Practice Start Test

Area Under Curves

JEE Mathematics · 107 questions · Page 1 of 11 · Click an option or "Show Solution" to reveal answer

Q1
The area of the region enclosed between the parabolas y2 = 2x −- 1 and y2 = 4x −- 3 is
A 13{1 \over {3}}
B 16{1 \over {6}}
C 23{2 \over {3}}
D 34{3 \over {4}}
Correct Answer
Option A
Solution

Area of the shaded region

=2∫01(y2+34−y2+12)dy= 2\int_0^1 {\left( {{{{y^2} + 3} \over 4} - {{{y^2} + 1} \over 2}} \right)dy}
=2∫01(14−y24)dy= 2\int_0^1 {\left( {{1 \over 4} - {{{y^2}} \over 4}} \right)dy}
=2[14−112]=13= 2\left[ {{1 \over 4} - {1 \over {12}}} \right] = {1 \over 3}
Q2
The area of the region {(x,y):∣x−y∣≤y≤4x}\{(x, y):|x-y| \leq y \leq 4 \sqrt{x}\} is
A 5123\dfrac{512}{3}
B 20483\dfrac{2048}{3}
C 512
D 10243\dfrac{1024}{3}
Correct Answer
Option D
Solution

∣x−y∣≤y≤4x|x-y| \leq y \leq 4 \sqrt{x}

 Area =∫064(4x−x2)dx=4x3/232−x24]064\begin{aligned} & \text{ Area }=\int_0^{64}\left(4 \sqrt{x}-\frac{x}{2}\right) d x \\ & \left.=\frac{4 x^{3 / 2}}{\frac{3}{2}}-\frac{x^2}{4}\right]_0^{64} \end{aligned}
Q3
Area (in sq. units) of the region outside ∣x∣2+∣y∣3=1{{\left| x \right|} \over 2} + {{\left| y \right|} \over 3} = 1 and inside the ellipse x24+y29=1{{{x^2}} \over 4} + {{{y^2}} \over 9} = 1 is :
A 6(4−π)6\left( {4 - \pi } \right)
B 3(4−π)3\left( {4 - \pi } \right)
C 6(π−2)6\left( {\pi - 2} \right)
D 3(π−2)3\left( {\pi - 2} \right)
Correct Answer
Option C
Solution

Area of Ellipse = π\piab = 6π\pi ∴\therefore Required area = Area of ellipse – 4 (Area of triangle OAB) = 6π\pi -

4(12×2×3)4\left( {{1 \over 2} \times 2 \times 3} \right)

= 6π\pi - 12 =

6(π−2)6\left( {\pi - 2} \right)

sq.units

Q4
The area (in sq. units) of the region, given by the set {(x,y)∈R×R∣x≥0,2x2≤y≤4−2x}\{ (x,y) \in R \times R|x \ge 0,2{x^2} \le y \le 4 - 2x\} is :
A 83{8 \over 3}
B 173{{17} \over 3}
C 133{{13} \over 3}
D 73{7 \over 3}
Correct Answer
Option D
Solution

Required area =

∫01(4−2x−2x2)dx=4x−x2−2x33∣01\left. {\int\limits_0^1 {\left( {4 - 2x - 2{x^2}} \right)dx = 4x - {x^2} - {{2{x^3}} \over 3}} } \right|_0^1
=4−1−23=73= 4 - 1 - {2 \over 3} = {7 \over 3}
Q5
The area of the region bounded by y −- x = 2 and x2 = y is equal to :
A 163{{16} \over 3}
B 23{{2} \over 3}
C 92{{9} \over 2}
D 43{{4} \over 3}
Correct Answer
Option C
Solution

y −- x = 2, x2 = y Now, x2 = 2 + x ⇒\Rightarrow x2 −- x −- 2 = 0 ⇒\Rightarrow (x + 1)(x −- 2) = 0 Area =

∫−12(2+x−x2)\int\limits_{ - 1}^2 {(2 + x - {x^2})}
=∣2x+x22−x33∣−12= \left| {2x + {{{x^2}} \over 2} - {{{x^3}} \over 3}} \right|_{ - 1}^2
=(4+2−83)−(−2+12+13)= \left( {4 + 2 - {8 \over 3}} \right) - \left( { - 2 + {1 \over 2} + {1 \over 3}} \right)
=6−3+2−12=92= 6 - 3 + 2 - {1 \over 2} = {9 \over 2}
Q6
Let the area of the region (x,y):2y≤x2+3, y+∣x∣≤3, y≥∣x−1∣ (x, y) : 2y \leq x^2 + 3,\ y + |x| \leq 3, \ y \geq |x - 1| be A A . Then 6A 6A is equal to :
A 14
B 18
C 16
D 12
Correct Answer
Option A
Solution

A⇒A \Rightarrow Rectangle ABDE −- Area of region EDC

A⇒4−2∫01(3−x)−(x2+32)dxA⇒4−2{3x−x22−x36−32x}01A⇒4−2{3−12−16−32}=73\begin{aligned} & A \Rightarrow 4-2 \int_0^1(3-x)-\left(\frac{x^2+3}{2}\right) d x \\ & A \Rightarrow 4-2\left\{3 x-\frac{x^2}{2}-\frac{x^3}{6}-\frac{3}{2} x\right\}_0^1 \\ & A \Rightarrow 4-2\left\{3-\frac{1}{2}-\frac{1}{6}-\frac{3}{2}\right\}=\frac{7}{3} \end{aligned}

So 6 A=146 \mathrm{~A}=14

Q7
The area enclosed between the curves y2=x{y^2} = x and y=∣x∣y = \left| x \right| is :
A 1/61/6
B 1/31/3
C 2/32/3
D 11
Correct Answer
Option A
Solution

The area enclosed between the curves

y2=x{y^2} = x

and

y=∣x∣y = \left| x \right|

From the figure, area lies between

y2=x{y^2} = x

and

y=xy = x

∴\therefore Required area

=∫01(y2−y1)dx= \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx
=∫01(x−x)dx=[x3/23/2−x22]01= \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1

∴\therefore Required area

=23[x3/2]01−12[x2]01= {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1
=23−12=16= {2 \over 3} - {1 \over 2} = {1 \over 6}
Q8
The area bounded by the curves y=ln⁡x,y=ln⁡∣x∣,y=∣ln⁡ x∣y = \ln x,y = \ln \left| x \right|,y = \left| {\ln {\mkern 1mu} x} \right| and y=∣ln⁡∣x∣∣y = \left| {\ln \left| x \right|} \right| is :
A 44sq. units
B 66sq. units
C 1010sq. units
D none of these
Correct Answer
Option A
Solution

First we draw each curve as separate graph NOTE : Graph of

y=∣f(x)∣y = \left| {f\left( x \right)} \right|

can be obtained from the graph of the curve

y=f(x)y = f\left( x \right)

by drawing the mirror image of the portion of the graph below

xx

-axis, with respect to

xx

-axis. Clearly the bounded area is as shown in the following figure. Required area

=4∫01(−ln⁡x)dx= 4\int\limits_0^1 {\left( { - \ln x} \right)} dx
=−4[x ln⁡x+−x01]=4  = - 4\left[ {x\,\ln x + - x_0^1} \right] = 4\,\,

sq. units

Q9
The area of the region bounded by the curves y=∣x−1∣y = \left| {x - 1} \right| and y=3−∣x∣y = 3 - \left| x \right| is :
A 66 sq. units
B 22 sq. units
C 33 sq. units
D 44 sq. units
Correct Answer
Option D
Solution
A=∫−10{(3+x)−(−x+1)}dx+A = \int\limits_{ - 1}^0 {\left\{ {\left( {3 + x} \right) - \left( { - x + 1} \right)} \right\}dx + }
            \,\,\,\,\,\,\,\,\,\,\,\,
∫01{(3−x)−(−x+1)}dx+\int\limits_0^1 {\left\{ {\left( {3 - x} \right) - \left( { - x + 1} \right)} \right\}dx + }
            \,\,\,\,\,\,\,\,\,\,\,\,
∫12{(3−x)−(x−1)}dx\int\limits_1^2 {\left\{ {\left( {3 - x} \right) - \left( {x - 1} \right)} \right\}dx}
=∫−10(2+2x)dx+∫012dx+∫12(4−2x)dx= \int\limits_{ - 1}^0 {\left( {2 + 2x} \right)dx + \int\limits_0^1 {2dx + \int\limits_1^2 {\left( {4 - 2x} \right)dx} } }
=[2x−x2]−10+[2x]01+[4x−x2]12= \left[ {2x - {x^2}} \right]_{ - 1}^0 + \left[ {2x} \right]_0^1 + \left[ {4x - {x^2}} \right]_1^2
=0−(−2+1)+(2−0)+(8−4)−(4−1)= 0 - \left( { - 2 + 1} \right) + \left( {2 - 0} \right) + \left( {8 - 4} \right) - \left( {4 - 1} \right)
=1+2+4−3=4= 1 + 2 + 4 - 3 = 4

sq. units

Q10
The area of the region bounded by the curves y=∣x−2∣,x=1,x=3y = \left| {x - 2} \right|,x = 1,x = 3 and the xx-axis is :
A 44
B 22
C 33
D 11
Correct Answer
Option D
Solution

The required area is shown by shaded region

A=∫13∣x−2∣dx=2∫23(x−2)dxA = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx
=2[x22−2x]23=1= 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1
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