To find the area bounded by the parabola [tex]\( y^2 = x \)[/tex] along the X-axis and the lines [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], we will follow these steps:
1. Express the curve [tex]\( y^2 = x \)[/tex] as [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
y = \sqrt{x}
\][/tex]
2. Set up the definite integral for the area:
The area under the curve [tex]\( y = \sqrt{x} \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex] can be found using the definite integral:
[tex]\[
\text{Area} = \int_{0}^{2} \sqrt{x} \, dx
\][/tex]
3. Evaluate the integral:
We need to find the antiderivative of [tex]\( \sqrt{x} \)[/tex]. Recall that [tex]\( \sqrt{x} \)[/tex] can be written as [tex]\( x^{1/2} \)[/tex]. The integral of [tex]\( x^{1/2} \)[/tex] is:
[tex]\[
\int x^{1/2} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2} + C
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
4. Apply the limits of integration:
[tex]\[
\left[ \frac{2}{3} x^{3/2} \right]_{0}^{2}
\][/tex]
5. Calculate the definite integral:
[tex]\[
\frac{2}{3}(2^{3/2}) - \frac{2}{3}(0^{3/2})
\][/tex]
Simplifying [tex]\( 2^{3/2} \)[/tex], we get:
[tex]\[
2^{3/2} = (\sqrt{2})^3 = 2\sqrt{2}
\][/tex]
Therefore:
[tex]\[
\frac{2}{3} \cdot 2\sqrt{2} - 0 = \frac{4\sqrt{2}}{3}
\][/tex]
So, the area bounded by the parabola [tex]\( y^2 = x \)[/tex] along the X-axis and the lines [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex] is:
[tex]\[
\boxed{\frac{4\sqrt{2}}{3}}
\][/tex]
This corresponds to option (B).
