1. **Problem Statement:** (i) Find the integral $\int \sqrt{4 + x} \, dx$ without a calculator. (ii) Given the curve $y = \sqrt{4 + x}$ and points $A$ on the y-axis and $B$ on the line $x=5$, find the area enclosed by the curve and line segment $AB$. 2. **Part (i): Evaluate the integral** We want to find: $$\int \sqrt{4 + x} \, dx = \int (4 + x)^{1/2} \, dx$$ 3. **Substitution:** Let $$u = 4 + x$$ Then $$du = dx$$ and the integral becomes: $$\int u^{1/2} \, du$$ 4. **Integrate using the power rule:** $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} + C = \frac{2}{3} u^{3/2} + C$$ 5. **Substitute back to $x$:** $$\int \sqrt{4 + x} \, dx = \frac{2}{3} (4 + x)^{3/2} + C$$ --- 6. **Part (ii): Finding area enclosed** - Point $A$ is on the y-axis where $x=0$, so $$y_A = \sqrt{4 + 0} = 2$$ - Point $B$ is at $x=5$, so $$y_B = \sqrt{4 + 5} = \sqrt{9} = 3$$ 7. **Equation of line $AB$:** - Slope: $$m = \frac{y_B - y_A}{x_B - x_A} = \frac{3 - 2}{5 - 0} = \frac{1}{5}$$ - Line passing through $A(0, 2)$: $$y = m x + 2 = \frac{1}{5} x + 2$$ 8. **Area enclosed by the curve and line $AB$ between $x=0$ and $x=5$: ** Area $= \int_0^5 \left( \sqrt{4 + x} - \left( \frac{x}{5} + 2 \right) \right) dx$ 9. **Calculate each integral separately:** - From part (i): $$\int_0^5 \sqrt{4 + x} \, dx = \left[ \frac{2}{3} (4+x)^{3/2} \right]_0^5 = \frac{2}{3} (9)^{3/2} - \frac{2}{3} (4)^{3/2}$$ Calculate powers: $$9^{3/2} = (9^{1/2})^3 = 3^3 = 27$$ $$4^{3/2} = (4^{1/2})^3 = 2^3 = 8$$ So: $$= \frac{2}{3} (27 - 8) = \frac{2}{3} (19) = \frac{38}{3}$$ 10. **Integral of line $AB$ term:** $$\int_0^5 \left( \frac{x}{5} + 2 \right) dx = \int_0^5 \frac{x}{5} dx + \int_0^5 2 dx = \frac{1}{5} \int_0^5 x dx + 2 \times 5$$ Calculate: $$\int_0^5 x dx = \left[ \frac{x^2}{2} \right]_0^5 = \frac{25}{2}$$ So: $$= \frac{1}{5} \times \frac{25}{2} + 10 = \frac{25}{10} + 10 = 2.5 + 10 = 12.5$$ 11. **Final area:** $$\text{Area} = \frac{38}{3} - 12.5 = \frac{38}{3} - \frac{25}{2} = \frac{76}{6} - \frac{75}{6} = \frac{1}{6}$$ **Answer:** The area enclosed is $\frac{1}{6}$ square units.