Area Under the Curve Questions - JEE

If the area enclosed between $f(x) = \min\left\{\cos^{-1}(\cos x), \cot^{-1}(\cot x)\right\}$ and the x-axis in $x \in \left(\frac{k\pi}{2}, \frac{(k+1)\pi}{2}\right)$ where $k \in \mathbb{N}$, then k is equal to

Let A be the area of the region $\{(x,y):y\geq x^2,\,y\geq(1-x)^2,\,y\leq2x(1-x)\}$. Then $540A$ is equal to ___.

Let the area of the region $\{(x,y):|2x-1|\leq y\leq|x^2-x|,\,0\leq x\leq1\}$ be $A$. Then $(6A+11)^2$ is equal to ___.

Let $A = \{(x, y) : y^2 \leq 4x,\ y - 2x \geq -4\}$. The area (in square units) of the region $A$ is

Let the line $x=-1$ divide the area of the region $\{(x,y):1+x^2\leq y\leq3-x\}$ in the ratio $m:n$, $\gcd(m,n)=1$. Then $m+n$ is equal to

Let $S(\alpha) = \{(x, y) : y^2 \leq x,\ 0 \leq x \leq \alpha\}$ and $A(\alpha)$ is area of the region $S(\alpha)$. If for a $\lambda$, $0 < \lambda < 4$, $A(\lambda) : A(4) = 2 : 5$, then $\lambda$ equals:

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

If ABC is an isosceles triangle inscribed in a circle of radius r. If AB = AC and h is altitude from A to BC then the triangle ABC has perimeter $P = 2(\sqrt{2hr - h^2} + \sqrt{2hr})$, calculate area A and $\lim_{h \to 0} \dfrac{A}{P^3}$.

The area between the parabola $x=4y-y^2$ and the line $x=y$ is: [MAU014]

The area of the region enclosed by $y^2\le4x$ and $x\le4$. [JEE Main 2022]

Let g(x) = cos x², f(x) = √x and α, β (α < β) be the roots of the quadratic equation 18x² − 9πx + π² = 0. Then, the area (in sq. units) bounded by the curve y = (g ∘ f)(x) and the lines x = α, x = β and y = 0 is

The area (in sq. units) of the region bounded by the curve x² = 4y and the straight line x = 4y − 2 is:

The area of the region of the xy plane defined by the inequality $|x| + |y| + |x + y| \leq 1$ is

The area bounded by the curve y = f(x), the coordinate axes and the line x = x₁ is given by x₁eˣ¹ - 1. Therefore f(x) equals:

Find the area of the region lying inside $x^2 + (y-1)^2 = 1$ and outside $c^2x^2 + y^2 = c^2$, where $c = \sqrt{2} - 1$.

The area bounded by y = x² + 2 and y = 2|x| – cos x is equal to

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

If area bounded by the line y = x, curve y = f(x) and lines x = 1, x = t, is t/2(2t – 1 + 1/t + t) then f(x) =

Suppose y = f(x) and y = g(x) are two functions whose graphs intersect at the three points (0, 4), (2, 2) and (4, 0) with f(x) > g(x) for 0 and f(x) for 2 . If ∫₀⁴ [f(x) - g(x)]dx = 10 and ∫₂⁴ [g(x) - f(x)]dx = 5, the area between two curves for 0 is:

The area enclosed between the curves $|x| + |y| \geq 2$ and $y^2 = 4\left(1 - \frac{x^2}{9}\right)$ is:

Let A be the area bounded by the curve $y=x|x-3|$, the x-axis and the ordinates $x=-1$ and $x=2$. Then $12A$ is equal to ___.

First draw the graph of $[x] + [y] = 3$, where $[\cdot]$ denotes the greatest integer function. Then find the area of the graph of $\lfloor |x| \rfloor + \lfloor |y| \rfloor = 3$.

The area of the region above the $x$-axis bounded by the curve $y = \tan x,\ 0 \leq x \leq \dfrac{\pi}{2}$ and the tangent to the curve at $x = \dfrac{\pi}{4}$ is

Area of $y=|x^2-4|$ from $x=-3$ to $x=3$. [JEE Main 2022]

Find the area of the region bounded by the square ABCD with area 2 sq units and a circle with radius $\frac{1}{2}$ sq units inscribed in it.

Area enclosed by the curve $|x + y - 1| + |2x + y + 1| = 1$ is

Let for $x\in\mathbb{R}$, $f(x)=\dfrac{x+|x|}{2}$ and $g(x)=\begin{cases}x, & x<0\\x^2, & x\geq0\end{cases}$. Then the area bounded by the curve $y=(f\circ g)(x)$ and the lines $y=0$, $2y-x=15$ is equal to ___.

Area bounded by $y=|x-1|$ and $y=1$. [JEE Main 2014]

Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-px+\dfrac{5}{4}p=0$ are rational. Then the area of the region $\{(x,y):0\leq y\leq(x-q)^2,\,0\leq x\leq q\}$ is:

The area bounded by $y=|x-1|+|x-3|$ and the $x$-axis between $x=0$ and $x=4$ is: [MAU011]

The area enclosed between the curves $y^2 = x$ and $y = |x|$ is

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

If the area of the region bounded by the curves $y^2-2y=-x$, $x+y=0$ is $A$, then $8A$ is equal to ___.

The area (in sq. units) of the region bounded by $y=x^2+2$ and the lines $y=x$, $x=0$ and $x=3$ is: [MAU004]

Let $y=p(x)$ be the parabola passing through the points $(-1,0)$, $(0,1)$ and $(1,0)$. If the area of the region $\{(x,y):(x+1)^2+(y-1)^2\leq 1,\ y\leq p(x)\}$ is $A$, then $12(\pi-4A)$ is equal to ________.

Given the equation of parabola $(y-2)^2 = (x-1)$, find the area bounded by the parabola and its tangent at $P(2, 3)$ and the line $x = 2y - 4$.\[\Delta = \int_0^3 \left[(y-2)^2 + 1 - (2y - 4)\right] dy\]

Let the area of the region {(x, y) : 2y \le x 2 + 3, y + ∣x∣ \le 3, y \ge ∣x - 1∣} be A. Then 6 A is equal to :

The minimum area bounded by $y = g(x)$ and $y = f(x)$ is:

If area bounded by $|x + 2y| + |2x - y| = p$ is P then area bounded by $|x + 3y| + |3x - y| = 2p$ is KP, then the value of [K] is, where [K] is G.I.F.

Line x = 0 divides the region mentioned above in two parts. The ratio of area of left hand side of line to that of right hand side of line is

The area of the region bounded by the curve and lines x = 0 and x = 1/2 is

The area of the region, inside the circle $(x-2\sqrt{3})^2+y^2 = 12$ and outside the parabola $y^2 = 2\sqrt{3}\,x$ is:

The area of the smaller portion enclosed between the curves $x^2 + y^2 = 4$ and $y^2 = 3x$ is

The area (in sq. units) of the region bounded by the curve $x = |y|\sqrt{1-y^2}$ and the curve $x = y^2 - 1$ is:

Area bounded by the curve $y = (x - 1)(x - 2)(x - 3)$ and X-axis lying between the ordinates $x = 0$ and $x = 3$ is equal to

Area of the region $\{(x,y):\ x^2+(y-2)^2\leq 4,\ x^2\geq 2y\}$ is

Let the area enclosed by the lines $x+y=2$, $y=0$, $x=0$ and the curve $f(x)=\min\!\left\{x^2+\dfrac{3}{4},\ 1+[x]\right\}$ where $[x]$ denotes the greatest integer $\leq x$, be $A$. Then the value of $12A$ is

Let the area of the region $\{(x,y): x-2y+4\ge0,\, x+2y^2\ge0,\, x+4y^2\le8,\, y\ge0\}$ be $\frac{m}{n}$, where $m$ and $n$ are coprime numbers. Then $m+n$ is equal to

The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ in two parts. The area of the smaller part is equal to:

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y'(X)(X-x)$ and the co-ordinate axes, where $(x,y)$ is any point on the curve, is always $\frac{-y^2}{2Y'(x)}+1$, $Y'(x)\ne0$. If $Y(1)=1$, then $12Y(2)$ equals

Area Under the Curve Questions (274)