Given that $y = \frac{1}{4 + \sqrt{(x-1)}}$, complete the table below with values of $y$ corresponding to $x = 3$ and $x = 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2011 - Paper 6

Using the trapezium rule, we can estimate the integral $I = \int_{2}^{5} f(x) \, dx$ as follows:

$I \approx \frac{1}{2} \times (b-a) \times (y_0 + y_n) + (y_1 + y_2 + y_3)$

Where:

• $y_0 = 0.2$ (for $x = 2$)
• $y_1 \approx 0.1847$ (for $x = 3$)
• $y_2 \approx 0.1745$ (for $x = 4$)
• $y_3 \approx 0.1667$ (for $x = 5$)

Calculating the estimate:

$I \approx \frac{1}{2} \times (5 - 2) \times (0.2 + 0.1667 + 0.1847 + 0.1745)$ $I \approx \frac{3}{2} \times (0.7260) = 1.0890$

Thus, the estimated value of $I$ is approximately $1.089$.

To evaluate the integral $I = \int_{2}^{5} \frac{1}{4 + \sqrt{(x-1)}} \, dx$, we make the substitution $x = (u-4)^{2} + 1$, which implies:

$dx = 2(u-4) \, du$

Now we re-evaluate the bounds:

• When $x = 2$, $u = 3$;
• When $x = 5$, $u = 5$.

Substituting into the integral, we have:

$I = \int_{3}^{5} \frac{2(u-4)}{4 + \sqrt{((u-4)^{2}+1-1)}} \, du = \int_{3}^{5} \frac{2(u-4)}{\sqrt{u}} \, du$

Evaluating yields:

$= 2 \int_{3}^{5} (2u - 8) \, du = 2 \left[\frac{u^2}{2} - 8u\right]_{3}^{5} = \left[(25 - 40) - (4.5 - 24)\right] = 2 + 8 \left(\frac{5 - 3}{6}\right) = \frac{3}{2} (\frac{25 -12}{5}) = 2 + \frac{8}{6} = 2 + \frac{4}{3} = 2.667$

Thus, the exact value of $I$ is:

$I = 2 + \frac{4}{3} = \frac{10}{3}$