Question

Find the area of the region bounded by \( y^2 = 4x \), \( x = 1 \), \( x = 4 \), and the x-axis in the first quadrant.

Hint:

To find the area under a curve between two points, integrate the function that describes the curve with respect to \( x \) over the given interval.

Answer 1

Step 1: Analyzing the curve. 
The given equation is \( y^2 = 4x \), which is a parabola opening to the right. The intersection of this curve with the x-axis occurs when \( y = 0 \), i.e., when \( x = 0 \). 
Step 2: Finding the points of intersection. 
The region we are concerned with is bounded by \( x = 1 \), \( x = 4 \), and the x-axis. These points correspond to the values of \( x \) between 1 and 4, where the parabola intersects the given lines. The equation of the curve gives \( y = \sqrt{4x} \) (considering the positive root, as we are in the first quadrant). 
Step 3: Setting up the integral. 
To find the area under the curve, we set up the integral of \( y = \sqrt{4x} = 2\sqrt{x} \) from \( x = 1 \) to \( x = 4 \): \[ A = \int_1^4 2\sqrt{x} \, dx \] Step 4: Solving the integral. 
Now, we solve the integral: \[ A = 2 \int_1^4 \sqrt{x} \, dx = 2 \int_1^4 x^{1/2} \, dx \] The integral of \( x^{1/2} \) is \( \frac{2}{3} x^{3/2} \), so we get: \[ A = 2 \left[ \frac{2}{3} x^{3/2} \right]_1^4 \] Evaluating the limits: \[ A = 2 \left( \frac{2}{3} \left[ (4)^{3/2} - (1)^{3/2} \right] \right) \] \[ A = 2 \times \frac{2}{3} \left( 8 - 1 \right) = \frac{4}{3} \times 7 = \frac{28}{3} \] Step 5: Conclusion. 
Thus, the area of the region is \( \frac{28}{3} \, \text{square units} \).