Figure 1 shows a sketch of part of the curve with equation $y = \frac{10}{2x + 5\sqrt{x}}$, $x > 0$ The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the x-axis, and the lines with equations $x = 1$ and $x = 4$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 7

Using the trapezium rule:

$A \approx \frac{1}{2} h (y_0 + 2y_1 + 2y_2 + y_3)$

where $h$ is the width (1 in this case), and the $y$ values are:

• $y_0 = 1.42857$
• $y_1 = 0.90326$
• $y_2 = 0.68212$
• $y_3 = 0.55556$

Thus:

$A \approx \frac{1}{2} (1) (1.42857 + 2(0.90326) + 2(0.68212) + 0.55556)$

Calculating this:

$A \approx \frac{1}{2} (1.42857 + 1.80652 + 1.36424 + 0.55556)$

Giving an estimate of:

$A \approx \frac{1}{2} (5.15459) \approx 2.57730$

Finalizing to 4 decimal places, the area is approximately $2.5773$.

The estimate in part (b) is an overestimate. This is because the trapezium rule assumes that the function is linear between the intervals. Given that the curve is concave down (as seen in the figure), the actual area under the curve will be less than the area calculated using the trapezium rule.

Using the substitution $u = \sqrt{x}$ gives us $x = u^2$ and $dx = 2u \, du$. Hence, changing the limits as $x$ goes from 1 to 4 implies $u$ goes from 1 to 2. We rewrite the integral:

$\int_1^2 \frac{10}{2u^2 + 5u} (2u \,. du) = \int_1^2 \frac{20u}{2u^2 + 5u} \, du$

This can be simplified:

$= \int_1^2 \frac{20}{2u + 5} \, du$

Integrating gives:

$= 20\ln(2u + 5)\bigg|_1^2$

Substituting the limits:

$= 20(\ln(9) - \ln(7)) = 20 \ln\left(\frac{9}{7}\right)$

Thus, the exact value of the integral is $20 \ln\left(\frac{9}{7}\right)$.