datavorous/entrance-exam-dataset huggingface dataset:
The area of the closed region bounded by y=sec-1â¡x, y=cosec-1 x and the line x-1=0 is

Options:
A logeâ¡3+22-Ï2 sq. units
B Ï2-logeâ¡3+22Â sq.Â units
C Ï-3logeâ¡3Â sq.Â units
D None of these

The correct answer is:
logeâ¡3+22-Ï2 sq. units

Solution:
Integrating along y-axis, we get
A=2∫0π4sec⁡y-1dy
=2logesec⁡y+tan⁡y-y0π4
=2loge⁡2+1-π4=loge⁡3+22-π2 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area of the director circle of the ellipse $\frac{x^2}{5}+\frac{y^2}{4}=1$ is

Options:
A $5 \pi$ sq. units
B $4 \pi$ sq. units
C $9 \pi$ sq. units
D $18 \pi$ sq. units

The correct answer is:
$9 \pi$ sq. units

Solution:
The correct option is 3 9 sq. units
Given : $\frac{x^2}{5}+\frac{y^2}{4}=1$
We know, equation of director circle of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2+b^2$
Here, $a^2=5, b^2=4$
$\therefore$ Equation of director circle is
$\begin{aligned}
& x^2+y^2=5+4 \\
& \Rightarrow x^2+y^2=9=r^2 \\
& \therefore \text { Area }=\pi r^2=9 \pi \text { sq. units }
\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The area of the figure bounded by the parabola $x=-2 y^{2}, x=1-3 y^{2}$ is

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

The correct answers are:
$\frac{4}{3}$ square unit

Solution:
Hint:
$\mathrm{A}=\int_{-1}^{1}\left(1-3 \mathrm{y}^{2}\right)-\left(-2 \mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}$
$\Rightarrow \mathrm{A}=2 \int_{0}^{1}\left(1-\mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}=\frac{4}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the parallelogram whose diagonals are \( \vec{a}=3 \hat{i}+\hat{j}-2 \hat{k} \) and \( \vec{b}=\hat{i}-3 \hat{j}+4 \hat{k} \) is

Options:
A \( 10 \sqrt{3} \)
B \( 5 \sqrt{3} \)
C \( 8 \)
D \( 4 \)

The correct answer is:
\( 5 \sqrt{3} \)

Solution:
Δ=12 a→ ×b→
But a→ ×b→= i^j^k^31-21-34= -2i^-14j^-10k^ .
Hence Δ=12 a→ × b→=12 4+196+100=53

Topic: Mathematics, Vector Algebra, JEE Main
The area of the plane region bounded by the curves $x+2 y^2=0$ and $x+3 y^2=1$ is equal to

Options:
A
$\frac{5}{3}$

B
$\frac{1}{3}$

C
$\frac{2}{3}$

D
$\frac{4}{3}$

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

Solution:
Solving the equations we get the points of intersection $(-2,1)$ and $(-2,-1)$
The bounded region is shown as shaded region.
$$
\begin{aligned}
& \text { The required area }=2 \int_0^1\left(1-3 y^2\right)-\left(-2 y^2\right) \\
& =2 \int_0^1\left(1-y^2\right) d y=2\left[y-\frac{y^3}{3}\right]_0^1=2 \times \frac{2}{3}=\frac{4}{3} .
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2008
The area of the polygon, whose vertices are the non-real roots of the equation z¯=iz2 is

Options:
A 332
B 334
C 34
D 32

The correct answer is:
334

Solution:
z¯=iz2⇒x-iy=ix+iy2

⇒ x-iy=x2-y2i-2xy

i.e., x=-2xy and -y=x2-y2

⇒ x=0,y=-12

When x=0; y=0,1

When y=-12; x=±32

0,0 will be rejected as vertices would be non-real roots.

So, the vertices will be 0,1, 32,-12 and -32,--12

Hence, area of ∆=12×3×32=334

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The area of the quadrilateral ABCD with vertices A(2,1,1), B(1,2,5), C(-2,-3,5) and D(1,-6,-7) is equal to

Options:
A 48
B 838
C 54
D 938

The correct answer is:
838

Solution:
Given,

The quadrilateral ABCD with vertices A(2,1,1), B(1,2,5), C(-2,-3,5) and D(1,-6,-7)

Now finding, AB→=-i^+j^+4k^ and AD→=-i^-7j^-8k^

And plotting the diagram we get,
Now we know that,

Area of ∆ABD is given by,

=12i^j^k^-114-1-78

=|10i^-6j^+4k^|=238

Now finding, CB→=3i^+5j^ and CD→=3i^-3j^-12k^
So, Area of ∆CBD=12i^j^k^3-3-12350

=|6(5i^-3j^-2k^)|

=638
∴ Area of quadrilateral ABCD=838 square units

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The area of the quadrilateral formed by the tangents and normals at the extremities of the latus rectum of the parabola y2-4y+4+12x=0 is

Options:
A 36 sq.units
B 72 sq.units
C 144 sq.units
D 96 sq.units

The correct answer is:
72 sq.units

Solution:
y-22=-12x
The quadrilateral is a square with diagonal length
=12
∴ Area =12×12×12=72 sq. units

Topic: Mathematics, Parabola, JEE Main
The area of the quadrilateral having vertices \( (0,4,1),(2,3,-1),(4,5,0) \) and \( (2,6,2) \) is:

Options:
A \( 27 \) sq. units
B \( 9 \) sq. units
C \( 81 \) sq. units
D None of these

The correct answer is:
\( 9 \) sq. units

Solution:
Let the vertices of the quadrilateral be A, B, C and D.

Then, we get, A=(0, 4, 1), B=(2, 3, -1), C=(4, 5, 0) and D=(2, 6, 2).

So, we get,

AB→=(2−0)i^+(3−4)j^+(−1−1)k^=2i^−j^−2k^

BC→=(4−2)i^+(5−3)j^+(0+1)k^=2i^+2j^+k^

CD→=(2−4)i^+(6−5)j^+(2−0)k^=−2i^+j^+2k^

DA→=(0−2)i^+(4−6)j^+(1−2)k^=−2i^−2j^−k^

As the magnitude of the opposite sides are same, we can say it is parallelogram.

So, area of the parallelogram=AB→×BC→=ijk2-1-2221

=i^-1+4-j^2+4+k^4+2

=3i^-6j^+6k^

=9+36+36

=81

=9

Topic: Mathematics, Vector Algebra, JEE Main
The area of the region A={x, y: 0≤y≤xx+1 and -1≤x≤1} in sq. units, is

Options:
A 43
B 2
C 13
D 23

The correct answer is:
2

Solution:
xx=x2, x≥0-x2, x<0

Area enclosed =∫-10-x2+1dx+∫01x2+1dx
=-x33+x-10+x33+x01
=2

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\frac{\pi}{4}$ is

Options:
A
$\int_0^{\sqrt{2}-1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

B
$\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

C
$\int_0^{\sqrt{2}+1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

D
$\int_0^{\sqrt{2}+1} \frac{t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

The correct answer is:
$\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t$

Solution:
$$
\begin{aligned}
& \text { Required area }=\int_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x \quad\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right] \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}} \sqrt{\frac{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}}\right) d x \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}\right) d x \\
&
\end{aligned}
$$
$$
=\int_0^{\pi / 4} \frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x=\int_0^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x
$$
Put $\tan \frac{x}{2}=t$
$$
\begin{aligned}
\Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x & =d t \\
\therefore \text { Required area } & =\int_0^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^2\right) \sqrt{1-t^2}} \\
& =\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \quad\left[\text { astan } \frac{\pi}{8}=\sqrt{2}-1\right]
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by \( f(x)=\sin \mathrm{x}, \mathrm{g}(\mathrm{x})=\cos \mathrm{x} \) from \( x=0 \) to \( x=\frac{\pi}{2} \) is

Options:
A \( 2(\sqrt{2}+1) \) sq. units
B \( (\sqrt{3}-1) \) sq. units
C \( 2(\sqrt{3}-1) \) sq. units
D \( 2(\sqrt{2}-1) \) sq. units

The correct answer is:
\( 2(\sqrt{2}-1) \) sq. units

Solution:
We have, cosx≥sinx, 0≤x≤π4 and sinx≥cosx, π4≤x≤π2

Area of the shaded region

∆=∫0π4cosx-sinxdx+∫π4π2sinx-cosxdx
=sinx+cosx0π4+-cosx-sinxπ4π2
=12+12-0+1-1-12+12
=42-2

=22-2

=22-1 sq. units.

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curve $9 x^2+4 y^2-36=0$ is

Options:
A $9 \pi$
B $4 \pi$
C $36 \pi$
D $6 \pi$

The correct answer is:
$6 \pi$

Solution:
$\quad 9 x^2+4 y^2=36 \Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1$, which is equation of an ellipse. Remember area enclosed by ellipse is $\pi a b$ i.e., $\pi 2.3=6 \pi$.

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curve $y=\left|x^{3}-4 x^{2}+3 x\right|$ and the $x$-axis, $0 \leq x \leq 3$, is $-$

Options:
A $\frac{37}{6}$
B $\frac{9}{4}$
C $\frac{37}{12}$
D 0

The correct answer is:
$\frac{37}{12}$

Solution:
$A=\int_{0}^{1} f(x) d x-\int_{1}^{3} f(x) d x=\int_{0}^{1}\left(x^{3}-4 x^{2}+3 x\right) d x-\int_{1}^{3}\left(x^{3}-4 x^{2}+3 x\right) d x=\frac{37}{12}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curve $y=x^3$, and the lines, $y=8$, and $x=0$, is

Options:
A
8

B
12

C
10

D
16

The correct answer is:
12

Solution:
Required Area $=\int_{y=0}^8 y^{1 / 3} d y$
$$
\begin{aligned}
& =\left.\frac{y^{1 / 3+1}}{\frac{1}{3}+1}\right|_0 ^8=\left.\frac{3}{4}\left(y^{4 / 3}\right)\right|_0 ^8 \\
& =\frac{3}{4}\left[(8)^{4 / 3}-0\right]=\frac{3}{4}\left[2^4\right] \\
& =\frac{3}{4} \times 16=12 \text { sq. unit. }
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (19 May Online)
The area of the region bounded by the curves $x=y^2-2$ and $x=y$ is

Options:
A $\frac{9}{4}$
B 9
C $\frac{9}{2}$
D $\frac{9}{7}$

The correct answer is:
$\frac{9}{2}$

Solution:
We have,
$$
x=y^2-2 \text { and } x=y
$$
Solving, $x=y^2-2$ and $x=y$, we get $(-1,-1)$ and $(2,2)$
Area of shaded region is
$$
\begin{aligned}
& \int_{-1}^2 y d y-\int_{-1}^2\left(y^2-2\right) d y \\
& \Rightarrow\left[\frac{y^2}{2}-\frac{y^3}{3}+2 y\right]_{-1}^2=\left(\frac{4}{2}-\frac{8}{3}+4\right)-\left(\frac{1}{2}+\frac{1}{3}-2\right) \\
&=\frac{10}{3}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curves $y=|x-1|$ and $y=3-|x|$ is

Options:
A
6 sq. units

B
2 sq. units

C
3 sq. units

D
4 sq. units

The correct answer is:
4 sq. units

Solution:
$A=\int_{-1}^0\{(3+x)-(-x+1)\} d x+\int_0^1\{(3-x)-(-x+1)\} d x+\int_1^2\{(3-x)-(-x-1)\} d x$
$=\int_{-1}^0(2+2 x) d x+\int_0^1 2 d x+\int_1^2(4-2 x) d x$
$=\left[2 x-x^2\right]_{-1}^0+[2 x]_0^1+\left[4 x-x^2\right]_1^2$
$=0-(-2+1)+(2-0)+(8-4)-(4-1)$
$=1+2+4-3=4$ sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2003
The area of the region bounded by the curves $y=x^{2}$ and $x=y^{2}$ is

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

The correct answer is:
$1 / 3$

Solution:
Given curves are $y=x^{2}$ and $x=y^{2}$, which is the form of parabola.

The point of intersection, $x=\left(x^{2}\right)^{2}$$
\begin{array}{l}
\Rightarrow \quad x=x^{4} \Rightarrow x\left(1-x^{3}\right)=0 \\
\Rightarrow \quad x=0 \text { and } 1=x^{3} \\
\Rightarrow \quad x=0 \text { and } x=1 \\
\text { When } x=0, \text { then } y=0 \\
\text { When } x=1, \text { then } y=1^{2}=1
\end{array}
$
$\therefore$ The point of intersection is (0,0) and (1,1)
$\therefore$ Area of shaded region
$$
\begin{array}{l}
=\int_{0}^{1}\left(y_{2}-y_{1}\right) d x \\
=\int_{0}^{1}\left[\sqrt{x}-x^{2}\right] d x=\left[\frac{x^{3 / 2}}{3 / 2}-\frac{x^{3}}{3}\right]_{0}^{1} \\
=\frac{2}{3}(1)^{3 / 2}-\frac{(1)^{3}}{3}-0-0 \\
=\frac{2}{3}-\frac{1}{3}=\frac{1}{3} \mathrm{sq} \text { units }
\end{array}
$$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curves $y=x^2$ and $y=|x|$ is

Options:
A $1/6$
B $1 / 3$
C $5/6$
D $5/3$

The correct answer is:
$1 / 3$

Solution:
Required area = 2 (shaded area in first quadrant)
$=2 \int_0^1\left(x-x^2\right) d x=2 \times \frac{1}{6}=\frac{1}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the curves $y=|x-2|, x=1, x=3$ and the $x$-axis is

Options:
A
1

B
2

C
3

D
4

The correct answer is:
1

Solution:
Area $=\int_1^2(2-x) d x+\int_2^3(x-2) d x=1$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2004
The area of the region bounded by the curves $y=x^{3}, y=\frac{1}{x}, x=2$ is

Options:
A $4-\log _{e} 2$
B $\frac{1}{4}+\log _{e} 2$
C $3-\log _{e} 2$
D $\frac{15}{4}-\log _{e} 2$

The correct answer is:
$\frac{1}{4}+\log _{e} 2$

Solution:
First of all we draw the graph $y=x^{3}, y=\frac{1}{x}, x=2$

Required area i.e., (OMPNO) $=\int_{0}^{1} x^{3} d x+\int_{1}^{2} \frac{1}{x} d x$
$=\left[\frac{x^{4}}{4}\right]_{0}+\left[\log _{4} x\right]_{1}^{2}=\frac{1}{4}+\log _{c} 2$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the lines x=1, x=2, and the curves xy-ex=sinx and 2xy=2sinx+x3 is

Options:
A e2-e-16
B e2-e-76
C e2-e+16
D e2-e+76

The correct answer is:
e2-e-76

Solution:
The area of region bounded by the curves

xy-ex=sinx or y=sinxx+ex and 2xy=2sinx+x3

or y=sinxx+x22

between lines x=1 and x=2 is

A=∫12sinxx+ex-sinxx+x22dx

=∫12ex-x22dx=ex-x3612=e2-e-76

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by the parabola (y-2)2=(x-1), the tangent to it at the point whose ordinate is 3 and the x-axis, is:

Options:
A 4
B 6
C 9
D 10

The correct answer is:
9

Solution:
Given, the parabola y-22=x-1 and y=3, x=2

So, point is (2,3) Differentiate given equation w.r.t. x, we get

2(y-2)y'=1

y′=12(y−2)

y′(2,3)=12

So, the equation of tangent at 2,3

y-3=12x-2

x-2y+4=0

Required area =∫03[(y-2)2+1]dy-∫03(2y-4)dy
Required area =∫03y2-4y+5dy+∫032y-4dy
Required area =y33-2y2+5y03-y2-4y03

Required area=9 sq units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The area of the region bounded by the parabola $(y-2)^2=x-1$, the tangent to the parabola at the point $(2,3)$ and the $x$-axis is

Options:
A
3

B
6

C
9

D
12

The correct answer is:
9

Solution:
Equation of tangent at $(2,3)$ to
$$
\begin{aligned}
& (y-2)^2=x-1 \text { is } S_1=0 \\
& \Rightarrow x-2 y+4=0
\end{aligned}
$$
Required Area $=$ Area of $\triangle O C B+$ Area of OAPD - Area of $\triangle P C D$
$$
\begin{aligned}
& =\frac{1}{2}(4 \times 2)+\int_0^3\left(y^2-4 y+5\right) d y-\frac{1}{2}(1 \times 2) \\
& =4+\left[\frac{y^3}{3}-2 y^2+5 y\right]_0^3-1=4-9-18+15-1 \\
& =28-19=9 \text { sq. units }
\end{aligned}
$$
$$
\text { (or) }
$$
$$
\text { Area }=\int_0^3\left(2 y-4-y^2+4 y-5\right) d y=\int_0^3\left(-y^2+6 y-5\right) d y=-\int_0^3(3-y)^2 d y=\left[\frac{(y-3)^3}{3}\right]_0^3=\frac{27}{3}=9 \text { sq.units }
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2009
The area of the region bounded by the parabola $y=x^{2}-4 x+5$ and the straight line $y=x+1$ is

Options:
A $\frac{1}{2}$
B 2
C 3
D $\frac{9}{2}$

The correct answer is:
$\frac{9}{2}$

Solution:
Given equation of parabola is
$$
y=x^{2}-4 x+5
$$
$\Rightarrow \quad y=(x-2)^{2}+1$
$\Rightarrow \quad(x-2)^{2}=(y-1)$
and equation of line is
$$
\begin{array}{r}
y=x+1 \\
\Rightarrow \quad x-y=-1
\end{array}
$$

On putting the value of $(y-1)$ from Eqs. (ii) in (1). we get
$$
(x-2)^{2}=x \Rightarrow x^{2}+4-4 x=x
$$
$\Rightarrow \quad x^{2}-5 x+4=0$
$\Rightarrow x^{2}-4 x-x+4=0$
$\Rightarrow \quad x(x-4)-1(x-4)=0$
$\Rightarrow \quad(x-1)(x-4)=0$ or $x=1,4$
then from is (ii) $y=2,5$
$\therefore$ Required area $=\int_{1}^{4}\left\{(x+1)-\left(x^{2}-4 x+5\right)\right\} d x$
$=\int_{1}^{4}\left(-x^{2}+5 x-4\right) d x=\left[\frac{-x^{3}}{3}+\frac{5 x^{2}}{2}-4 x\right]_{1}^{4}$
$=\left[-\frac{64}{3}+40-16+\frac{1}{3}-\frac{5}{2}+4\right]$
$=\left(-21-\frac{3}{2}+28\right)=\frac{9}{2}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by \( x=0, y=0, x=2, y=2, y \leq e^{x} \) and \( y \geq \ln x \) is

Options:
A \( 6-4 \ln 2 \) sq. units
B \( 4 \ln 2-2 \) sq. units
C \( 2 \ln 2-4 \) sq. units
D \( 6-2 \ln 2 \) sq. units

The correct answer is:
\( 6-4 \ln 2 \) sq. units

Solution:
A=∫12lnx dx
=xlnx-x12
=2ln2-1
⇒ Required area =4-22ln2-1=6-4ln2 sq.units

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by $y=2 x-x^{2}$ and the $X$-axis is

Options:
A $\frac{8}{3}$ sq units
B $\frac{4}{3} \mathrm{sq}$ units
C $\frac{7}{3}$ sq units
D $\frac{2}{3} \mathrm{sq}$ units

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

Solution:
We have,
$$
\begin{aligned}
y &=2 x-x^{2} \\
\Rightarrow \quad y &=-\left(x^{2}-2 x\right) \\
\Rightarrow \quad y &=-\left[(x-1)^{2}-1\right] \\
\Rightarrow \quad y &=-(x-1)^{2}+1 \\
\Rightarrow \quad(x-1)^{2}=-(y-1)
\end{aligned}
$\begin{aligned} \text { Required area } &=\int_{0}^{2} y d x \\ &=\int_{0}^{2}\left(2 x-x^{2}\right) d x \\ &=\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2}=4-\frac{8}{3} \\ &=\frac{4}{3} \operatorname{sq} \text { unit } \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region bounded by y-x=2 and x2=y is equal to :-

Options:
A 163
B 23
C 92
D 43

The correct answer is:
92

Solution:
y-x=2,x2=y

Now, x2=2+x

⇒x2-x-2=0

⇒x+1x-2=0⇒x=-1,2

Required area=∫-122+x-x2dx

=2x+x22-x33-12

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

=6-3+2-12=92 square units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The area of the region covered by the TV broadcast, by a TV tower of 100 m height is (Take, the radius of the earth, Re=6.4Ã106 m )

Options:
A 12.8π×108 km2
B 1.28π×103 km2
C 0.64π×103 km2
D 1.28 ×103 km2

The correct answer is:
1.28π×103 km2

Solution:
Given, the height of the tower,
h=100 m
The radius of the earth, Re=6.4×106 m
Since we know,
the area covered by the TV tower,
A=2πReh
=2π×6.4×106×100
=12.8π×108 m2
=1.28π×103 km2

Topic: Physics, Communication System, JEE Main
The area of the region enclosed between the parabolas y2=2x-1 and y2=4x-3 is.

Options:
A 13
B 16
C 23
D 34

The correct answer is:
13

Solution:
We have, y2=2x-1 ...1 and y2=4x-3 ...2

We first find the points of intersection of the given parabolas by solving the equations 1 & 2simultaneously.

This gives, 2x-1=4 x-3

⇒ 2x=2 ⇒x=1

Now, put x=1 in equation 1, we get y=±1.

Now, the equations 1 & 2 can be written as

y2=2x-12

y2=4x-34 respectively.

The graphs of these two curves are shown in figure.

We have to find the area of the shaded region.

Now, the equations 1 & 2 can be written as

x=y2+12 & x=y2+34 respectively.

It is clear from the figure, required area A=2∫01x2-x1dy (∵ Area is symmetrical about x- axis)

where x1=y2+12 and x2=y2+34

⇒A=2∫01y2+34-y2+12dy =24∫011-y2dy

=12y-y3301 =121-13

=1223=13 sq.units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The area of the region enclosed by fx=-2xex and the x -axis is

Options:
A 1 sq. unit
B 2 sq. units
C 12 sq. unit
D not defined

The correct answer is:
2 sq. units

Solution:
fx=-2xe-x
f'x=2xe-x-2e-x
=2e-xx-1
Area =∫0∞-2xe-xdx=2∫0∞xe-xdx
=-xe-x0∞+∫0∞e-xdx
=20-e-x0∞=2×1=2

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region enclosed by the curve fx=max{sinx,cosx},-π≤x≤π and the x-axis is

Options:
A 22(2+1)
B 4
C 42
D 22+1

The correct answer is:
4

Solution:
Plotting the required diagram we get,

Now from diagram we can say that,

sinx is maximum for x∈-π,-3π4∪π4,π and cosx is maximum for x∈-3π4,-π2∪-π2,π4

So, area of the region is given by,

A=∫-π-3π4sinxdx+∫π4πsinxdx+∫-3π4-π2cosxdx+∫-π2π4cosxdx

⇒A=12-1+1+12+-1+12+12+1

⇒A=2+22+2-22=4

Hence this is the required option.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (13 Apr Shift 1)
The area of the region enclosed by the curve y=x3 and its tangent at the point (–1, –1) is

Options:
A 194
B 234
C 314
D 274

The correct answer is:
274

Solution:
Given y=x3 ... i

⇒dydx=3x2

dydx-1, -1=3

Equation of tangent at (–1, –1)

(y+1)=3(x+1)

y=3x+2 … (ii)

Solving i and ii

⇒x3=3x+2

⇒x3-3x+2=0

⇒x=-1,-1,2

Another point of intersection is Q(2, 8)

So, now plotting the diagram we get,

Required area from the above diagram will be,

=∫-123x+2-x3dx

=324-1+22+1-1416-1

=274

Hence this is the required option.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $x$-axis is

Options:
A
1 square units

B
$\frac{3}{2}$ square units

C
$\frac{5}{2}$ square units

D
$\frac{1}{2}$ square units

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

Solution:
Area $=\int_0^1 x d x+\int_1^e \frac{1}{x} d x=\frac{1}{2}+1=\frac{3}{2}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2011
The area of the region enclosed by the parabola (y-2)2=x-1, the line x-2 y+4=0 and the positive coordinate axes is __________.

Options:

The correct answer is:
5

Solution:
Given,

(y-2)2=x-1 and x-2y+4=0

Now, plotting the graph of the above equation, we get,

Now, from above diagram we get,

Required area =∫03xdy- Area of ΔABC.
=∫03(y-2)2+1dy-12×1×2

=∫03y2-4y+5dy-1

=y33-2y2+5y03-1

=9-18+15-1=5

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola y=4x−x2 and 3y=x−42 is equal to

Options:
A 329
B 4
C 6
D 143

The correct answer is:
6

Solution:
Given: y=4x-x2 and 3y=x-42

⇒4x-x2=x-423

⇒12x-3x2=x2+16-8x

⇒4x2-20x+16=0

⇒x2-5x+4=0

⇒x-4x-1=0

⇒x=4, 1

⇒y=0, 3

So, the intersection points are 4,0 and 1,3.

Area =∫144x−x2−x−423dx

Area =4x22−x33−x−43914

=642−643−42+13−279

=27−21=6

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The area of the region enclosed by y≤4x2,x2≤9y and y≤4, is equal to

Options:
A 403
B 563
C 1123
D 803

The correct answer is:
803

Solution:
Plotting the required region

Enclosed area =2∫043y-y2dy

=2∫045y2dy

=523y3204

=103(4)32=803

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The area of the region given by A=x,y:x2≤y≤minx+2,4-3x is

Options:
A 318
B 176
C 196
D 278

The correct answer is:
176

Solution:
The graph of y=x2 & y=minx+2,4-3x will be

From the graph,

required area A=∫-112x+2-x2dx+∫1214-3x-x2dx

=x22+2x-x33-112+4x-3x22-x33121=176

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The area of the region given by x,y:xy≤8,1≤y≤x2 is :

Options:
A 8 loge2-133
B 16 loge2-143
C 8 loge2+76
D 16 loge2+73

The correct answer is:
16 loge2-143

Solution:
Given,

xy≤8 & 1≤y≤x2

On solving above curve we get intersecting point as 2,4

Now on plotting the diagram of the given curves we get,

Required area

A=∫12x2-1dx+∫288x-1dx

⇒A=∫12x2dx+∫288xdx-∫18dx

⇒A=x3312+8lnx28-x18

⇒A=8ln4-143

⇒A=16ln2-143 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The area of the region (in sq. units), enclosed by the circle x2+y2=2 which is not common to the region bounded by the parabola y2=x and the straight line y=x, is

Options:
A 1624π-1
B 136π-1
C 1312π-1
D 1612π-1

The correct answer is:
1612π-1

Solution:
Required area

=2π-∫01x-xdx

=2π-2x323-x2201

=2π-23-12

=2π-16

=12π-16 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The area of the region (in sq. units), in the first quadrant bounded by the parabola $y=9 x^2$ and the lines $x=0, y=1$ and $y=4$, is :

Options:
A
$7 / 9$

B
$14 / 3$

C
$7 / 3$

D
$14 / 9$

The correct answer is:
$14 / 9$

Solution:
$$
\begin{aligned}
& \text { Required area }=\int_{y=1}^4 \sqrt{\frac{y}{9}} d y \\
& =\frac{1}{3} \int_{y=1}^4 y^{1 / 2} d y=\frac{1}{3} \times\left.\frac{2}{3}\left(y^{3 / 2}\right)\right|_1 ^4 \\
& =\frac{2}{9}\left[\left(4^{1 / 2}\right)^3-\left(1^{1 / 2}\right)^3\right]=\frac{2}{9}[8-1]
\end{aligned}
$$
$$
=\frac{2}{9} \times 7=\frac{14}{9} \text { sq. units. }
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2013 (22 Apr Online)
The area of the region (in square units) above the x-axis bounded by the curve y=tanx, 0≤x≤π2 and the tangent to the curve at x=π4 is

Options:
A 12log2-12
B 121+log2
C 121-log2
D 12log2+12

The correct answer is:
12log2-12

Solution:
The given curve is y=tanx and at x=π4, y=tanπ4=1

Also, dydx=sec2x and dydxx=π4=sec2π4=22=2

We know that the equation of the tangent to a curve y=fx at a point x1, y1 is y-y1=dydxx=π4x-x1

Hence, the equation of the tangent to y=tanx at Pπ4, 1 is y-1=2x-π4

⇒y-1=2x-π2

⇒y-1+π2=2x

To find the point where this line cuts the x-axis, put y=0, to get

2x=π2-1

⇒x=π4-12

Thus, the point A≡π4-12, 0.

Now, the graph for the given information is

And, the required Area
=∫0π4tanxdx-ar∆PAB

=logsecx0π4-12×PB×AB
=logsecπ4-logsec0-12×1×π4-π4-12

=log2-log1-14

=log212-0-14

=12log2-14=12log2-12 sq units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2014 (19 Apr Online)