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 curve is \( y = \sqrt{x} \), which is the upper half of a parabola. The area under this curve from \( x = 0 \) to \( x = 4 \) can be found using the following definite integral: \[ A = \int_0^4 \sqrt{x} \, dx \] We know that: \[ \sqrt{x} = x^{1/2} \] So, the integral becomes: \[ A = \int_0^4 x^{1/2} \, dx \] Now, we can integrate: \[ A = \left[ \frac{2}{3} x^{3/2} \right]_0^4 \] Substituting the limits of integration: \[ A = \frac{2}{3} \left( 4^{3/2} - 0^{3/2} \right) \] Since \( 4^{3/2} = 8 \), we get: \[ A = \frac{2}{3} \times 8 = \frac{16}{3} \] Therefore, the area of the region is \( \frac{16}{3} \, \text{square units}. \)