Calculus

Application of Integrals

Medium

common

The area of the region (in square units) bounded by $x=1, x=2$ and the curve $y^2 = 4x$ in the first quadrant is

$\frac{4}{3}[2\sqrt{2} - 1]$

$\frac{2}{3}[2\sqrt{2} - 1]$

$\frac{2}{3}[2\sqrt{2} + 1]$

$\frac{4}{3}[2\sqrt{2} + 1]$

✅ Correct Option: 1

The equation of the curve is $y^2 = 4x$ , which represents a rightward-opening parabola.

Since the region is restricted to the first quadrant, we take the positive root for $y$ :

$y = 2\sqrt{x}$

The area bounded by the curve between the lines $x = 1$ and $x = 2$ is given by the definite integral:

$\text{Area} = \int_{1}^{2} y \, dx$

$\text{Area} = \int_{1}^{2} 2\sqrt{x} \, dx$

Integrating with respect to $x$ :

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

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

Now, substitute the upper and lower limits:

$\text{Area} = \frac{4}{3} \left( 2^{3/2} - 1^{3/2} \right)$

$\text{Area} = \frac{4}{3} (2\sqrt{2} - 1) \text{ sq. units}$

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