Figure 1 shows a sketch of part of the curve with equation $y = \frac{10}{2x + 5\sqrt{x}}$, where $x > 0$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 7

The trapezium rule states:

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

For our table, we have:
h = 1 (the width between each $x$ value),

$A \approx \frac{1}{2}(1.42857 + 2(0.90326) + 2(0.62312) + 0.55556)$

Calculating gives:

Since the trapezium rule approximates the area under the curve by using straight lines to connect the values of $y$, if the actual curve is concave (bending downwards) like in Figure 1, then our estimation will be an overestimate. This is because the trapezoids will extend above the curve in the regions $1 < x < 4$.

Using the substitution $u = \sqrt{x}$, hence $x = u^2$ and $dx = 2u \, du$. The limits of integration change from $x=1$ (where $u=1$) to $x=4$ (where $u=2$).

The integral becomes:

$\int_1^2 \frac{10}{2u^2 + 5u} \times 2u \, du$

This simplifies to:

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

Now integrating:

$= 20 \int \frac{1}{2u + 5} \, du = 20 \times \frac{1}{2} \ln |2u + 5| + C = 10 \ln |2u + 5| + C$

Calculating from $u=1$ to $u=2$:
$= 10 [\ln(2(2)+5) - \ln(2(1)+5)] \approx 10 [\ln(9) - \ln(7)] = 10 \ln\left(\frac{9}{7}\right).$

Thus, the exact value is:
$10 \ln\left(\frac{9}{7}\right).$