The area bounded by the x-axis and the parabola y = 3x-x^2 is: Option 1: 27/2 Sq.unit Option 2: 9/2 Sq.unit Option 3: 15/2 Sq.unit Option 4: 11/2 Sq.unit

Answer: 9/2 Sq.unit. Solution: The area bounded by the x-axis and the parabola is the region between the curve and the x-axis. The parabola touches the x-axis when y = 0. 3x - x^2 = 0 x(3 - x) = 0 This gives x = 0 and x = 3. The area is between x = 0 and x = 3. Area = _0^3 (3x - x^2) dx (3x - x^2) dx = 3x^2/2 - x^3/3 Area = [3x^2/2 - x^3/3]_0^3 At x = 3: = (3(3)^2)/2 - ((3)^3)/3 = (3 x 9)/2 - 27/3 = 27/2 - 9 At x = 0: = (3(0)^2)/2 - ((0)^3)/3 = 0 Area = 27/2 - 9 - 0 = 27/2 - 18/2 = 9/2 square units Therefore, the area bounded by the x-axis and the parabola is 9/2 square units.

Calculus

Application of Integrals

Medium

applied

The area bounded by the x-axis and the parabola $y = 3x-x^2$ is:

$\frac{27}{2}$ Sq.unit

$\frac{9}{2}$ Sq.unit

$\frac{15}{2}$ Sq.unit

$\frac{11}{2}$ Sq.unit

✅ Correct Option: 2

The area bounded by the x-axis and the parabola is the region between the curve and the x-axis.

The parabola touches the x-axis when $y = 0$ .

$3x - x^2 = 0$

$x(3 - x) = 0$

This gives $x = 0$ and $x = 3$ .

The area is between $x = 0$ and $x = 3$ .

$\text{Area} = \int_0^3 (3x - x^2) \, dx$

$\int (3x - x^2) dx = \frac{3x^2}{2} - \frac{x^3}{3}$

$\text{Area} = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3$

At $x = 3$ :

$= \frac{3(3)^2}{2} - \frac{(3)^3}{3}$

$= \frac{3 \times 9}{2} - \frac{27}{3}$

$= \frac{27}{2} - 9$

At $x = 0$ :

$= \frac{3(0)^2}{2} - \frac{(0)^3}{3}$

$= 0$

$\text{Area} = \frac{27}{2} - 9 - 0$

$= \frac{27}{2} - \frac{18}{2}$

$= \frac{9}{2}$ square units

Therefore, the area bounded by the x-axis and the parabola is $\frac{9}{2}$ square units.

2026: 15 May Shift 1

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:

2023: 15 June Shift 2

Area of the region bounded by the curve $|x| + |y| = 1$ and x-axis is :

2023: 30 May Shift 3

The area enclosed between the curve $y = x^2 + 2$ and x-axis between $x = 0$ and $x = 3$ is :

2025: 26 May Shift 2

Area (in sq. units) of the region bounded by the curves $y = -1$ , $y = 2$ , $x = y^3$ and $x = 0$ is

2026: 19 May Shift 2

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

2022: 16 July Shift 2

The line $y = mx$ ( $m > 0$ ) partitions the area of the circle $x^2 + y^2 = a^2$ ( $a > 0$ ) in the ratio:

The area bounded by the x-axis and the parabola y=3x−x2y = 3x-x^2y=3x−x2 is:

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The area bounded by the x-axis and the parabola $y = 3x-x^2$ is: