Area Under The Curve – JEE Main Previous Year Question with Solutions

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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]

Sol. (1)

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]

Sol. (1)

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]

Sol. (1)

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]

Sol. (3)

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]

Sol. (4)

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]

Sol. (1)

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]

Sol. (3)

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]

Sol. (3)

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]

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]

Sol. (4)

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]

Sol. (1)

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]

Sol. (3)

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]

Sol. (1)

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]

Sol. (4)