CUET Mathematics: CalculusApplication of Integrals. Free, no login required.

The area (in sq. units) of the bounded region $\{(x, y): x^2 \le y \le x\}$ is

The area of region $R = \{(x,y): x^2 + y^2 \le 16, y \le \sqrt{3}x, x \ge 0, y \ge 0\}$ is: (in Sq. units)

The area (in square units) of the region bounded by the curve $y^2 = 4ax\ (a > 0)$ and the lines $y = 2a, x = 0$ is equal to:

The area in the first quadrant bounded by $y = 9x^2$, $x = 0$, $y = 1$ and $y = 4$ is :

Area bounded by the curves $y = x|x|$, $x = -1$, $x = 1$ is

The supply function of a producer is given by $p = \frac{3}{5}e^{3x}$, where $x$ denotes a thousand units. If sales are 4000 units, then the producers' surplus is

The area bounded by the curve defined by $x = 2\cos\theta$ and $y = 3\sin\theta$; is [where $\theta$ is a parameter]

The area bounded by the curve $|-x|+|-y|=2$ is

Area bounded by the curves $y = \cos^2 x, x = 0, y = 0$ and $x = \pi$, is:

Area bounded by the curves $y = 3^x, y = x$ and $x = 1, x = 2$ is:

The area (in sq. units) of region bounded by the curves $y^2 = 4x$ and $y = x$ is:

The area of the region bounded by the line $3y = -2x + 7$, x-axis and the lines $x = 0$ and $x = 4$, is (in Sq. units)

The area bounded by the curve $y^2 = 2x$ and the lines $y = 2, y = 4$ and y-axis, is

The supply function for a commodity is $p = x^2 + x + 3$. The producer's surplus when $x = 3$ is,

The area of the region bounded between the line $x=3$ and the parabola $y^2=6x$ is:

The area of the smaller region in first quadrant bounded by $x^2+y^2=4$ and $x=\dfrac{3}{2}$ is:

The area bounded by $2x+y=8, y=a$ and $y=a+2$ with y-axis is 5 sq.units where $a>0$, then the value of $a$ is:

Area bounded by the curve $y^2 = 4x$ and the lines $y = 2$ and $x = 0$ is

If the area bounded by curves $y = 2^{kx}$, $x = 0$, $x = 2$ in first quadrant is $\frac{3}{\log 2}$, then $k$ is ($k \neq 0$)

The area of the region bounded by curves $y = |x-4|$, $x = 1$, $x = 5$ and the x-axis is

The area (in sq. units) bounded by a parabola $y^2=4x$ and its latus rectum, is

The area (in sq. units) bounded by the parabola $y^2 = x$ and line $x = 4$ in first quadrant is equal to $A$, then which of the following statements is/are TRUE ?

A. $A = \dfrac{16}{3}$

B. $A = \displaystyle\int_0^4 \sqrt{x}\, dx$

C. $A = 2\displaystyle\int_0^4 \sqrt{x}\, dx$

D. $A = \displaystyle\int_0^2 y^2\, dy$

Choose the correct answer from the options given below:

The area (in sq. units) bounded by the curves $x = y^2$ and $x = y$ in the first quadrant is:

The area of the region bounded by the curves $y = 1+|x-1|, x=-1, x=2, y=0$ is

The area (in sq. units) bounded between the curves $y = x^2$ and $y = 1$ in the first quadrant is

Area of the region bounded by the curves $y = x^2$, $x=0$, $x=2$ and $x-axis$ is

The area bounded by the line $y=x$, the $x-axis$ and the ordinates $x=-1$ and $x=2$ is

The area of the region bounded by the curve $y=x^3, y=8$ and $x=0$ is equal to :

The area of region bounded by the curves $4y = x^3, y = 0$ and $x = 4$ in the first quadrant is :

The area of the region bounded by the curves $x^2=4y$ and $x=4y$ is