Area Under the Curve Questions - JEE

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 bounded by $y=xe^x$, $y=xe^{-x}$ and $x=1$. [JEE Main 2018]

The area bounded by the axes of reference and the normal to y = loge x at (1, 0), is

Area of $A=\{(x,y)\,:\,|x|+|y|\le1,\,2y^2\ge x\}$. [JEE Main 2021]

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

The area (in sq. units) of the region described by $\{(x,y): y^2\leq2x\text{ and }y\geq4x-1\}$ is:

The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is:

The value of α – β is equal to

Area of the region enclosed between the locus of M and the pair of tangents on it from the origin, is

The area bounded by $y = |x| - 1$ and $y = 1 - |x|$ is ______

The area bounded by the curve $y = x^2$ and $y = \frac{-2}{(1 + x)}$ is

Area of the smaller region bounded by $x^2+y^2=4$ and the line $x+y=2$ is: [MAU019]

Find the area enclosed by the curve $[|x|] + [|y|] = 3$ where [.] denotes the greatest integer function.

If the area of the region $\left\{(x,y):\dfrac{a}{x^2}\leq y\leq\dfrac{1}{x},\ 1\leq x\leq2,\ 0<a<1\right\}$ is $(\log_e 2)-\dfrac{1}{7}$, then the value of $7a-3$ is equal to:

The area (in sq. units) of the region described by $\{(x, y) : y^2 \leq 2x$ and $y \geq 4x - 1\}$ is

If K is the area enclosed by two functions f and g then find the sum of digits in K, where$f = \max(|x|, |y|) = 8$$g = \min(|x|, |y|) = 3$

Let $f(x)$ be a positive function such that the area bounded by $y=f(x)$, $y=0$ from $x=0$ to $x=a>0$ is $e^{-a}+4a^2+a-1$. Then the differential equation whose general solution is $y=c_1f(x)+c_2$, where $c_1,c_2$ are arbitrary constants, is:

The area (in square units) of the region enclosed by the ellipse $x^2+3y^2=18$ in the first quadrant below the line $y=x/\sqrt{3}$ is:

The area (in sq. units) of the region $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid 0 \leq x \leq 3, 0 \leq y \leq 4, y \leq x^2 + 3x\}$ is ______ (up to three decimal places).

The area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y|\sqrt{1 - y^2}$ is

The area bounded by the curves $y = \left(x - 1\right)^2$, $y = \left(x + 1\right)^2$ and $y = \frac{1}{8}$ is

If $\begin{vmatrix} 4a^2 & 4a & 1 \\ 4b^2 & 4b & 1 \\ 4c^2 & 4c & 1 \end{vmatrix} \begin{bmatrix} f(-1) \\ f(1) \\ f(2) \end{bmatrix} = \begin{bmatrix} 3a^2 + 3a \\ 3b^2 + 3b \\ 3c^2 + 3c \end{bmatrix}$, $f(x)$ is a quadratic function and its maximum value occurs at a point $V$. $A$ is a point of intersection of $y = f(x)$ with $x$-axis and point $B$ is such that chord $AB$ subtends a right angle at $V$. Find the area enclosed by $f(x)$ and chord $AB$ (up to two decimal places).

The area bounded by the two curves $y = x^2 + 3x + 5$ and $y = -x^2 + 5x + 9$ for $x$ between $-1$ and $4$, is ______ (up to two decimal places).

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

If the area of the region $S=\{(x,y):2y-y^2\leq x^2\leq 2y,\ x\geq y\}$ is equal to $\dfrac{n+2}{n+1}-\dfrac{\pi}{n-1}$, then the natural number $n$ is equal to _______.

Area Under the Curve Questions (274)