Practice Start Test

Area Under Curves

JEE Mathematics · 107 questions · Page 2 of 11 · Click an option or "Show Solution" to reveal answer

Q11
The area enclosed between the curve y=log⁡e(x+e)y = {\log _e}\left( {x + e} \right) and the coordinate axes is :
A 11
B 22
C 33
D 44
Correct Answer
Option A
Solution

The graph of the curve

y=log⁡e(x+e)y = {\log _e}\left( {x + e} \right)

is as shown in the fig. Required area

A=∫1−e0ydx=∫1−e0log⁡e(x+e)dxA = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx

put

x+e=t⇒dx=dtx + e = t \Rightarrow dx = dt

also At

x=1−e,t=1x = 1 - e,t = 1

At

x=0,  t=ex = 0,\,\,t = e

∴\therefore

A=∫1elog⁡e tdt=[t log⁡et−t1e]A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]
e−e−0+1=1e - e - 0 + 1 = 1

Hence the required area is

11

square unit.

Q12
The parabolas y2=4x{y^2} = 4x and x2=4y{x^2} = 4y divide the square region bounded by the lines x=4,x=4, y=4y=4 and the coordinate axes. If S1,S2,S3{S_1},{S_2},{S_3} are respectively the areas of these parts numbered from top to bottom ; then S1,S2,S3{S_1},{S_2},{S_3} is :
A 1:2:11:2:1
B 1:2:31:2:3
C 2:1:22:1:2
D 1:1:11:1:1
Correct Answer
Option D
Solution

Intersection points of

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

and

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

are

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

and

(4,4).\left( {4,4} \right).

The graph is as shown in the figure. By symmetry, we observe

S1=S3=∫04ydx=∫04x24dx{S_1} = {S_3} = \int\limits_0^4 {ydx = \int\limits_0^4 {{{{x^2}} \over 4}dx} }
=[x312]04=163= \left[ {{{{x^3}} \over {12}}} \right]_0^4 = {{16} \over 3}

sq. unit Also

S2=∫04(2x−x24)dx{S_2} = \int\limits_0^4 {\left( {2\sqrt x - {{{x^2}} \over 4}} \right)} dx
            \,\,\,\,\,\,\,\,\,\,\,\,
=[2x3232−x312]04= \left[ {{{2{x^{{3 \over 2}}}} \over {{3 \over 2}}} - {{{x^3}} \over {12}}} \right]_0^4
            \,\,\,\,\,\,\,\,\,\,\,\,
=43×8−163=163= {4 \over 3} \times 8 - {{16} \over 3} = {{16} \over 3}

∴\therefore

S1:S2:S3=1:1:1{S_1}:{S_2}:{S_3} = 1:1:1
Q13
Let f(x)f(x) be a non - negative continuous function such that the area bounded by the curve y=f(x),y=f(x), xx-axis and the ordinates x=π4x = {\pi \over 4} and x=β>π4x = \beta > {\pi \over 4} is (βsin⁡β+π4cos⁡β+2β).\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right). Then f(π2)f\left( {{\pi \over 2}} \right) is
A (π4+2−1)\left( {{\pi \over 4} + \sqrt 2 - 1} \right)
B (π4−2+1)\left( {{\pi \over 4} - \sqrt 2 + 1} \right)
C (1−π4−2)\left( {1 - {\pi \over 4} - \sqrt 2 } \right)
D (1−π4+2)\left( {1 - {\pi \over 4} + \sqrt 2 } \right)
Correct Answer
Option D
Solution

Given that

∫π/4βf(x)dx=βsin⁡β+π4cos⁡ β+2β\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta

Differentiating

w.r.tw.r.t

β\beta

f(β)=βcos⁡β+sin⁡β−π4sin⁡β+2f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2
f(π2)=(1−π4)sin⁡π2+2f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2
=1−π4+2= 1 - {\pi \over 4} + \sqrt 2
Q14
The area of the plane region bounded by the curves x+2y2=0x + 2{y^2} = 0 and  x+3y2=1\,x + 3{y^2} = 1 is equal to :
A 53{5 \over 3}
B 13{1 \over 3}
C 23{2 \over 3}
D 43{4 \over 3}
Correct Answer
Option D
Solution
x+2y2=0⇒y2=−x2x + 2{y^2} = 0 \Rightarrow {y^2} = - {x \over 2}
[\left[ {} \right.

Left handed parabola with vertex at

(0,0)\left( {0,0} \right)
]\left. {} \right]
x+3y2=1⇒y2=−13(x−1)x + 3{y^2} = 1 \Rightarrow {y^2} = - {1 \over 3}\left( {x - 1} \right)
[\left[ {} \right.

Left handed parabola with vertex at

(1,0)\left( {1,0} \right)
]\left. {} \right]

Solving the two equations we get the points of intersection as

(−2,1),(−2,−1)\left( { - 2,1} \right),\left( { - 2, - 1} \right)

The required area is

ACBDA,ACBDA,

given by

=∣∫−11(1−3y2−2y2)dy∣= \left| {\int\limits_{ - 1}^1 {\left( {1 - 3{y^2} - 2{y^2}} \right)dy} } \right|
=∣[y−5y33]−11∣= \left| {\left[ {y - {{5{y^3}} \over 3}} \right]_{ - 1}^1} \right|
=∣(1−53)−(−1+53)∣= \left| {\left( {1 - {5 \over 3}} \right) - \left( { - 1 + {5 \over 3}} \right)} \right|
=2×23=43  = 2 \times {2 \over 3} = {4 \over 3}\,\,

sq. units.

Q15
The area of the region bounded by the parabola (y−2)2=x−1,{\left( {y - 2} \right)^2} = x - 1, the tangent of the parabola at the point (2,3)(2, 3) and the xx-axis is :
A 66
B 99
C 1212
D 33
Correct Answer
Option B
Solution

The given parabola is

(y−2)2=x−1{\left( {y - 2} \right)^2} = x - 1

Vertex

(1,2)\left( {1,2} \right)

and it meets

xx

-axis at

(5,0)\left( {5,0} \right)

Also it gives

y2−4y−x+5=0{y^2} - 4y - x + 5 = 0

So, that equation of tangent to the parabola at

(2,3)\left( {2,3} \right)

is

y.3−2(y+3)−12(x+2)+5=0y.3 - 2\left( {y + 3} \right) - {1 \over 2}\left( {x + 2} \right) + 5 = 0

or

x−2y+4=0x - 2y + 4 = 0

which meets

xx

-axis at

(−4,0).\left( { - 4,0} \right).

In the figure shaded area is the required area. Let us draw

PDPD

perpendicular to

yy

-axis. Then required area

=Ar  ΔBOA+Ar  (OCPD) − Ar(ΔAPD)= Ar\,\,\Delta BOA + Ar\,\,\left( {OCPD} \right)\, - \,Ar\left( {\Delta APD} \right)
=12×4×2+∫03xdy−12×2×1= {1 \over 2} \times 4 \times 2 + \int_0^3 {xdy - {1 \over 2}} \times 2 \times 1
=3+∫03(y−2)2+1 dy= 3 + \int_0^3 {{{\left( {y - 2} \right)}^2} + 1\,dy}
=3+[(y−2)33+y]03= 3 + \left[ {{{{{\left( {y - 2} \right)}^3}} \over 3} + y} \right]_0^3
=3+[13+3+83]=3+6=9= 3 + \left[ {{1 \over 3} + 3 + {8 \over 3}} \right] = 3 + 6 = 9

Sq. units

Q16
The area bounded by the curves y=cos⁡xy = \cos x and y=sin⁡xy = \sin x between the ordinates x=0x=0 and x=3π2x = {{3\pi } \over 2} is
A 42+24\sqrt 2 + 2
B 42−14\sqrt 2 - 1
C 42+14\sqrt 2 + 1
D 42−24\sqrt 2 - 2
Correct Answer
Option D
Solution

∴\therefore Required area

=[∫0π4(cos⁡ x−sin⁡x)dx+= \left[ {\int\limits_0^{{\pi \over 4}} {\left( {\cos \,x - \sin x} \right)dx + } } \right.
             ∫π45π4(sin⁡x−cos⁡x)dx+\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{\pi \over 4}}^{{5\pi \over 4}} {\left( {\sin x - \cos x} \right)dx + }
             ∫5π43π2(cos⁡x−sin⁡x)dx\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{{5\pi \over 4}}^{{3\pi \over 2}} {\left( {\cos x - \sin x} \right)dx }
=42−2= 4\sqrt 2 - 2
Q17
The area of the region enclosed by the curves y=x,x=e,y=1xy = x,x = e,y = {1 \over x} and the positive xx-axis is :
A 11 square unit
B 32{3 \over 2} square units
C 52{5 \over 2} square units
D 12{1 \over 2} square unit
Correct Answer
Option B
Solution

Area of required region

AOCBAOCB
=∫01xdx+∫1e1xdx=12+1=32= \int\limits_0^1 {xdx} + \int\limits_1^e {{1 \over x}dx = {1 \over 2}} + 1 = {3 \over 2}

sq. units

Q18
The area between the parabolas x2=y4{x^2} = {y \over 4} and x2=9y{x^2} = 9y and the straight line y=2y=2 is :
A 20220\sqrt 2
B 1023{{10\sqrt 2 } \over 3}
C 2023{{20\sqrt 2 } \over 3}
D 10210\sqrt 2
Correct Answer
Option C
Solution

Given curves

x2=y4{x^2} = {y \over 4}

and

x2=9y{x^2} = 9y

are the parabolas whose equations can be written as

y=4x2y = 4{x^2}

and

y=19x2.y = {1 \over 9}{x^2}.

Also, given

y=2.y=2.

Now, shaded portion shows the required area which is symmetric. ∴\therefore Area

=2∫02(9y−y4)dy= 2\int\limits_0^2 {\left( {\sqrt {9y} - \sqrt {{y \over 4}} } \right)} dy

Area

=2∫02(3y−y2)dy= 2\int\limits_0^2 {\left( {3\sqrt y - {{\sqrt y } \over 2}} \right)} dy
=2[23×3.y32−12×23.y32]02= 2\left[ {{2 \over 3} \times 3.{y^{{3 \over 2}}} - {1 \over 2} \times {2 \over 3}.{y^{{3 \over 2}}}} \right]_0^2
=2[2y32−13y32]=2×53y32∣02= 2\left[ {2{y^{{3 \over 2}}} - {1 \over 3}{y^{{3 \over 2}}}} \right] = \left. {2 \times {5 \over 3}{y^{{3 \over 2}}}} \right|_0^2
=2.5322=2023= 2.{5 \over 3}2\sqrt 2 = {{20\sqrt 2 } \over 3}
Q19
The area (in square units) bounded by the curves y=x,y = \sqrt {x,} 2y−x+3=0,2y - x + 3 = 0, xx-axis, and lying in the first quadrant is :
A 99
B 3636
C 1818
D 274{{27} \over 4}
Correct Answer
Option A
Solution

Given curves are

y=xy = \sqrt x
                                    ...(1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

and

2y−x+3=0         ...(2)2y - x + 3 = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)

On solving both we get

y=−1,3y=-1,3

Required area

=∫03{(2y+3)−y2}dy= \int\limits_0^3 {\left\{ {\left( {2y + 3} \right) - {y^2}} \right\}} dy
                                  =y2+3y−y33∣03=9.\left. {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {y^2} + 3y - {{{y^3}} \over 3}} \right|_0^3 = 9.
Q20
The area of the region described by A={(x,y):x2+y2≤1A = \left\{ {\left( {x,y} \right):{x^2} + {y^2} \le 1} \right. and y2≤1−x}\left. {{y^2} \le 1 - x} \right\} is :
A π2−23{\pi \over 2} - {2 \over 3}
B π2+23{\pi \over 2} + {2 \over 3}
C π2+43{\pi \over 2} + {4 \over 3}
D π2−43{\pi \over 2} - {4 \over 3}
Correct Answer
Option C
Solution

Given curves are

x2+y2=1{x^2} + {y^2} = 1

and

y2=1−x.{y^2} = 1 - x.

Intersection points are

x=0,1x = 0,1

Area of shaded portion is the required area. So, Required Area == Area of semi-circle ++ Area bounded by parabola

=πr22+2∫011−xdx= {{\pi {r^2}} \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} dx}
=π2+2∫011−x dx= {\pi \over 2} + 2\int\limits_0^1 {\sqrt {1 - x} \,dx}

( As radius of circle

=1=1

)

=π2+2[(1−x)3/2−3/2]01= {\pi \over 2} + 2\left[ {{{{{\left( {1 - x} \right)}^{{{3}/{2}}}}} \over { - {{3}/{2}}}}} \right]_0^1
=π2−43(−1)=π2+43= {\pi \over 2} - {4 \over 3}\left( { - 1} \right) = {\pi \over 2} + {4 \over 3}

Sq. unit

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