datavorous/entrance-exam-dataset huggingface dataset:
The area (in sq. units) bounded by the parabola y=x2-1, the tangent at the point 2,3 to it and the y-axis is

Options:
A 143
B 83
C 323
D 563

The correct answer is:
83

Solution:
We have, dydx=2x

∴ Slope of tangent at 2,3 is

m=4

∴ Equation of tangent is y-3=4x-2

∴ Point T≡0,-5

∴ Area of ΔPTM=12×8×2=8

Area of curve PAMP=∫-13y+1dy

=23y+132-13=163

∴ Required shaded area =8-163=83

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The area (in sq. units) bounded by x2+y2=1 and the curve y2â¥x2, above the x-axis is

Options:
A 14
B π4
C 16
D π6

The correct answer is:
π4

Solution:
The required area is
A=2â«0121-x2-xdx
âA=2x1-x22+12sin-1â¡x-x221/20
=212212+12Ï4-14
=Ï4 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by x+y=2 in the first and third quadrant is (where, . is the greatest integer function),

Options:
A 4
B 3
C 6
D 10

The correct answer is:
6

Solution:
x+y=2 (Let x,yâ¥0)
âx,y=2,0 or 1,1 or 0,2
If x=2 and y=0
âxâ2,3 and yâ0,1 and so on
Since, the curve is symmetric in the Ist and IIIrd quadrant

Hence, the required area =61Ã1=6 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by x+y=2 with the co-ordinate axes is equal to (where, . denotes the greatest integer function)

Options:

The correct answer is:
2.5

Solution:
As, x+y=2âx+yâ2,3
The required area =12â3â3-12â2â2
=9-42=52 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=1x and y=12x-1 between x=1 and x=2 is

Options:
A ln⁡23
B ln⁡23
C ln⁡23
D ln⁡2

The correct answer is:
ln⁡23

Solution:
Required area =∫121x-12x-1dx
=ln⁡x-12ln⁡2x-112
=ln⁡2-12ln⁡3-ln⁡1-12ln⁡1 =ln⁡23 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=2-x-2 and the x-axis is

Options:

The correct answer is:
4

Solution:
Required area =12⋅4.2=4sq.units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=2x and y=2x-x2 from x=1 to x=2 is klog2e-l, then the value of kl is equal to

Options:

The correct answer is:
3

Solution:
Required area is
A=â«122x-2x-x2dx
=2xlnâ¡2-x2+x3312
=4lnâ¡2-4+83-2lnâ¡2-1+13
=2lnâ¡2-3+73
=2lnâ¡2-23=2log2e-23
â´k=2 & l=23
âkl=3

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=ex:x≥0e-x:x≤0 with the x-axis from x=-1 to x=1 is

Options:
A e
B 2e
C 2e-2
D 2e+2

The correct answer is:
2e-2

Solution:
Required area =∫-10e-xdx+∫01exdx
=-e-x-10+ex01
=-1+e+e-1
=2e-2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=ln⁡x from x=e to x=π with the x-axis is

Options:
A ln⁡π
B πln⁡π
C eln⁡π
D πln⁡π-π

The correct answer is:
πln⁡π-π

Solution:
Required area is =∫eπln⁡xdx
=∫eπln⁡x dx
=xln⁡x-xeπ
=πln⁡π-π-eln⁡e-e
=πln⁡π-π

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=lnâ¡x, y=xe and the y-axis is equal to

Options:
A e2-1
B e2
C 5e2
D 3e2-1

The correct answer is:
e2

Solution:
Note that y=xe is a tangent to y=lnâ¡x at e,1
Required Area =â«0exe-lnâ¡xdx=x22e-xlnâ¡x+x0e

=e2-e+e-0=e2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=maxâ¡sin2â¡x,sin4â¡x, xâ0,Ï2 with the x-axis, from x=0 to x=Ï2 is

Options:
A π
B π2
C π4
D π6

The correct answer is:
π4

Solution:
As xâ0,Ï2, sinâ¡xâ0,1
âsin2â¡x>sin4â¡x
So, the required area is
A=â«0Ï2sin2â¡xdx
=â«0Ï21-cosâ¡2x2dx
=x2-sinâ¡2x40Ï2
=Ï4-0-0-0
=Ï4 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) bounded by y=xx and the line y=x is equal to

Options:
A 1
B 2
C 12
D 13

The correct answer is:
13

Solution:
xx=x2x≥0-x2 x<0
∴ Required area =2∫01x-x2dx
=2x22-x3301
=212-13
=13 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) covered by x-y=-3 with the coordinate axes is (where, ⋅ is the greatest integer function)

Options:
A 2
B 4
C 52
D 114

The correct answer is:
52

Solution:
As, x-y=-3
âx-yâ-3,-2
Hence, the required area is 1232-22=52 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) enclosed between the curve x=1-t21+t2, y=2t1+t2, ∀t∈R and the line y=x+1 above the line is

Options:
A π4
B 12
C 3π4+12
D π4-12

The correct answer is:
π4-12

Solution:
Let, t=tan⁡θ
⇒x=cos⁡2θ, y=sin⁡2θ
⇒x2+y2=1
Area of the shaded region =π4-12×1×1=π4-12 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) enclosed by the graphs of |x+y|=2 and |x|=1 is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
8

Solution:
Given equation | x+y |=2 represent two parallel lines x+y=2 and x+y=â2 .
Also, the equation | x |=1 represent two parallel lines x=1 and x=â1
Hence, the four lines combined together represent a parallelogram.

Hence, required area = 2 1 + 3 2 Â· 2 = 8

Topic: Mathematics, Straight Lines, JEE Main
The area (in sq. units) in the first quadrant bounded by the parabola, $y=x^{2}+1$, the tangent to it at the point (2,5) and the coordinate axes is :

Options:
A $\frac{8}{3}$
B $\frac{37}{24}$
C $\frac{187}{24}$
D $\frac{14}{3}$

The correct answer is:
$\frac{37}{24}$

Solution:
The equation of parabola $x^{2}=y-1$

The equation of tangent at (2,5) to parabola is

$y-5=\left(\frac{d y}{d x}\right)_{(2,5)}(x-2)$

$y-5=4(x-2)$

$4 x-y=3$

Then, the required area

$=\int_{0}^{2}\left\{\left(x^{2}+1\right)-(4 x-3)\right\} d x-$ Area of $\Delta A O D$

$=\int_{0}^{2}\left(x^{2}-4 x+4\right) d x-\frac{1}{2} \times \frac{3}{4} \times 3$

$=\left[\frac{(x-2)^{3}}{3}\right]_{0}^{2}-\frac{9}{8}=\frac{37}{24}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The area (in sq. units) of an equilateral triangle inscribed in the parabola y2=8x, with one of its vertices on the vertex of this parabola is

Options:
A 643
B 2563
C 1923
D 1283

The correct answer is:
1923

Solution:
y2=8x,a=2

A≡2t2,2(2)t≡2t2,4t

tan30°=4t2t2=2t=13

t=23

Area of ΔOAB=2·ΔOMA=2·12·2t2(4t)=8t3=8(23)3=1923

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The area (in sq. units) of the circle touching the line x+y=4 at 1,3 and intersecting x2+y2=4 orthogonally is equal to

Options:
A 9π8
B 7π8
C 5π4
D 4π3

The correct answer is:
9π8

Solution:
Let equation of the circle is x-12+y-32+αx+y-4=0
⇒x2+y2+α-2x+α-6y+10-4α=0
Which is orthogonal to x2+y2=4
Now, 2g1g2+2f1f2=c1+c2

⇒0+0=10-4α-4⇒α=32
⇒ Equation of the circle is
x2+y2-x2-9y2+4=0
Radius =116+8116-4=1816=322 units

⇒ area of the circle is 9π8 sq. units

Topic: Mathematics, Circle, JEE Main
The area (in sq units) of the equilateral triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3 y^2=3$ with the pair of asymptotes of the hyperbola is

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

The correct answer is:
$\sqrt{3}$

Solution:
Given equation of hyperbola is $x^2-3 y^2=3$ $\therefore$ Equation of tangent at point $(\sqrt{3}, 0)$ is
$\begin{aligned} & S_1=0 \\ \Rightarrow \quad & x \sqrt{3}-3 y \times 0=3\end{aligned}$
$\begin{aligned} \Rightarrow & & x \sqrt{3} & =3 \\ \Rightarrow & & x & =\sqrt{3}\end{aligned}$
The asympotes of given hyperbolas are
$x+\sqrt{3} y=0$
and
$x-\sqrt{3} y=0$
On solving Eqs. (i), (ii) and (iii), we get
$(0,0),(\sqrt{3},-1)$ and $(\sqrt{3}, 1)$
$\therefore$ Area of triangle formed by joining the above points
$\begin{aligned} & =\frac{1}{2}\left|\begin{array}{rrr}0 & 0 & 1 \\ \sqrt{3} & -1 & 0 \\ \sqrt{3} & 1 & 0\end{array}\right| \\ & =\frac{1}{2}[1(\sqrt{3}+\sqrt{3})]=\frac{1}{2} \times 2 \sqrt{3}=\sqrt{3}\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y=x2-1 below the x-axis, is :

Options:
A 233
B 133
C 43
D 433

The correct answer is:
433

Solution:
A(α,0),β(-α,0)

⇒Dα,α2-1

Area (ABCD) = (AB) (AD)

⇒S=(2α)1-α2=2α-2α3

dsdα=2-6α2=0⇒α2=13⇒α=13

Area =2α-2α3=23-233=433

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The area (in sq. units) of the locus of the point at which the two circles x2+y2=1 and x-42+y2=4 subtend equal angles is

Options:
A 329π
B 323π
C 643π
D 649π

The correct answer is:
649π

Solution:
sin⁡θ=1h2+k2=2h-42+k2
⇒h2-8h+16+k2=4h2+4k2
⇒3x2+3y2+8x-16=0
Radius of the circle is 432+163=83 units

Hence, the area of the circle is equal to 64π9 sq. units

Topic: Mathematics, Straight Lines, JEE Main
The area (in sq. units) of the parallelogram whose diagonals are along the vectors 8i^-6j^ and 3i^+4j^-12k^, is:

Options:
A 20
B 65
C 52
D 26

The correct answer is:
65

Solution:
Let the given diagonals be d 1 → and d 2 → , then
d1×d2=i^j^k^8-6034-12
=72i^--96j^+50k^
d1×d2=722+-962+502=5184+9216+2500
d2×d2= 16900=130
Area=12d1×d2=12×130
=65

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2017 (08 Apr Online)
The area (in sq. units) of the part of the circle x2+y2=36, which is outside the parabola y2=9x, is equal to

Options:
A 12π+33
B 24π+33
C 24π-33
D 12π-33

The correct answer is:
24π-33

Solution:
Required area

=π×62-2∫039xdx+∫3636-x2dx

=36π-6x3/23/203-2x236-x2+18sin-1x636

=36π-123-20-932+9π-3π

=24π-33 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The area (in sq. units) of the portion lying above the $X$-axis and enclosed between the curves $y^2=2 a x-x^2$ and $y^2=a x$ is

Options:
A $a^2\left(\frac{-\pi}{2}+\frac{2}{3}\right)$
B $a^2\left(\frac{2}{3}-\frac{\pi}{4}\right)$
C $a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$
D $a^2\left(\frac{\pi}{4}+\frac{2}{3}\right)$

The correct answer is:
$a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$

Solution:
Equation of given curves $x^2+y^2-2 a x=0$ and $y^2=a x$
$\because$ Area of the portion lying above the $X$-axis and enclosed between the given curves, from the diagram is
$\int_0^a\left(\sqrt{2 a x-x^2}-\sqrt{a x}\right) d x$
$=\int_0^a \sqrt{a^2-(x-a)^2} d x-\sqrt{a} \int_0^a x^{1 / 2} d x$
$=\left[\frac{x-a}{2} \sqrt{2 a x-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x-a}{a}\right)-\sqrt{a} \frac{2 x^{3 / 2}}{3}\right]_0^a$
$=-\frac{2}{3} a^2+\frac{\pi}{2} \frac{a^2}{2}=a^2\left(\frac{\pi}{4}-\frac{2}{3}\right)$ sq units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus ractum to the ellipse x29+y25=1, is

Options:
A 27
B 2 7 4
C 18
D 2 7 2

The correct answer is:
27

Solution:
x29+y25=1

e=1-59=23

a2=9, b2=5

Foci =±ae,0=±2,0

Ends of latus rectum=±ae,±b2a

=±2,±53

Tangent at 'L' is T=0

2×x9+53×y5=1

It cuts coordinate axes at P92,0 & Q0,3

Area of quadrilateral PQRS=4 (Area of triangle OPQ)

=4 12×92×3=27 sq. units

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2015 (04 Apr)
The area (in sq. units) of the region A=x,y∈R×R0≤x≤3, 0≤y≤4,y≤x2+3x is

Options:
A 263
B 8
C 536
D 596

The correct answer is:
596

Solution:
The area of the region OCD is
=∫014-x2+3xdx=4x-x33-3x2201=4-13-32=136
So, the area of the given region OABC is
= Area of OABD - Area of OCD
=3×4-136
=596

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The area (in sq. units) of the region A=x,y:x2 ≤y≤x+2 is

Options:
A 136
B 316
C 92
D 103

The correct answer is:
92

Solution:
The two given curves are y=x+2 and y=x2

To find their points of intersection, we equate the value of y to get

x2=x+2

⇒x2-x-2=0

⇒x-2x+1=0

⇒x=2 or x=-1.

The graphs of the curves are given below.

Required area is the area of the shaded portion.

Area =∫-12yline-yparaboladx
=∫-12x+2-x2dx

Using ∫xndx=xn+1n+1

=x22+ 2x-x33-12

=2+4-83-12-2+13

=6-83-12+ 2-13

=8-3-12

=92 sq units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The area (in sq. units) of the region A=x, y:y22≤x≤y+4 is:

Options:
A 30
B 18
C 533
D 16

The correct answer is:
18

Solution:
We have to find the area of the region A=x, y:y22≤x≤y+4

Now, to find the point of intersection of the curves x=y22 and x=y+4, we equate the value of x, to get

y22=y+4

⇒y2=2y+8

⇒y2-2y-8=0

⇒y-4y+2=0

⇒y=4 or y=-2

And, the graph of the given functions is as

The shaded area is the required area, and

Area=∫-24xline-xparabolady

=∫-24y+4-y22dy

Now, using ∫xndx=xn+1n+1, we get

Area=y22+4y-y36-24

=8+16-323-2-8+43

=18 sq units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The area (in sq. units) of the region bounded by the circle $x^2+y^2=64$, positive $x$-axis and the line $y=\sqrt{3} x$ is

Options:
A $\frac{16 \pi}{3}$
B $\frac{8 \pi}{3}$
C $\frac{64 \pi}{3}$
D $\frac{32 \pi}{3}$

The correct answer is:
$\frac{32 \pi}{3}$

Solution:
$x^2+y^2=64 \Rightarrow y=\sqrt{3} x$
$\begin{aligned} & y^2=3 x^2 \Rightarrow x= \pm 4 \\ & \therefore \quad y=\sqrt{48}\end{aligned}$
Area $=\int_0^4 y d x+\int_4^8 y_2 d x$
$\begin{aligned} & =\int_0^4(\sqrt{3} x) d x+\int_4^8 \sqrt{64-x^2} d x \\ & =\left[\sqrt{3} \frac{x^2}{2}\right]_0^4+\left[\frac{x}{2} \sqrt{64-x^2}+\frac{64}{2} \sin ^{-1}\left(\frac{x}{8}\right)\right]_4^8\end{aligned}$
$\begin{aligned} & =8 \sqrt{3}+\left[32 \frac{\pi}{2}-2 \sqrt{48}-32 \cdot \frac{\pi}{6}\right] \\ & =16 \pi-\frac{16 \pi}{3}=\frac{32 \pi}{3}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region bounded by the curve $x^{2}=4 y$ and the straight line $x=4 y-2$ is :

Options:
A $\frac{5}{4}$
B $\frac{9}{8}$
C $\frac{7}{8}$
D $\frac{3}{4}$

The correct answer is:
$\frac{9}{8}$

Solution:
Let points of intersection of the curve and the line be

$P$ and $Q$

$x^{2}=4\left(\frac{x+2}{4}\right)$

$x^{2}-x-2=0$

$x=2,-1$

Point are (2,1) and $\left(-1, \frac{1}{4}\right)$

$\text { Area }=\int_{-1}^{2}\left[\left(\frac{x+2}{4}\right)-\left(\frac{x^{2}}{4}\right)\right] d x=\left[\frac{x^{2}}{8}+\frac{1}{2} x-\frac{x^{3}}{12}\right]_{-1}^{2}$

$=\left(\frac{1}{2}+1-\frac{2}{3}\right)-\left(\frac{1}{8}-\frac{1}{2}+\frac{1}{12}\right)=\frac{9}{8}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The area (in sq. units) of the region bounded by the curves y=2-x2 and y=x is k, then the value of 3k is

Options:

The correct answer is:
7

Solution:
2-x2=x
x2+x-2=0
x≠-2
x=1
Required area =2∫012-x2-xdx
⇒k=22x-x33-x2201
⇒k=22-13-12=73sq.unit
⇒3k=7

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region bounded by the curves y=2x and y=x+1, in the first quadrant is

Options:
A 32-1loge⁡2
B 12
C loge⁡2+32
D 32

The correct answer is:
32-1loge⁡2

Solution:
We know that x= x,x≥0-x,x<0

Hence, x+1= x+1,x≥-1-x+1,x<-1
The graph of the given functions is

The shaded area is the required area.

Area =∫01yline-ycurvedx

Area =∫01x+1-2xdx

Using ∫xndx=xn+1n+1 & ∫axdx=axlogea, we get

Area =x22+x-2xloge201

Area =12+1-2loge2- 0+0-1loge2

Area =32-1loge2 sq units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (10 Apr Shift 2)
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

Options:
A 174
B 212
C 152
D 154

The correct answer is:
152

Solution:
Area of region ABCDA=∫03x2+2-x+1dx

=∫03x2-x+1dx=x33-x22+x03

=273-92+3

=12-92=152

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The area (in sq units) of the region bounded by $x=-1, x=2, y=x^2+1$ and $y=2 x-2$ is

Options:
A $10$
B $7$
C $8$
D $9$

The correct answer is:
$9$

Solution:
Given curve is $y=x^2+1 \Rightarrow x^2=y-1$ and line $\quad y=2 x-2$
The intersection point of curve and line is
$\begin{aligned} & \qquad x^2=2 x-2-1 \\ & \begin{array}{l}\text { Now, } \quad x^2-2 x+3=0 \\ \text { Hence, there is no point of intersection }\end{array} \\ & \therefore \text { Required area }=\int_{-1}^2\left(y_2-y_1\right) d x \\ & \qquad=\int_{-1}^2\left[\left(x^2+1\right)-(2 x-2)\right] d x \\ & \qquad=\left[\frac{x^3}{3}+x\right]_{-1}^2-\left[x^2-2 x\right]_{-1}^2 \\ & =\left[\frac{8}{3}+2-\left(-\frac{1}{3}-1\right)\right]-[4-4-(1+2)] \\ & =\frac{14}{3}+\frac{4}{3}-[-3] \\ & =6+3=9\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region described by A=x,y| y≥x2-5x+4, x+y≥1, y≤0 is

Options:
A 196
B 176
C 72
D 136

The correct answer is:
196

Solution:
The given equation is

A=x,y| y≥x2-5x+4,x+y≥1,y≤0 is

The equation x2-5x+4 is cutting the x- axis at x=1,4

Required area =12×2×2+∫34x2-5x+4dx

=2+x33-5x22+4x34

=2+76=196 sq.units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region described by A=x,y| y≥x2-5x+4, x+y≥1, y≤0 is

Options:
A 196
B 176
C 72
D 136

The correct answer is:
196

Solution:
The given equation is

A=x,y| y≥x2-5x+4,x+y≥1,y≤0 is

The equation x2-5x+4 is cutting the x- axis at x=1,4

Required area =12×2×2+∫34x2-5x+4dx

=2+x33-5x22+4x34

=2+76=196 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2016 (09 Apr Online)
The area (in sq. units) of the region described by $\left\{(x, y): y^2 \leq 2 x\right.$, and $\left.y \geq 4 x-1\right\}$ is

Options:
A $\frac{11}{32}$
B $\frac{8}{9}$
C $\frac{11}{12}$
D $\frac{9}{32}$

The correct answer is:
$\frac{9}{32}$

Solution:
$\begin{aligned} & \text { Shaded area }=\int_{-\frac{1}{2}}^1\left(x_{\text {Right }}-x_{\text {Left }}\right) d y \\ & y^2=2 x \\ & y=4 x-1 \quad \text { Solve } \\ & y=1, y=-\frac{1}{2} \\ & \text { Shaded area }=\int_{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y \\ & =\left(\frac{1}{4}\left(\frac{y^2}{2}+y\right)-\frac{y^3}{6}\right)_{-\frac{1}{2}}^1=\frac{9}{32}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (04 Apr Shift 2)
The area (in sq. units) of the region described by x,y: y2≤2x and y≥4x-1 is

Options:
A 932 sq.units
B 732 sq.units
C 564 sq.units
D 1564 sq.units

The correct answer is:
932 sq.units

Solution:
Let us find the points intersections of y2=2x and y=4x-1

⇒4x-12=2x

⇒16x2-10x+1=0

⇒8x-12x-1=0

⇒ x=12,18

x=12 ⇒y=412-1=1 ⇒A12,1

x=18 ⇒y=418-1=-12⇒B18,-12

A1=∫0182xdx + 12×18×12

A1=2x3232018 + 132

A1=124 + 132 sq.units

A2=∫0122xdx - 12×1×14

A2=2x3232012 - 18

A2=524 sq.units

Therefore, the required area is,

A=124+132+524=932 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2015 (04 Apr)
The area (in sq. units) of the region enclosed between the parabola y2=2x and the line x+y=4 is ______.

Options:

The correct answer is:
18

Solution:
y2=2x ⇒y2=24-y

⇒y2+2y-8=0 ⇒y=-4,2

The point of intersection of y2=2x and x+y=4 are 8,-4 and 2,2

Area =∫-424-y-y22dy =4y-y22-y36-42

=8-2-43--16-8+323

=543=18 Sq.units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (24 Jun Shift 2)
The area (in sq. units) of the region enclosed by the curves y=x2-1 and y=1-x2 is equal to:

Options:
A 43
B 83
C 72
D 163

The correct answer is:
83

Solution:
Given curves are

y=x2-1

and

y=1-x2

so intersection point are ±1,0

∴ bounded area

=4∫011-x2dx

=4x-x3301

=41-1383 sq·units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The area (in sq. units) of the region, given by the set x,y∈R×R∣x≥0,2x2≤y≤4-2x is :

Options:
A 83
B 173
C 133
D 73

The correct answer is:
73

Solution:
Given curves are y=2x2 & y=4-2x

So, 2x2=4-2x

⇒2x2+2x-4=0⇒x-1x+2=0

So, the x-coordinate of the intersection point is 1 as x=-2<0

Required area ∫014-2x-2x2dx=4x-x2-2x3301

=4-1-23=73

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The area (in sq. units) of the region in the first quadrant bounded by y=x2, y=2x+3 and the y-axis is

Options:
A 23
B 6
C 9
D 443

The correct answer is:
9

Solution:
In first quadrant y=x2 and y=2x+3 meet at x=3
Hence, the required area =∫032x+3-x2dx
=x2+3x-x3303
=9 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$-axis, the straight line $x-\sqrt{3} y=0$ and the circle $x^2+y^2=4$, is

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

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

Solution:
Equation of given circle
$\begin{aligned} & \Rightarrow x^2+y^2=2^2 \\ & \therefore \text { radius }=2\end{aligned}$
and line $x-\sqrt{3} y=0$
$$
\begin{aligned}
(\sqrt{3} y)^2+y^2 & =4 \\
\Rightarrow \quad & 4 y^2=4 \Rightarrow y^2=1 \Rightarrow y= \pm 1
\end{aligned}
$$
If $y=+1$
$$
\therefore \quad x=\sqrt{3}
$$
If $y=-1$
$$
\therefore \quad x=-\sqrt{3}
$$
As point $C$ is in first quadrant
$$
\therefore C \text { is }(\sqrt{3}, 1)
$$
Area of shaded region
$$
\begin{aligned}
& =\int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} d x+\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x \\
& =\frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x d x+\int_{\sqrt{3}}^2 \sqrt{2^2-x^2} d x
\end{aligned}
$$
$\begin{aligned} & =\frac{1}{\sqrt{3}}\left[\frac{x^2}{2}\right]_0^{\sqrt{3}}+\left\lfloor\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right]_{\sqrt{3}}^2 \\ & {\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}+c\right\rceil} \\ & =\frac{1}{\sqrt{3}} \cdot \frac{3}{2}+\left\lceil\left(0+2 \sin ^{-1} 1\right)-\left(\frac{\sqrt{3}}{2}\right.\right. \\ & \quad=\frac{\sqrt{3}-3}{2}+\left[2 \times \frac{\pi}{2}-\frac{\sqrt{3}}{2}-2 \frac{\pi}{3}\right\rceil \\ & \quad=\frac{\sqrt{3}}{2}+\pi-\frac{\sqrt{3}}{2}-\frac{2 \pi}{3}=\pi-\frac{2 \pi}{3}=\frac{\pi}{3} \\ & \therefore \text { Required area }=\frac{\pi}{3} \text { sq. units }\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area (in sq. units) of the region {x∈R: x≥0, y≥0, y≥x-2 and y≤x} is

Options:
A 133
B Â 83
C 103
D Â 53

The correct answer is:
103

Solution:
y=x

x=x-2

(x)2-(x)-2=0

(x-2)(x+1)=0

x=2⇒x=4
Therefore, the area of the region is
∫04xdx- Area of âABC

=23x3/2â04-12×2×2

=23×43/2-2
=163-2=103sq. unit

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2018 (15 Apr)
The area (in sq. units) of the region $\{x \in R: x \geq 0, y \geq 0, y \geq x-2$ and $y \leq \sqrt{x}\}$, is

Options:
A
$\frac{13}{3}$

B
$\frac{10}{3}$

C
$\frac{5}{3}$

D
$\frac{8}{3}$

The correct answer is:
$\frac{10}{3}$

Solution:
The intersection point of $y=x-2$ and $y=\sqrt{x}$ is $(4,2)$.
The required area
$$
=\int_0^4 \sqrt{x} d x-\frac{1}{2} \times 2 \times 2=\frac{16}{3}-2=\frac{10}{3}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The area (in sq. units) of the region x, y:0≤y≤x2+1, 0≤y≤x+1, 12≤x≤2 is

Options:
A 2316
B 7924
C 7916
D 236

The correct answer is:
7924

Solution:
0≤y≤x2+1, 0≤y≤x+1, 12≤x≤2

Drawing the graph according to the given information,

Area is, A=∫121x2+1dx+∫12x+1dx

A=x33+x121+x22+x12

A=13+1-123·13+12+42+2-12+1

A=43-124+12+2+2-32

A=43-1324+4-32

A=32-13+96-3624

A=7924 sq. units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The area (in sq. units) of the region x,y∈R24x2≤y≤8x+12 is

Options:
A 1253
B 1283
C 1243
D 1273

The correct answer is:
1283

Solution:
4x2=y
y=8x+12
4x2=8x+12
x2-2x-3=0
x=-1,3
A=∫-138x+12-4x2dx
A=8x22+12x-4x333-1=49+36-36-4-12+43=36+8-43
=44-43=132-43=1283

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The area (in sq. units) of the region x, y∈R2:x2≤y≤3-2x, is.

Options:
A 323
B 343
C 293
D 313

The correct answer is:
323

Solution:
Point of intersection of y=x2 & y=-2x+3, is obtained by x2+2x-3=0⟹x=-3,1.

So, area of the required region is

∫-313-2x-x2 dx

=3x-x2-x33-31=323.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The area (in sq. units) of the region x, y: x≥0, x+y≤3, x2≤4y and y≤1+x is

Options:
A 5912 sq. units
B 32 sq. units
C 73 sq. units
D 52 sq. units

The correct answer is:
52 sq. units

Solution:
Required Area
=∫011+xdx+∫123-xdx-∫02x24dx
=x+23x3201+3x-x2212-x31202
=1+23+4-52-23
=52 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2017 (02 Apr)