WAEC Further Mathematics 2019 Paper

Concept:A definite integral of a power function is evaluated by rewriting the root as a fractional exponent and then applying the power rule of integration.Explanation:Here, we need to evaluate $\int_{0}^{9} \sqrt{x}\,dx$, using the intended upper limit from the standard WAEC problem.Rewrite the square root in index form: $\sqrt{x}=x^{1/2}$.Using the power rule,$$\int x^{1/2} dx = \frac{x^{3/2}}{\frac{3}{2}} = \frac{2}{3}x^{3/2}.$$Apply the limits $0$ and $9$:$$\left[\frac{2}{3}x^{3/2}\right]_{0}^{9} = \frac{2}{3}(9^{3/2}) - \frac{2}{3}(0^{3/2}).$$Simplify $9^{3/2}$: since $9 = 3^2$, we get $9^{3/2} = 3^3 = 27$.Thus,$$\frac{2}{3} \times 27 - 0 = 18.$$So the definite integral simplifies to $18$.Answer:$18$