Calculus

Application of Integrals

Easy

applied

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

$\dfrac{2}{3}$

$\dfrac{4}{3}$

$\dfrac{1}{3}$

$\dfrac{8}{3}$

✅ Correct Option: 1

In the first quadrant the curves meet at $x=1$ , and between $x=0$ and $x=1$ the line $y=1$ lies above the parabola. So the area is $\int_0^1(1-x^2)dx=1-\frac{1}{3}=\frac{2}{3}$ .

2023: 22 May Shift 3

The area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is given by :

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The area of the region bounded by the curve $y = x + 1$ , $x = axis$ and the lines $x = 2$ and $x = 3$ is

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The area (in Sq. units) of the region bounded by $y = -2$ , $y = 2$ , $x = y^3$ and $x = 0$ is equal to

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The area (in square units) of the region bounded by the curve $x = y^3, y = -1, y = 2$ and $x = 0$ is:

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Area of the region bounded by the curve $y = \sin x$ and x-axis between $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ is

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The area (in sq. units) bounded by the parabola $y^2 = 4ax$ , its latus rectum and the $x$ -axis in the first quadrant is:

The area (in sq. units) bounded between the curves y=x2y = x^2y=x2 and y=1y = 1y=1 in the first quadrant is

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

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