Question

Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1$, $x = 4$ and the $x$-axis.

Hint:

Remember: For $y^2 = x$, use $y = \sqrt{x}$ when area is above $x$-axis.

Answer 1

Concept: The curve $y^2 = x$ represents a right-opening parabola. Since the region is bounded by the $x$-axis, we take the upper branch: \[ y = \sqrt{x} \] Step 1: {Set up the integral.}
Area between curve and $x$-axis from $x=1$ to $x=4$: \[ A = \int_{1}^{4} \sqrt{x} \, dx \]
Step 2: {Integrate.}
\[ \int \sqrt{x} \, dx = \int x^{1/2} dx = \frac{2}{3} x^{3/2} \]
Step 3: {Apply limits.}
\[ A = \left[\frac{2}{3} x^{3/2}\right]_{1}^{4} \] \[ = \frac{2}{3} \left(4^{3/2} - 1^{3/2}\right) \] \[ 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \] \[ A = \frac{2}{3}(8 - 1) = \frac{2}{3} \times 7 = \frac{14}{3} \]
Step 4: {Conclusion.}
\[ \boxed{\text{Area} = \frac{14}{3} \text{ square units}} \]