \le x ≤ \le 2 } is :
A 7916{{79} \over {16}}
B 7924{{79} \over {24}}
C 236{{23} \over {6}}
D 2316{{23} \over {16}}
Correct Answer
Option B
Solution
A=∫121(x2+1)dx+A = \int\limits_{{1 \over 2}}^1 {({x^2} + 1)dx + }
∫12\int\limits_1^2 {}
(x+1)dx{(x + 1)dx}

=

[x33+x]121+[x22+x]12{\left[ {{{{x^3}} \over 3} + x} \right]_{{1 \over 2}}^1 + \left[ {{{{x^2}} \over 2} + x} \right]_1^2}
=(43−1324)+(4−32){ = \left( {{4 \over 3} - {{13} \over {24}}} \right) + \left( {4 - {3 \over 2}} \right)}
=1924+52{ = {{19} \over {24}} + {5 \over 2}}
=7924{ = {{79} \over {24}}}
Q39
Given : f(x)={x     ,0≤x<1212    ,x=121−x   ,12<x≤1f(x) = \left\{ \begin{array}{ll}{x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \\ {{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \\ {1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \end{array} \right. and g(x)=(x−12)2,x∈Rg(x) = \left( {x - {1 \over 2}} \right)^2,x \in R Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = 3\sqrt 3 , is :
A 12+34{1 \over 2} + {{\sqrt 3 } \over 4}
B 12−34{1 \over 2} - {{\sqrt 3 } \over 4}
C 13+34{1 \over 3} + {{\sqrt 3 } \over 4}
D 34−13{{\sqrt 3 } \over 4} - {1 \over 3}
Correct Answer
Option D
Solution

Required area = Area of trepezium ABCD – Area of parabola between x =

12{1 \over 2}

and x =

32{{\sqrt 3 } \over 2}

= Area of trepezium ABCD -

∫1232(x−12)2dx\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{\left( {x - {1 \over 2}} \right)}^2}dx}

=

12(32−12)(12+1−32){1 \over 2}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)\left( {{1 \over 2} + 1 - {{\sqrt 3 } \over 2}} \right)

-

13[(x−12)3]1232{1 \over 3}\left[ {{{\left( {x - {1 \over 2}} \right)}^3}} \right]_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}

=

12(3−12)(3−32){1 \over 2}\left( {{{\sqrt 3 - 1} \over 2}} \right)\left( {{{3 - \sqrt 3 } \over 2}} \right)
−13[(3−12)3−0]- {1 \over 3}\left[ {{{\left( {{{\sqrt 3 - 1} \over 2}} \right)}^3} - 0} \right]

=

34−13{{\sqrt 3 } \over 4} - {1 \over 3}
Q40
Let A1 be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = π2{\pi \over 2} in the first quadrant. Then,
A A1:A2=1:2{A_1}:{A_2} = 1:\sqrt 2 and A1+A2=1{A_1} + {A_2} = 1
B A1=A2{A_1} = {A_2} and A1+A2=2{A_1} + {A_2} = \sqrt 2
C 2A1=A22{A_1} = {A_2} and A1+A2=1+2{A_1} + {A_2} = 1 + \sqrt 2
D A1:A2=1:2{A_1}:{A_2} = 1:2 and A1+A2=1{A_1} + {A_2} = 1
Correct Answer
Option A
Solution
A1+A2=∫0π/2cos⁡x. dx=sin⁡x∣0π/2=1{A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1
A1=∫0π/4(cos⁡x−sin⁡x)dx=(sin⁡x+cos⁡x)∣0π/4=2−1{A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1

∴\therefore

A2=1−(2−1)=2−2{A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2

∴\therefore

A1A2=2−12(2−1)=12{{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}
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