In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 9 - 2022 - Paper 2

The area $A$ under the curve from $x = 2$ to $x = 4$ can be expressed as: $A = \int_{2}^{4} \frac{(x - 2)(x - 4)}{4\sqrt{x}} \, dx$

First, we simplify the integrand: $\frac{(x - 2)(x - 4)}{4\sqrt{x}} = \frac{x^2 - 6x + 8}{4\sqrt{x}} = \frac{1}{4} \left( x^{\frac{1}{2}}(x - 6 + \frac{8}{x})\right)$ Now apply the integral: $A = \frac{1}{4} \int_{2}^{4} (x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 8x^{-\frac{1}{2}}) \, dx$

Solving the integral term by term: $\int x^{\frac{3}{2}} \, dx = \frac{2}{5}x^{\frac{5}{2}}, \quad \int x^{\frac{1}{2}} \, dx = \frac{2}{3}x^{\frac{3}{2}}, \quad \int x^{-\frac{1}{2}} \, dx = 2\sqrt{x}$ Evaluating these from $2$ to $4$, we have: $\left[ \frac{2}{5}x^{\frac{5}{2}} - 6 \cdot \frac{2}{3}x^{\frac{3}{2}} + 16\sqrt{x} \right]_{2}^{4}$

Now, substituting the limits: At $x = 4$: $\frac{2}{5}(32) - 6 \cdot \frac{2}{3}(8) + 16(2) = \frac{64}{5} - 32 + 32 = \frac{64}{5}$ At $x = 2$: $\frac{2}{5}(8) - 6 \cdot \frac{2}{3}(4) + 16(\sqrt{2}) = \frac{16}{5} - 16 + 16\sqrt{2} = \frac{16}{5} - 16 + 16\sqrt{2} = \frac{16 - 80 + 80\sqrt{2}}{5}$

Combining the results from $x = 4$ and $x = 2$ gives the area: $$ A = \left(\frac{64}{5} - \left(\frac{16}{5} - 16 + 16\sqrt{2}\right)\right) = \sqrt{2} \text{ terms combine to give the area} $