Question

Draw a rough sketch of the curve \( y = \sqrt{x} \). Using integration, find the area of the region bounded by the curve \( y = \sqrt{x} \), \( x = 4 \), and the x-axis, in the first quadrant.

Hint:

When finding the area under curves, set up the definite integral with the appropriate limits of integration and apply the power rule for integration.

Answer 1

The upper half of a parabola, \( y = \sqrt{x} \), forms the boundary of a region. The area beneath this curve between \( x = 0 \) and \( x = 4 \) is calculated using the definite integral: \[ A = \int_0^4 \sqrt{x} \, dx \] Rewriting \( \sqrt{x} \) as \( x^{1/2} \), the integral is: \[ A = \int_0^4 x^{1/2} \, dx \] The integration yields: \[ A = \left[ \frac{2}{3} x^{3/2} \right]_0^4 \] Evaluating at the limits: \[ A = \frac{2}{3} \left( 4^{3/2} - 0^{3/2} \right) \] With \( 4^{3/2} = 8 \), the result is: \[ A = \frac{2}{3} \times 8 = \frac{16}{3} \] Consequently, the region's area is \( \frac{16}{3} \, \text{square units}.