Calculus

Application of Integrals

Medium

core

The area (in square units) bounded by the curve $y = x^2|x|$ , x-axis and the lines $x=-2$ and $x=2$ , is

8/3

✅ Correct Option: 4

$y=x^2|x|$ is non-negative everywhere and even, so the curve lies above the x-axis on $[-2,2]$ .

Area $=2\int_0^2x^3dx=2\left[\frac{x^4}{4}\right]_0^2=2(4)=8$ square units.

2026: 18 May Shift 2

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

2022: 17 Aug Shift 2

The area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ is :

2022: 23 Aug Shift 1

The area bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is :

2022: 16 July Shift 2

The smaller area enclosed by the curve $y = |x|$ and the circle $(x-a)^2 + y^2 = a^2$ is:

2025: 3 June Shift 1

The area of the smaller region of the circle $x^2 + y^2 = 8$ cut off by the line $x = 2$ is

2024: 16 May Shift 1

The area of the region bounded by the lines $\frac{x}{7 \sqrt{3} a}+\frac{y}{b}=4, x=0$ and $y=0$ is :

The area (in square units) bounded by the curve y=x2∣x∣y = x^2|x|y=x2∣x∣, x-axis and the lines x=−2x=-2x=−2 and x=2x=2x=2, is