Figure 3 shows a sketch of part of the curve with equation y = 7x^2(5 - 2 \sqrt{x}), where x > 0 The curve has a turning point at the point A, where x > 0, as shown in Figure 3 - Edexcel - A-Level Maths Pure - Question 1 - 2018 - Paper 4

The curve crosses the x-axis at point B when y = 0. Starting from:

y = 7x^2(5 - 2 \sqrt{x})

Setting this equal to zero gives:

$7x^2(5 - 2 \sqrt{x}) = 0$

This results in two possibilities: either $x = 0$ or:

$5 - 2 \sqrt{x} = 0$

From the second equation, solving for \sqrt{x} leads to:

$\sqrt{x} = \frac{5}{2}$

Squaring both sides yields:

$x = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25$

Thus, the x-coordinate of point B is 6.25.

To find the area of region R, we use integration between the x-coordinates of points A and B. The area A can be calculated as follows:

$A = \int_{x=4}^{x=6.25} (7x^2(5 - 2\sqrt{x})) \, dx$

Carrying out the integration:

1. First, we can simplify the equation and integrate:

$\int (35x^2 - 14x^{5/2}) \, dx$

2. This integral results in:

$= \left[\frac{35x^3}{3} - \frac{14x^{7/2}}{\frac{7}{2}}\right]_{4}^{6.25}$

3. Evaluating this between the limits gives us:

• First evaluate at $x = 6.25$:

$\frac{35(6.25)^3}{3} - \frac{4(6.25)^{7/2}}{7}$

• Then evaluate at $x = 4$:

$\frac{35(4)^3}{3} - \frac{14(4)^{7/2}}{7}$

4. Subtract to find the area:

Finally, compute these values. The final answer when evaluated will yield:

$A \approx 172.23$

Thus, the area of the region R is approximately 172.23 square units.