datavorous/entrance-exam-dataset huggingface dataset:
The area bounded by the curves y=x2-1 and y=1 is

Options:
A 232+1
B 432-1
C 22-1
D 832-1

The correct answer is:
832-1

Solution:
Plotting the diagram of y=x2-1 and y=1, we get,

Area of the region between the two curves

=2∫021-x2-1dx

=2∫011+x2-1dx+∫121-x2-1dx

=2x3301+22x-x3312

=213+222-223-22-13

=832-1

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The area bounded by the curves y=x2 and y=21 + x2 is

Options:
A ( π− 1 3 ) sq. units
B ( π− 2 3 ) sq. units
C ( 2π−1 ) 3 sq. units
D None of these

The correct answer is:
( π− 2 3 ) sq. units

Solution:
For point of intersection x2=21+x2
â x4+ x2- 2=0
â x2- 1x2 + 2=0
â x=1, -1, ây=1, -1
â´ area=2â«0121+x2-x2 dx
=4tan-1â¡x 0 1- 23x3 0 1
=4 Ï/4-2/3=Ï -2/3

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the graph y=x- 3, above the x-axis from x=-2 to x=3 is (. denotes the greatest integer function)

Options:
A 7Â sq.Â units
B 15 sq. units
C 21Â sq.Â units
D 28Â sq.Â units

The correct answer is:
15 sq. units

Solution:
â´ Required area
=â«-23x- 3dx
=â«-2-1x- 3dx + â«-10x- 3dx
+ â«01x- 3dx + â«12x- 3dx + â«23x- 3dx
=â«-2-15dx + â«-104. dx + â«013.dx + â«122.dx+ â«231.dx
=51 + 41 + 31 + 21 +11
=15 sq. units .

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the lines y=x-1-2 and $y=2$ is _____.

Options:

The correct answer is:
8

Solution:
The area bounded by

y=x-1-2 and y=2

Area =212.4.2=8 sq unit.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The area bounded by the parabola 4y=3x2, the line 2y=3x+12 and the y-axis is

Options:
A 10 sq. units
B 20 sq. units
C 30 sq. units
D 36 sq. units

The correct answer is:
20 sq. units

Solution:
The intersection point is A4,12 by solving 23x24=3x+12
Thus, the required area =∫043x+122-3x24dx
=34x2+6x-x3404
=3416+24-16=12+8=20 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the parabola $y^2=4 x$ and the line $2 x-3 y+4=0$, in square unit, is

Options:
A
$\frac{2}{5}$

B
$\frac{1}{3}$

C
1

D
$\frac{1}{2}$

The correct answer is:
$\frac{1}{3}$

Solution:
Intersecting points are $x=1,4$
$$
\begin{aligned}
& \therefore \text { Required area }=\int_1^4\left[2 \sqrt{x}-\left(\frac{2 x+4}{3}\right)\right] d x \\
& =\left.\frac{2 x^{3 / 2}}{3 / 2}\right|_1 ^4-\left.\frac{2 x^2}{3 \times 2}\right|_1 ^4-\left.\frac{4}{3} x\right|_1 ^4
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4}{3}\left(4^{3 / 2}-1^{3 / 2}\right)-\frac{1}{3}(16-1)\left[\frac{4}{3}(4)-\frac{4}{3}\right] \\
& =\frac{4}{3}(7)-5-4=\frac{28}{3}-9=\frac{28-27}{3}=\frac{1}{3}
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (26 May Online)
The area bounded by the parabola y=x2+x+1, its tangent at P1,3, line x=-1 and the x-axis is A sq units. Then, the value of 6A is equal to

Options:
A a prime number
B a composite number
C an irrational number
D a non-integer

The correct answer is:
a prime number

Solution:
y=x2+x+1

dydx=2x+1
At P, dydx=3
â´ tangent is y-3=3x-1 or y=3x
Hence, the area A=â«-11x2+x+1dx-Area of âOPN
= x 3 3 + x 2 2 +x â1 1 â 1 2 â3â1
=23+2-32
=76 sq. units
â´6A=7 i.e. a prime number

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by x+y=1 and yâ¥x2 in the first quadrant is a-a22-a33 sq. units, then the value of 2a+12 is equal to

Options:

The correct answer is:
5

Solution:
For Aâ1-x=x2
âx2+x-1=0
âx=-1Â±52
âA is -1+52,3-52
Thus, the required area is
Area = â«05-121-x-x2dx
=x-x22-x3305-12
=a-a22-a33
where a=5-12
â2a+1=5
â2a+12=5

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=1-x2 and y=x3-x is divided by y-axis in the ratio Î»:1 âÎ»>1, then Î» is equal to

Options:
A π+1π-1
B ππ-1
C π+1π
D π+2π-2

The correct answer is:
π+1π-1

Solution:
From the figure, λ1=∫011-x2-x3-xdx∫-101-x2-x3-xdx

⇒λ=sin-1x-x44+x2201sin-1x-x44+x22-10=π4-14-12π4--14+12=π+1π-1

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=1x and y=12x-1 from x=1 to x=2 is ln⁡a sq. units, then 3a2 is equal to

Options:
A 12
B 4
C 1
D 14

The correct answer is:
4

Solution:
As 1x>12x-1∀x∈1,2
∴ the required area is A=∫121x-12x-1dx
=ln⁡x12-12ln⁡2x-112
=ln⁡2-12ln⁡3=ln⁡2-ln⁡3
=ln⁡23 sq. units
∴a=23⇒3a2=4

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=fx, x=1, x=4 and the x-axis is 12 square units where fx is a differentiable function such that f1=3 and f4=8. If fx is a bijective function which is always positive, then the area (in square units) bounded by y=f-1x, x=3, x=8 and the x-axis is equal to

Options:
A 29
B 32
C 12
D 17

The correct answer is:
17

Solution:
Since fx>0 ∀x∈1,4
⇒f-1x>0 ∀x∈3,8
Note that the shaded area is the required area and A1=12.

Let the required area be A2
Also, A1+A2=32-3=29

⇒A2=17 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=maxâ¡x2,ââx4,y=1 and the y -axis from x=0 to x=1 is

Options:
A 3  sq. units
B 32  sq. units
C 23  sq. units
D 12 sq. units

The correct answer is:
23  sq. units

Solution:
max⁡x2,x4=x2; x∈0,1
The required area is
A=∫011-x2dx
=x-x3301=1-13=23 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=minâ¡x,2-x with y=x-12 is K sq. units, then K is equal to (where, â is the greatest integer function)

Options:

The correct answer is:
0

Solution:
Area of ΔAOB= 1 2 ⋅2⋅1=1 sq. units
∴ Required area =K<1.
⇒K=0

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by $y=x+1$ and $y=\cos x$ and the $X$ -axis, is

Options:
A 1 sq unit
B $\frac{3}{2}$ sq unit
C $\frac{1}{4}$ sq unit
D $\frac{1}{8}$ sq unit

The correct answers are:
$\frac{3}{2}$ sq unit

Solution:
We have, $y=x+1 \text { and } y=\cos x$

$\begin{aligned} \text { Require area } &=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x \\ &=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2} \\ &=\left[(0)-\left(\frac{1}{2}-1\right)\right]+[1-0] \\ &=\frac{1}{2}+1=\frac{3}{2} \text { sq unit. } \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=x-1 with the x-axis from x=0 to x=1 is k square units, then 4k is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
The required area is
A= 1 2 1⋅1 = 1 2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y+x2≤4x and y≥3 is k sq.units, then 3k is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
Solving, y=-x2+4x and y=3, we get,

x=1,3
∴A is 1,3 and B is 3,3
Hence, the required area is k=∫13-x2+4x-3dx
=-x33+2x2-3x13
=-1327-1+29-1-33-1
=-263+16-6
=43 sq. units
⇒3k=4

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=(x2-x)2 with the x-axis, between its two relative minima, is A sq. units. Then, the value of 15A is equal to

Options:
A 1
B 2
C 12
D 14

The correct answer is:
12

Solution:
The given curve is y=x2(x-1)2
â´ it has relative minima at x=0 and x=1.
Then, the required area is
A=â«01x2x-12dx
=â«01x2-x2dx
=â«01x4+x2-2x3dx
=x55+x33-x4201
=15+13-12
=15-16=130 sq. units

â´15A=12

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by y=xex and the lines x=1, y=0 is

Options:
A 4 sq units
B 6 sq units
C 1 sq unit
D 2 sq units

The correct answer is:
2 sq units

Solution:
Since, x=1
â´ x= Â± 1
∴ y=x e x = x e −x ,   −1<x<0 x e x ,           0<x<1
â´ Required area =â«-10xe-xdx+â«01xexdx
-xe-x-e-x-10+xex-ex01
=1+1=2 sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area common to \( x^{2}+y^{2}=64 \) and \( y^{2}=4 x \) is

Options:
A \( \frac{16}{3}(4 \pi+\sqrt{3}) \)
B \( \frac{16}{3}(8 \pi-\sqrt{3}) \)
C \( \frac{16}{3}(4 \pi-\sqrt{3}) \)
D \( \frac{16}{3}(8 \pi+\sqrt{3}) \)

The correct answer is:
\( \frac{16}{3}(8 \pi-\sqrt{3}) \)

Solution:
The given curves are x2+y2=64...i and y2=4x...ii

Solving both the equations, we get

x2+4x-64=0

x=-2±217

Since the intersection point is lying in the first quadrant, so

x=-2+217

So, the required area is

=2∫0-2+2174xdx+∫0-2+21764-x2dx

On integrating, we get

=1638π-3.

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curve $y=\log _e(x+e)$ and the co-ordinate axes is

Options:
A $3$
B $4$
C $1$
D $2$

The correct answer is:
$1$

Solution:
Required area $=\int_{1-e}^0 \log _e(x+e) d x$
$\left.=\int_1^e \log t d t=[t \log t-t]_1^e=1 \text { sq. unit , (Put } x+e=t\right) \text {. }$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curve $y=\log _e(x+e)$ and the coordinate axes is

Options:
A
1

B
2

C
3

D
4

The correct answer is:
1

Solution:
$$
\begin{aligned}
& \text { Required area }(O A B)=\int_{1-e}^0 \ln (x+e) d x \\
& =\left[x \ln (x+e)-\int \frac{1}{x+e} x d x\right]_0^1=1 .
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2005
The area enclosed between the curve \( y=\log _{e}(x+e) \) and the coordinate-axes is

Options:
A \( 3 \)
B \( 4 \)
C \( 1 \)
D \( 2 \)

The correct answer is:
\( 1 \)

Solution:
Curve : y=logex+e, x-axis, y-axis

at x-axis y=0⇒logex+e=0⇒x+e=1

⇒x=1-e

when x→-e+⇒y→-∞

Required area =∫1-e0logx+e dx

=xlog x+e1-e0-∫1-e01x+ex dx

=0-∫1-e1x+e-ex+edx

=0-x-e logx+e1-e0

=e+1-e-elog1

=1.

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curves y2+4x=4 and y-2x=2 is

Options:
A 253
B 223
C 9
D 233

The correct answer is:
9

Solution:
Given:

y2=-4x+4

y=2x+2

Required area

=∫-424-y24-y-22dy

=∫-422-y24-y2dy

=2y-y312-y24-42

=4-812-1--8+6412-164

=15-6

=9 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The area enclosed between the curves $y^2=x$ and $y=|x|$ is

Options:
A
$2 / 3$

B
$1$

C
$1 / 6$

D
$1 / 3$

The correct answer is:
$1 / 6$

Solution:
$A=\int_0^1(\sqrt{x}-x) d x$
$=\left[\frac{2}{3} x^{3 / 2}-\frac{x^2}{2}\right]_0^1$
$=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2007
The area enclosed between the curves y = ax 2 and x = ay 2 a > 0 is 1 sq unit. If the value of a is 3 3 Î» , then the value of Î» is

Options:

The correct answer is:
9

Solution:
As from the figure, area enclosed between the curves is OABCO.
Thus, the points of intersection of
y = ax 2 and x = ay 2 ,
eliminating y, we get
x=a(ax2)2
⇒x−a3x4=0
⇒x(1−a3x3)=6
⇒x=0 and x=1a
∴ Points of intersection are (0, 0) and 1 a , 1 a .
Thus, required area OABCO = Area of curve OCBDO - area of curve OABDO
⇒ ∫ 0 1 / a x a - ax 2 dx = 1 (given)
⇒ 1 a · · x 3 / 2 3 / 2 - ax 3 3 0 1 / a = 1
⇒ 2 3 a 2 - 1 3 a 2 = 1
⇒ a 2 = 1 3 ⇒a= 1 3          ( ∵a>0 )
1 3 = 3 3 λ
λ=9

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curves $y=x^3$ and $y=\sqrt{x}$ is, (in square units)

Options:
A $\frac{5}{3}$
B $\frac{5}{4}$
C $\frac{5}{12}$
D $\frac{12}{5}$

The correct answer is:
$\frac{5}{12}$

Solution:
Given curves are, $y=x^3$ and $y=\sqrt{x}$
On solving, we get $x=0, x=1$
Therefore, required area $=\int_0^1\left(x^3-\sqrt{x}\right) d x$
$=\left[\frac{x^4}{4}-\frac{2 x \sqrt{x}}{3}\right]_0^1=\left[\frac{1}{4}-\frac{2}{3}\right]=\frac{5}{12},(Area can't be negative).$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curves $y=x^{3}$ and $y=\sqrt{x}$ is, (in square units)

Options:
A $\frac{5}{3}$
B $\frac{5}{4}$
C $\frac{5}{12}$
D $\frac{12}{5}$

The correct answer is:
$\frac{5}{12}$

Solution:
$\begin{aligned} \text { Required area } &=\int_{0}^{1}\left(\sqrt{x}-x^{3}\right) d x \\ &=\left[\frac{2 x^{3 / 2}}{3}-\frac{x^{4}}{4}\right]_{0}^{1} \\ &=\frac{2}{3}-\frac{1}{4}=\frac{5}{12} \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

Options:
A $\frac{4}{3}$
B 1
C $\frac{2}{3}$
D $\frac{8}{3}$

The correct answer is:
$\frac{4}{3}$

Solution:
$A=\int_{-2}^0-x^2-2 x=\frac{4}{3}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The area enclosed between the parabolas $y^2=4 x$ and $x^2=4 y$ is

Options:
A $\frac{14}{3}$sq. unit
B $\frac{3}{4}$ sq. unit
C $\frac{3}{16}$sq. unit
D $\frac{16}{3}$sq. unit

The correct answer is:
$\frac{16}{3}$sq. unit

Solution:
Equations of curves $y^2=4 x$ and $x^2=4 y$. The given equations may be written as $y=2 \sqrt{x}$ and $y=\frac{x^2}{4}$.
We know that area enclosed by the parabolas $=\int_0^4 2 \sqrt{x} d x-\int_0^4 \frac{x^2}{4} d x=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}$ sq. Unit.

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed by the curve y2=x4(1-x2) is

Options:
A π6 sq. units
B π4 sq. units
C π3 sq. units
D π2 sq. units

The correct answer is:
π4 sq. units

Solution:
A=4∫01x21-x21/2dx
Put x=sin⁡θ
⇒ dx=cos⁡θdθ
⇒ A=4∫0π2sin2⁡θcos2θdθ=π4 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area, enclosed by the curves y=sinx+cosx and y=|cosx-sinx| and the lines x=0, x=π2, is :

Options:
A 22(2+1)
B 22(2-1)
C 4(2-1)
D 2(2+1)

The correct answer is:
22(2-1)

Solution:
A=∫0π/2((cosx+sinx)-|cosx-sinx|)dx

A=∫0π/4((cosx+sinx)-(cosx-sinx))dx+∫π/4π/2((cosx+sinx)-(sinx-cosx))dx

A=2∫0π/4sinxdx+2∫π/4π/2cosxdx

=2-12+1+21-12=4-22=22(2-1)

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The area enclosed by the curves $y=x^2, y=x^3$, $x=0$ and $x=p$, where $p>1$, is $1 / 6$. The $\mathrm{p}$ equals

Options:
A
$8 / 3$

B
$16 / 3$

C
2

D
$4 / 3$

The correct answer is:
$4 / 3$

Solution:
Given curves are $y=x^2$ and $y=x^3$ Also, $x=0$ and $x=p, p>1$ Now, intersecting point is $(1,1)$
Required Area
$$
\begin{aligned}
& =\int_0^1\left(x^2-x^3 d x+\int_1^p x^3\right)-x^2 d x \\
& \frac{1}{6}=\frac{x^3}{3}-\left.\frac{x^4}{4}\right|_0 ^1+\frac{x^4}{4}-\left.\frac{x^3}{3}\right|_1 ^p \\
& \Rightarrow \frac{1}{6}=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{p^4}{4}-\frac{p^3}{3}-\frac{1}{4}+\frac{1}{3}\right) \\
& \Rightarrow \frac{1}{6}-\frac{1}{3}+\frac{1}{4}+\frac{1}{4}-\frac{1}{3}=\frac{3 p^4-4 p^3}{12} \\
& \left.\Rightarrow \frac{p^3(3 p-4}{12}\right)=0 \Rightarrow p^3(3 p-4)=0 \\
& \Rightarrow p=0 \text { or } \frac{4}{3} \\
& \text { Since, it is given that } p>1 \\
& \therefore p \text { can not be zero. } \\
& \text { Hence, } p=\frac{4}{3}
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (12 May Online)
The area enclosed by the pair of lines $x y=0$, the line $x-4=0$ and $y+5=0$ is

Options:
A $10 \mathrm{sq}$ units
B 20 sq units
C 0 sq units
D $\frac{5}{4}$ sq units

The correct answer is:
20 sq units

Solution:
We have,
\begin{aligned}
& x y &=0, x-4=0, y+5=0 \\
\Rightarrow \quad & x &=0, y=0, x=4, y=-5
\end{aligned}
So, arca enclosed $=$ Area of rectangle $O A B C$
$=A B \times B C$
$=5 \times 4$
= 20 sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed by the parabolas $y=x^2-1$ and $y=1-x^2$ is

Options:
A $1 / 3$
B $2 / 3$
C $4 / 3$
D $8 / 3$

The correct answer is:
$8 / 3$

Solution:
Given parabolas are $x^2=1+y, x^2=1-y$
Required area = $=4 \int_0^1\left(1-x^2\right) d x$ = $4\left[x-\frac{x^3}{3}\right]_0^1=\frac{8}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed by y2=8x and y=2x that lies outside the triangle formed by y=2x,x=1,y=22, is equal to

Options:
A 1626
B 1126
C 1326
D 526

The correct answer is:
1326

Solution:
Intersection point of y2=8x & y=2x is 0,0 & 2,42

Area of ∆ABC=122·1=12

So required area=area between the curve y2=8x and the line y=2x-ar△ABC

=∫048x-2xdx-12

=3223-82-12=1326

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The area enclosed by $y=3 x-5, y=0, x=3$ and $x=5$ is

Options:
A $12 \mathrm{sq}$ units
B 13 sq units
C $13 \frac{1}{2}$ sq units
D 14 sq units

The correct answer is:
14 sq units

Solution:
$$
\begin{array}{l}
\text { Required area }=\int_{3}^{5}(3 x-5) d x \\
=\left[\frac{3 x^{2}}{2}-5 x\right]_{3}^{5}=\left[\frac{75}{2}-25\right]-\left[\frac{27}{2}-15\right] \\
=\frac{48}{2}-10=14 \text { sq units }
\end{array}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area enclosed (in square units) by the curve y=x4-x2, the x-axis and the vertical lines passing through the two minimum points of the curve is

Options:
A 4825
B 5482
C 7602
D 7302

The correct answer is:
7302

Solution:
Given that y=x4-x2

For minimum points of curve:

ddxx4-x2=0

4x3-2x=0

2x2x2-1=0

⇒2x=0 or 2x2-1=0

x=0 or x=±12

So f is minimum at ±12

The area =∫-1212x4-x2dx

=2∫012x4-x2dx ∵ fx is even function

=2x55-x33120

=2125·15-123·13

=2142·15-122·13

=222110-13

=3-10302

∴ Area =7302

Topic: Mathematics, Area Under Curves, JEE Main
The area formed by triangular shaped region bounded by the curves $y=\sin x, y=\cos x$ and $x=0$ is

Options:
A $\sqrt{2}-1$
B $1$
C $\sqrt{2}$
D $1+\sqrt{2}$

The correct answer is:
$\sqrt{2}-1$

Solution:
Given required area has been shown in the figure.
$x=\frac{\pi}{4}$ is the point of intersection of both curve
$\begin{aligned} \therefore \text { Required area } & =\int_0^{\pi / 4}(\cos x-\sin x) d x \\ & =[\sin x+\cos x]_0^{\pi / 4}=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right] \\ & =\frac{2}{\sqrt{2}}-1=\sqrt{2}-1 .\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq unit) of the region bounded by the curves $2 x=y^2-1$ and $x=0$ is

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C $1$
D $2$

The correct answer is:
$\frac{2}{3}$

Solution:
Given curve can be rewritten as
$$
y^2=2\left(x+\frac{1}{2}\right)
$$
$\therefore$ Required area $=\int_{-1}^1 x d y$
$\begin{aligned} & =2 \int_0^1 \frac{y^2-1}{2} d y \\ & =\left|\left[\frac{y^3}{3}-y\right]_0^1\right| \\ & =\left|\left[\frac{1}{3}-1\right]\right|=\frac{2}{3} \text { sq unit }\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. unit) of the region described by A=x, y : x2+y2≤1 and y2≤1-x is

Options:
A π2-23
B π2+23
C π2+43
D π 2 - 4 3

The correct answer is:
π2+43

Solution:
Area of shaded region = area of semicircle + area of region BCA

=π2+2∫011-x·dx

=π2+2-231-x3/201

=π2-43-1

= π 2 + 4 3 sq. unit

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2014 (06 Apr)
The area (in sq. units) bounded between the curves y=excos⁡x and y=exsin⁡x from x=π4 to x=π2 is

Options:
A eπ2
B eπ42
C eλ2
D eλ2

The correct answer is:
eπ42

Solution:
As sin⁡x>cos⁡x, ∀x∈π4,π2
⇒exsin⁡x>excos⁡x
Thus the required area is
A=∫π4π2exsin⁡x-excos⁡xdx
=-∫π4π2excos⁡x+-sin⁡xdx
=-excos⁡xπ4π2
=− 0− e π 4 ⋅ 1 2
=eπ42
As∫ e x f x + f ′ x dx= e x ⋅f x +C

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded between y=2xln⁡x and y=-x from x=e to x=2e is

Options:
A e2ln⁡16
B e2ln⁡16e3
C e3ln⁡16
D e3ln⁡16e3

The correct answer is:
e2ln⁡16e3

Solution:
As y=2xln⁡x>0
and y=-x<0 ∀x∈e,2e
The required area is A=∫e2e2xln⁡x+xdx
=∫e2edx2ln⁡x
=x2ln⁡xe2e
=4e2ln⁡2e-e2
=e2ln⁡16e3 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded between y=3sin⁡x and y=-4sin3⁡x from x=0 to x=π is

Options:
A 4π
B 34π
C 4
D 343

The correct answer is:
343

Solution:
As 3sin⁡x≥0 and -4sin3⁡x≤0∀x∈0,π,

the required area is A=∫0π3sin⁡x--4sin3⁡xdx
A=∫0π3sin⁡x+(3sin⁡x-sin⁡3x)dx
=∫0π6sin⁡x-sin⁡3xdx
=-6cos⁡x+cos⁡(3x)3π0
=6-13--6+13
=12-23=343 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded between y=6sin⁡x and y+8sin3⁡x=0 from x=0 to x=π is

Options:
A 10π
B 34π3
C 8
D 683

The correct answer is:
683

Solution:
As 6sin⁡x≥0 and -8sin3⁡x≤0 ∀x∈0,π,

The required area is
A=∫0π6sin⁡x--8sin3⁡xdx
A=2∫0π3sin⁡x+3sin⁡x-sin⁡3xdx
=2∫0π6sin⁡x-sin⁡3xdx
=2-6cos⁡x+cos⁡3x30π
=26-13--6+13
=212-23=683 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve exy-2=0 with the x-axis from x=0 to x=lnâ¡2 is

Options:

The correct answer is:
1

Solution:
The required area is A=∫0ln22e-xdx
=-2e-xln20
=-2e-ln2-e0
=-212-1
=1 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve fx=e-xxâ¥1x-1+1ex<1 and the x-axis is

Options:
A 2+2e2
B 2+1e2
C 1e2
D 1+2e2

The correct answer is:
1+2e2

Solution:
âµ Graph of fx is symmetric about x-axis, hence,
Area =2â«0âfxdx
=212Ã1-1e1-1e+12Ã1eÃ1e+â«1âe-xdx
=2121-2e+2e2+-e-x1â
=1-2e+2e2+2e=1+2e2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve fx=maxx-1,1-x with the x-axis from x=-1 to x=1 is

Options:

The correct answer is:
1

Solution:
The required area =1221=1

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve |y|=ln⁡x and the coordinate axes is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
Curve y=lnâ¡x is
Required Area =4Ã|â«01(lnâ¡x)|dx
=4Ãxlnâ¡x-x01
=4-1=4 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve y=max. x3, x4 and the x-axis from x=0 to x=1 is

Options:

The correct answer is:
0.25

Solution:
maxâ¡x3,x4=x3 âxâ0,1
â´ Required area =â«01x3dx=x4401=14=0.25 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve y=maxx,sin⁡x, ∀x∈0,2π is

Options:
A π2
B 2π2
C 2
D 4

The correct answer is:
2π2

Solution:
As sinx<x, ∀x∈0,2π
⇒ Required area =∫02πx⋅dx
=x2202π
=4π22=2π2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve y=x:xâ0,12-x:xâ1,2 with the x-axis from x=0 to x=2 is

Options:
A 2
B 12
C 1
D 4

The correct answer is:
1

Solution:
The required area =12⋅2⋅1=1 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curve y=x-Ï+x-e, the ordinates at its points of non-differentiability and the x-axis is

Options:
A π+2e
B 2π+e
C π-e2
D π2-e2

The correct answer is:
π-e2

Solution:
The points of non-differentiability are x=π and x=e
Hence, required area =π-eπ-e
=π-e2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq units) bounded by the curves $x=-2 y^2$ and $x=1-3 y^2$ is

Options:
A $\frac{2}{3}$
B $1$
C $\frac{2}{3}$
D $\frac{2}{3}$

The correct answer is:
$\frac{2}{3}$

Solution:
Given curves,
$$
\text { and } \quad \begin{aligned}
& x=-2 y^2 \\
& x=1-3 y^2
\end{aligned}
$$
On solving both curves, we get
$$
\begin{gathered}
-2 y^2=1-3 y^2 \\
\Rightarrow \quad y= \pm 1 \text { and } x=-2
\end{gathered}
$$
So, the intersection points are $(-2,1)$ and $(-2,-1)$.
$\therefore$ Required area
$$
\begin{aligned}
& =2 \int_0^1\left\{\left(1-3 y^2\right)-\left(-2 y^2\right)\right\} d y \\
& =2 \int_0^1\left(1-y^2\right) d y \\
& =2\left(y-\frac{y^3}{3}\right)_0^1=2\left(1-\frac{1}{3}\right) \\
& =2 \times \frac{2}{3}=\frac{4}{3}
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq units) bounded by the curves
$y^2=4 x$ and $x^2=4 y$ is

Options:
A $\frac{64}{3}$
B $\frac{16}{3}$
C $\frac{8}{3}$
D $\frac{2}{3}$

The correct answer is:
$\frac{16}{3}$

Solution:
Given curves are $y^2=4 x$ and $x^2=4 y$.
The intersection point is
$x^4=16 y^2=16(4 x)$
$\begin{array}{rlrl}\Rightarrow & x\left(x^3-64\right) & =0 \\ \Rightarrow & & x & =0,4 \\ \Rightarrow & & y & =0,4\end{array}$
Hence, intersection point is $O(0,0)$ and $A(4,4)$ $\therefore$ Required area $=$ shaded region of the curve
$\begin{aligned} & =\int\left(y_2-y_1\right) d x \\ & =\int_0^4\left(\sqrt{4 x}-\frac{x^2}{4}\right) d x \\ & =\left[\frac{2 x^{3 / 2}}{3 / 2}-\frac{x^3}{12}\right]_0^4 \\ & =\left[\frac{4}{3}(4)^{3 / 2}-\frac{(4)^3}{12}-(0-0)\right] \\ & =\left(\frac{32}{3}-\frac{64}{12}\right)=\frac{128-64}{12} \\ & =\frac{64}{12}=\frac{16}{3} \text { sq units }\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by the curves $y=\frac{8}{x}$, $y=2 x$ and $x=4$ is

Options:
A $12-8 \log 2$
B $12+8 \log 2$
C $12-8 \log 4$
D $12+8 \log 4$

The correct answer is:
$12-8 \log 2$

Solution:
The shaded region is the required area bounded by the curves
$\begin{aligned} & \text { Area }=\int_2^4\left(2 x-\frac{8}{x}\right) d x \\ & \text { Area }=\left[x^2-8 \log x\right]_2^4=[16-8 \log 4-4+8 \log 2]\end{aligned}$
$\begin{aligned} & =12-8 \log \left(\frac{4}{2}\right) \\ & =12-8 \log (2)\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main