Q. The area of the region bounded by the parabola $(\mathrm{y}-2)^{2}$ = x – 1, the tangent to the parabola at the point (2, 3) and the x–axis is :-
(1) 9 (2) 12 (3) 3 (4) 6

**[AIEEE-2009]**
Q. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = $\frac{3 \pi}{2}$ is :-
(1) $4 \sqrt{2}-2$
(2) $4 \sqrt{2}+2$
(3) $4 \sqrt{2}-1$
(4) $4 \sqrt{2}+1$

**[AIEEE-2010]**
Q. The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $x$ -axis is:-
(1) $\frac{3}{2}$ square units
(2) $\frac{5}{2}$ square units
(3) $\frac{1}{2}$ square units
(4) 1 square units

**[AIEEE-2011]**
Q. The area bounded by the curves $y^{2}=4 x$ and $x^{2}=4 y$ is :-
(1) 0 (2) $\frac{32}{3}$ (3) $\frac{16}{3}$ (4) $\frac{8}{3}$

**[AIEEE-2011]**
Q. The area bounded between the parabolas $x^{2}=\frac{y}{4}$ and $x^{2}=9 y,$ and the straight line $y=2$ is :
(1) $10 \sqrt{2}$
(2) $20 \sqrt{2}$
(3) $\frac{10 \sqrt{2}}{3}$
(4) $\frac{20 \sqrt{2}}{3}$

**[AIEEE-2012]**
Q. The area (in square units) bounded by the curves $y=\sqrt{x}, 2 y-x+3=0,$ x-axis and lying in the first quadrant is :
(1) 9 (2) 36 (3) 18 (4) $\frac{27}{4}$

**[JEE (Main)-2013]**
Q. The area bounded by the curve y =

*ln*(*x*) and the lines*y*= 0,*y*=*ln*(3) and*x*= 0 is equal to : (1) 3*ln*(3) – 2 (2) 3 (3) 2 (4) 3*ln*(3) + 2**[JEE-Main (On line)-2013]**
Q. 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 :-
(1) 7/9 (2) 14/3 (3) 14/9 (4) 7/3

**[JEE-Main (On line)-2013]**
Q. The area under the curve $y=|\cos x-\sin x|, 0 \leq x \leq \frac{\pi}{2},$ and above $x$ -axis is :
(1) $2 \sqrt{2}$
(2) $2 \sqrt{2}+2$
(3) 0
(4) $2 \sqrt{2}-2$

**[JEE-Main (On line)-2013]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**(4) Area $=2 \int_{0}^{\pi / 4}(\cos x-\sin x) d x=-2+2 \sqrt{2}$

Q. Let $f:[-2,3] \rightarrow[0, \infty)$ be a continuous function such that $f(1-x)=f(x)$ for all $x \in[-2,3] .$ If $R_{1}$ is the numerical value of the area of the region bounded by $y=f(x), x=-2, x=3$ and the axis of $x$ and $R_{2}=\int_{-2}^{3} x f(x) d x,$ then :
(1) $2 \mathrm{R}_{1}=3 \mathrm{R}_{2}$
(2) $\mathrm{R}_{1}=\mathrm{R}_{2}$
(3) $3 \mathrm{R}_{1}=2 \mathrm{R}_{2}$
(4) $\mathrm{R}_{1}=2 \mathrm{R}_{2}$

**[JEE-Main (On line)-2013]**
Q. The area of the region described by $\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^{2}+\mathrm{y}^{2} \leq 1 \text { and } \mathrm{y}^{2} \leq 1-\mathrm{x}\right\}$ is :
(1) $\frac{\pi}{2}+\frac{4}{3}$
(2) $\frac{\pi}{2}-\frac{4}{3}$
(3) $\frac{\pi}{2}-\frac{2}{3}$
(4) $\frac{\pi}{2}+\frac{2}{3}$

**[JEE(Main)-2014]**
Q. The area (in sq.units) of the region $\left\{(\mathrm{x}, \mathrm{y}): \mathrm{y}^{2} \geq 2 \mathrm{x} \text { and } \mathrm{x}^{2}+\mathrm{y}^{2} \leq 4 \mathrm{x}, \mathrm{x} \geq 0, \mathrm{y} \geq 0\right\}$ is :-
(1) $\frac{\pi}{2}-\frac{2 \sqrt{2}}{3}$
(2) $\pi-\frac{4}{3}$
(3) $\pi-\frac{8}{3}$
(4) $\pi-\frac{4 \sqrt{2}}{3}$

**[JEE(Main)-2016]**
Q. The area (in sq. units) of the region
$\left\{(\mathrm{x}, \mathrm{y}\}: \mathrm{x} \geq 0, \mathrm{x}+\mathrm{y} \leq 3, \mathrm{x}^{2} \leq 4 \mathrm{y} \text { and } \mathrm{y} \leq 1+\sqrt{\mathrm{x}}\right\}$ is :
(1) $\frac{5}{2}$
(2) $\frac{59}{12}$
(3) $\frac{3}{2}$
(4) $\frac{7}{3}$

**[JEE(Main)-2017]**
Q. Let $g(x)=\cos x^{2}, f(x)=\sqrt{x}$ and $\alpha, \beta(\alpha<\beta)$ be the roots of the quadratic equation $18 x^{2}-9 \pi x+\pi^{2}=0 .$ Then the area (in sq. units) bounded by the curve $y=(\operatorname{gof})$ (x) and the lines $x=\alpha, x=\beta$ and $y=0$ is-
( 1)$\frac{1}{2}(\sqrt{3}+1)$
(2) $\frac{1}{2}(\sqrt{3}-\sqrt{2})$
(3) $\frac{1}{2}(\sqrt{2}-1)$
( 4)$\frac{1}{2}(\sqrt{3}-1)$

**[JEE (Main)-2018]**
really helps revising

Excellent work bro

tq

thanku

Nice…..its very useful

It’s very helpful

Math

Is page per previous year question paper ke solution kyon nahin hai