Area Under the Curve Questions - JEE

The area of the region $\{(x,y):\ x^2\leq y\leq 8-x^2,\ y\leq 7\}$ is

If the area of the larger portion bounded between the curves $x^2+y^2 = 25$ and $y = |x-1|$ is $\dfrac{1}{4}(b\pi+c)$, $b,c\in\mathbb{N}$, then $b+c$ is equal to

The area (in sq. units) of the region $\{(x, y) : y^2 \geq 2x$ and $x^2 + y^2 \leq 4x,\ x \geq 0,\ y \geq 0\}$ is

Area enclosed by the curves $y = x^2 + 1$ and a normal drawn to it with gradient $–1$ is equal to:(a) $\frac{2}{3}$(b) $\frac{1}{3}$(c) $\frac{3}{4}$(d) $\frac{4}{3}$

Let the area of the region $\{(x,y): 2y\leq x^2+3,\; y+|x|\leq 3,\; y\geq|x-1|\}$ be $A$. Then $6A$ is equal to:

If the area of the region $\{(x,y): -1\leq x\leq 1,\; 0\leq y\leq a+e^{|x|}-e^{-x},\; a>0\}$ is $\dfrac{e^2+8e+1}{e}$, then the value of $a$ is:

Consider the region $R = \left\{(x,y): x\leq y\leq 9-\dfrac{11}{3}x^2,\; x\geq 0\right\}$. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:

The area of the region $\{(x,y): x^2+4x+2\leq y\leq |x+2|\}$ is equal to

96. Area bounded by the curve $f(x) = \dfrac{x^2 - 1}{x^2 + 1}$ and the line $y = 1$ is:

The area bounded by the curve $y = 2x - x^2$ and the straight line $y = -x$ is given by

If the area of the region bounded by the curves, y = x², $y = \dfrac{1}{x}$ and the lines y = 0 and x = t (t > 1) is 1 sq. unit, then t is equal to

The area of the region $A = \{(x, y) : 0 \leq y \leq x|x| + 1 \text{ and } -1 \leq x \leq 1\}$ in sq. units, is:

Given region $A = \{(x, y): 0 \leq y \leq x|x| + 1 \text{ and } -1 \leq x \leq 1\}$. The area of region $A$ is:

Area bounded by the parabola $(y-2)^2 = x - 1$, the tangent to it at the point P (2, 3) and the x-axis is equal to

The area of region $R$ that is completely bounded by the graph of $f(x) = 2x - 1$ and $g(x) = x^2 - 4$ is ______ (up to two decimal places).

If $x = a(1-t^2)$, $y = a(t - t^3/3)$ (or similar parametric form related to the solution shown), then the area enclosed by the loop of the curve is:

The area of the region described by $A - \{(x, y) : x^2 + y^2 \leq 1$ and $y^2 \leq 1 - x\}$ is

Consider a square with vertices at $(1, 1)$, $(-1, 1)$, $(1, -1)$ and $(-1, -1)$. Let S be the region consisting of all points inside the square which are nearer to the origin than to any edges. Sketch the region S and find its area.

Area of the region bounded by $y=\dfrac{|4x-x^2|}{2}$ and $y=x-1$ above x-axis. [JEE Main 2019]

Area bounded by $y=(x-2)^2$ and $y=4$. [JEE Main 2021]

Area bounded by $y=x^3-3x^2+2x$ and $y=0$. [JEE Main 2021]

The area of the region bounded by the ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{9}=1$ in the first quadrant is: [MAU010]

The area (in sq. units) bounded by $y=x^2-1$, tangent at $(2,3)$ and x-axis. [JEE Main 2019]

The area of the region $R=\{(x,y)\,:\,x^2\le y\le |x|\}$ is: [MAU016]

Using matrix multiplication, consider the quadratic equation $4x^2 f(-1) + 4x f(1) + f(2) = 3x^2 + 3x$. It is given that this equation has three roots $x = a, b, c$; hence it is an identity. Therefore $f(-1) = \dfrac{3}{4}$, $f(1) = \dfrac{3}{4}$ and $f(2) = 0$, giving $f(x) = \dfrac{4 - x^2}{4}$. Let point $A$ be $(-2, 0)$ and $B$ be $(2t, -t^2 + 1)$ and maximum value of $f(x) = 1$ at $x = 0$. Now, as $AB$ subtends a right angle at the vertex $V(0, 1)$, find the required area (in sq. units) given by $A = \displaystyle\int_{-2}^{8} \left(\dfrac{4 - x^2}{4} + \dfrac{3x + 6}{2}\right) dx$.

The area of the region $A = \{(x, y) \in R \times R \mid 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}$ is equal to:

The area bounded by the parabola $y=x^2-4x+3$ and the $x$-axis is: [MAU012]

Calculate the area bounded by the curve y = x(3 − x)2, the x-axis and the ordinates of the maximum and minimum points of the curve.

Area bounded by $y=\sqrt{x}$ and $y=x^2$ on $[0,1]$. [JEE Main 2022]

The area enclosed by the curves $y^2+4x=4$ and $y-2x=2$ is:

The area (in sq. units) of the region described by $A = \{(x, y)\mid y \geq x^2 - 5x + 4,\ x + y \geq 1,\ y \leq 0\}$ is

Let $f(x) = \max\left\{\sin x,\, \cos x,\, \dfrac{1}{2}\right\}$ then determine the area of region bounded by the curves $y = f(x)$, $x$-axis, $y$-axis and $x = 2\pi$.

Find the area of the region containing the points satisfying $|y| + \dfrac{1}{2} \leq e^{-|x|}$; $\max(|x|, |y|) \leq 2$.

The area bounded by the curves $y = \cos x$ and $y = \sin x$ between the coordinates $x = 0$ and $x = \dfrac{3\pi}{2}$ is

Let $R = \{(x, y) : 0 \le y \le x^2 + 1,\; 0 \le y \le x + 1,\; 0 \le x \le 2\}$Find the area of region $R$.

The area of the region bounded by $y=\sqrt{1-x^2}$ and $y=1-x$ in the first quadrant is: [MAU013]

The area bounded by $y^2=4x$ and $x=1$ is: [MAU017]

If (a, 0); a > 0 is the point where the curve y = sin 2x – 3 sinx cuts the x-axis first, A is the area bounded by this part of the curve, the origin and the positive x-axis, then

The area bounded by the parabolas $y=(x+1)^2$ and $y=(x-1)^2$ and the line $y=\frac{1}{4}$ is: [MAU002]

Area of $\{(x,y)\,:\,0\le y\le x^2+1,\,0\le y\le x+1,\,0\le x\le2\}$. [JEE Main 2021]

A function y = f(x) satisfies the differential equation dy/dx – y = cos x – sin x, with initial condition that y is bounded when x → ∞. The area enclosed by y = f(x), y = cos x and the y-axis in the 1st quadrant is

The area bounded by the curve y = f(x), x-axis and the ordinates x = 1 and x = b is (b − 1) sin(3b + 4). Find f(x).

The area of the region $\{(x,y)\,:\,y^2\le4x,\,4x^2+4y^2\le9\}$. [JEE Main 2020]

Find the area bounded by the curve satisfying $\frac{dy}{dx} = 2x + 1$ that passes through $(1, 2)$, the x-axis from $x = 0$ to $x = 1$.

The area (in sq units) of the region $A = \{(x, y):|x| + |y| \leq 1, 2y^2 \geq |x|\}$ is

The area of the region bounded by x^2 = 4y, y = 2, y = 4 and the Y-axis in the first quadrant is

The graphs of f(x) = x² and g(x) = cx³ (c > 0) intersect at the points (a, 0) & (1/c, 1/c²). If the region which lies between these & over the interval [0, 1/c] has the area equal to 2/3 then the value of c is

The area between curve y = 2x⁴ – x², x-axis and the ordinates of the two minima of the curve is

The area of the region bounded in first quadrant by $y = x^{1/3}$, $y = -x^2 + 2x + 3$, $y = 2x - 1$ and the axis of ordinates is

The area (in square units) of the region bounded by the parabola $y^2=4(x-2)$ and the line $y=2x-8$

Area Under the Curve Questions (274)