Calculus

Application of Integrals

Medium

common

The area of the region bounded by parabola $x^2 = 4y$ , straight line $x = 2$ and $x$ -axis, is

$\frac{4}{3}$ Sq. units

$\frac{2}{3}$ Sq. units

1 Sq. units

$\frac{5}{3}$ Sq. units

✅ Correct Option: 2

The equation of the parabola is $x^2 = 4y$ , which gives $y = \frac{x^2}{4}$ .

The area bounded by this curve, the line $x = 2$ , and the $x$ -axis (from $x = 0$ to $x = 2$ ) is given by:

$\text{Area} = \int_{0}^{2} y \, dx$

$\text{Area} = \int_{0}^{2} \frac{x^2}{4} \, dx$

$\text{Area} = \frac{1}{4} \left[ \frac{x^3}{3} \right]_{0}^{2}$

$\text{Area} = \frac{1}{12} \left[ 2^3 - 0^3 \right] = \frac{8}{12} = \frac{2}{3}\text{ sq. units}$

2025: 30 May Shift 1

The area (in sq. units) of the region bounded by the parabola $y^2 = 8x$ and the line $x = 2$ is

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$y(x)$ is strictly increasing in the interval

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Shown below is the graph of parabola $y^2=x$ , the area (in sq. units) of the shaded region is:

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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 :

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The area enclosed between $y^2 = 4x$ , $x = 1$ , $x = 4$ in first quadrant is :

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

The area of the region bounded by parabola x2=4yx^2 = 4yx2=4y, straight line x=2x = 2x=2 and xxx-axis, is