Calculus

Application of Integrals

Easy

core

The area of the region $\{(x,y): x^2 \le y \le x\}$ is:

1 sq.units

$\dfrac{5}{6}$ sq.units

$\dfrac{1}{3}$ sq.units

$\dfrac{1}{6}$ sq.units

✅ Correct Option: 4

The curves $y = x^2$ and $y = x$ meet where $x^2 = x$ , i.e. at $x = 0$ and $x = 1$ .

Area $= \int_0^1 (x - x^2)dx = \dfrac{1}{2}-\dfrac{1}{3} = \dfrac{1}{6}$ sq. units.

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

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Area of the region bounded by $y = -1$ , $y = 2$ , $x = y^3$ and $x = 0$ is :

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The area of the region bounded by the lines $x+2 y=12, x=2, x=6$ and $x$ -axis is :

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Area of the region bounded by the curve $y = \sqrt{x}$ and lines $x + y = 2$ , $y = 0$ is

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

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Area bounded by the curves

$y=\begin{cases}x+2, & x\ge-1\\-x, & x<-1\end{cases}$ , $x=-2, x=3$ and $y=0$ is:

The area of the region {(x,y):x2≤y≤x}\{(x,y): x^2 \le y \le x\}{(x,y):x2≤y≤x} is: