Area Under the Curve Questions - JEE

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 $y=2x-8$

If f(x) = min{x2, sin(x/2), (x − 2π)2}, the area bounded by the curve y = f(x), x-axis, x = 0 and x = 2π is given by:Note: x1 is the point of intersection of the curves x2 and sin(x/2); x2 is the point of intersection of the curves sin(x/2) and (x − 2π)2

The area (in sq. units) bounded by the parabola $y = x^2 - 1$, the tangent at the point (2, 3) to it and the $y$-axis is __________ (up to three decimal places).

The area of the region $\left\{(x,y): y^2\le4x,\, x<4,\, \frac{xy(x-1)(x-2)}{(x-3)(x-4)}>0,\, x\ne3\right\}$ is

The area enclosed by the curves $xy+4y=16$ and $x+y=6$ is equal to:

The area of the region $A=\{(x,y):4x^2+y^2\leq8\text{ and }y^2\leq4x\}$ is:

The line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x − x2. Find m.

The area bounded between the parabolas $x^2 = \dfrac{y}{4}$ and $x^2 = 9y$, and the straight line $y = 2$ is

The parabolas $y^2 = 4x$ and $x^2 = 4y$ divide the square region bounded by the lines $x = 4$, $y = 4$ and the coordinate axes. If $S_1, S_2, S_3$ are, respectively, the areas of these parts numbered from top to bottom; then $S_1 : S_2 : S_3$ is

The area enclosed by the curve $x = 3a\cos^3 t$, $y = b\sin^3 t$ is:

The required area bounded by the curves $y = x^2 + 3x + 5$ and $y = -x^2 + 5x + 9$ between appropriate limits is (in square units):

Area of region $\{x \in R; x \geq 0, y \geq 0, y \geq x - 2 \text{ and } y \leq \sqrt{x}\}$ is equal to:

The area (in sq. units) of the region bounded by the curves y = 2x and y = |x + 1|, in the first quadrant is:

Given curves $y = kx^2$ and $x = ky^2$. If the area bounded by $y = kx^2$ and $x = ky^2$ is 1, then $k$ equals:

Given y2 = 4λx and y = λx, λ > 0. Find the value of the enclosed area (the answer is 24).

Find the area bounded by the curve $y = 2x - x^2$ and the straight line $y = -x$.

The area (in sq. units) of the region enclosed by the curves $C_1: y + 2x^2 = 0$ and $C_2: y + 3x^2 = 1$ is

The area of the region represented by $|x - y| \leq 2$ and $|x + y| \leq 2$ is:

The area (in sq. units) of the region $\{x \in \mathbb{R} : x \geq 0,\ y \geq 0,\ y \geq x - 2 \text{ and } y \leq \sqrt{x}\}$ is:

The area of the region bounded by the curves $y = |x - 1|$ and $y = 3 - |x|$ is

Find the required area bounded by $y^2 = x + 2$, $y = x - 2$, and the $y$-axis where the region passes through $(-2, 0)$ and $(0, 0)$ with vertex at $(2, 2)$ marked.

Find $f'(x) = 2\cos(2x) = 0 \Rightarrow 2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$. Hence find the required area.

Given $f(x) = -2xe^{-x}$ and $f'(x) = -2e^{-x} - 2e^{-x}\cdot(-x) = 2e^{-x}(x-1)$. Find the area as 2 sq. units.

The required area is $A = \int_{0}^{1} x^2(z - 1)^2 dx$

The curve is $y = \frac{x(x-a)}{x}$ which is a cubic polynomial.

Area Under the Curve Questions (274)