Area Under the Curve Questions - JEE

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.

The length of sub-normal at any point P(x, y) on the curve, which is passing through M(0, 1) is unity. The area bounded by the curves satisfying this condition is equal to

The area (in sq. units) of the region bounded by $y^2 - 2x \geq 0$, $x^2 + y^2 - 4x \leq 0$, $x \geq 0$, $y \geq 0$ is

The area enclosed between the curves $y = \log_e(x + e)$, $x = \log_e\left(\frac{1}{y}\right)$ and the x-axis is

The area of the region bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b, where -\infty , is

Given, S(a) = \{(x, y) : y^2 \leq x, 0 \leq x \leq a\} and A(a) is the area of the region S(a). If \frac{A(l)}{A(4)} = \frac{2}{5} for 0 , find the value of l.

The area bounded by the curve y = \sin x between x = 0 and x = 2\pi is

The area enclosed by the curve \ y = \dfrac{x^2-1}{x^2+1} and the line \ y = 1 is:

Given region $S(\alpha) = \{(x, y) : y^2 \leq x,\ 0 \leq x \leq \alpha\}$ If for a $\lambda$, \(0

The ratio of areas of the figures bounded by line segments A1A2, A2A3 and the graph of the polynomial is

The area bounded by $y=x|x|$, $x$-axis and the ordinates $x=-1$ and $x=1$ is: [MAU003]

95. Let a function $f(x)$ be defined in $[-2, 2]$ as \(f(x) = \begin{cases} \{x\}, & -2 \leq x

Area bounded by $y=x(x-1)(x-2)$ and $y=0$ over $[0,2]$. [JEE Main 2020]

The area bounded by $y = xe^{|x|}$ and lines $|x| = 1, y = 0$ is

If the abscissa $x = a$ divides the area bounded by the X-axis part of the curve $y = 1 + \frac{8}{x^2}$ and the abscissa $x = 2, x = 4$ into two equal parts, then $a$ is equal to

Area bounded by the curve $y = x \sin x$ and X-axis between $x = 0$ and $x = 2\pi$ is

(A) Drawing graphs of $y = x^2$ and $y = \frac{4}{x^2-1}$

The area between the curve $y = 2x^2 - s^2$, the $x$-axis and the ordinates of the two minima of the curve is

The area (in sq units) of the largest rectangle ABCD whose vertices A and B lie on the X-axis and vertices C and D lie on the parabola, $y = x^2 - 1$ below the X-axis, is

If the area bounded by $f(x) = \frac{x^2}{3} - x + a$ and the straight lines $x = 0$, $x = 2$ and the X-axis is minimum, then the value of $a$ is

The area of the region, enclosed by the circle $x^2 + y^2 = 2$ which is not common to the region bounded by the parabola $y^2 = x$ and the straight line $y = x$, is

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

A triangle has one vertex at (0, 0) and the other two on the graph of $y = -2x^2 + 54$ at $(x, y)$ and $(-x, y)$ where \(0

Consider the following regions in the plane:$R_1 = \{(x, y) : 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1\}$$R_2 = \{(x, y) : x^2 + y^2 \leq \frac{4}{3}\}$The area of the region $R_1 \cap R_2$ can be expressed as $\frac{a\pi}{9} + \frac{b}{3}$, where $a$ and $b$ are integers. Find the value of $a + b$.

Let \[ f(x) = \begin{cases} 2x, & -1 \le x \le 1 \\ x^2 + ax + b, & x > 1,\; x f(x) is continuous, find the area of the region bounded by the curves y = f(x), x = −2y2, and relevant boundaries (in sq. units).

The area bounded by $y=\sin x$, $y=\cos x$ and the $x$-axis in $[0,\pi/2]$ is: [MAU015]

The area bounded by the curves $y = \ln z$, $y = |x|z$, $|x| \ln z$ and $y = |\ln z|$, for $z \in (-1, 1)$ is

Area Under the Curve Questions (274)