15 (a) Show that $$ \sin x - \sin x \cos 2x \approx 2x^3 $$ for small values of $x$ - AQA - A-Level Maths Pure - Question 15 - 2021 - Paper 1

Given the equation:

$$y = \sqrt{8(\sin x - \sin x \cos 2x)} \approx \sqrt{8(2x^3)} = 4\sqrt{2} x^{\frac{3}{2}}.$$

To find the area between the curve and the x-axis from $x = 0$ to $x = 0.25$, we calculate:

$$\text{Area} = \int_0^{0.25} 4\sqrt{2} x^{\frac{3}{2}} dx$$

Calculating the integral:

$$= 4\sqrt{2} \left[ \frac{2}{5} x^{\frac{5}{2}} \right]_0^{0.25} = 4\sqrt{2} \cdot \frac{2}{5} \cdot \left(0.25^{\frac{5}{2}}\right) = 4\sqrt{2} \cdot \frac{2}{5} \cdot \frac{1}{32}.$$

This simplifies to:

$$\text{Area} = 2^m \times 5^n\text{ with } m=1\text{ and } n=5.$$

The integral

$$\int_{6.4}^{6.3} 2x^3 dx$$

is not a suitable approximation because it is evaluated over a decreasing interval ($6.4 > 6.3$), leading to a negative value. Also, the approximation of the original integral is only valid for small values of $x$, and this range is not small.

The integral

$$\int_{6.4}^{6.3} (\sin x - \sin x \cos 2x) dx$$

is periodic due to the nature of the sine and cosine functions. By evaluating the integral over a suitable interval, we can find that:

$$\int_{6.4}^{6.4-2\pi} (\sin x - \sin x \cos 2x) dx$$

will yield the same result. Thus, we can adjust the limits of integration to fit a manageable range for valid approximations. For suitable values, we find that $a=6.3-2\pi$ and $b=6.4-2\pi$ will allow the approximation by

$$\int_{a}^{b} 2x^3 dx.$$