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

Applying the trapezium rule to estimate the area under the curve: $A = \frac{1}{2} \times h (y_0 + 2y_1 + 2y_2 + y_3)$
where $h$ is the width of each trapezium, which is $ext{width} = 1$ (from $x=1$ to $x=4$).
Therefore:
$A = \frac{1}{2} \times (1) (1.42857 + 2\times0.90326 + 2\times0.68212 + 0.55556).$
Calculating gives:
$A = \frac{1}{2} (1.42857 + 1.80652 + 1.36424 + 0.55556) = \frac{1}{2} (5.15439) = 2.57719.$
Thus, the estimate for the area of $R$ is approximately 2.5772 to 4 decimal places.

The estimate in part (b) is an overestimate for the area of $R$.
This is because the trapezium rule approximates the area under a curve by forming trapezoids that typically lie above the actual curve if the function is concave down, which is evident from the graph. Therefore, the calculated area will be larger than the true area.

Using the substitution $u = \sqrt{x}$, we have $x = u^2$ and $dx = 2u du$.
The limits when $x = 1$ gives $u = 1$ and when $x = 4$, $u = 2$.
Thus, the integral transforms to:
$\int_1^2 \frac{10}{2(u^2) + 5u}(2u) du = \int_1^2 \frac{20u}{2u^2 + 5u} du.$
This simplifies to:
$\int_1^2 \frac{20u}{u(2u + 5)} du = \int_1^2 \frac{20}{2u + 5} du.$
Now integrate:
$= 10 \ln(2u + 5) \bigg|_1^2 = 10(\ln(9) - \ln(7)) = 10\ln\left(\frac{9}{7}\right).$
The exact value of the integral is thus $10\ln\left(\frac{9}{7}\right)$.