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

Using the result from part (a), we can revise the expression under the square root:

$y = \sqrt{8(2x^3)} = \sqrt{16x^3} = 4x^{3/2}.$

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

$\text{Area} = \int_0^{0.25} 4x^{3/2} \, dx.$

We first find the antiderivative:

$\int 4x^{3/2} \, dx = \frac{4}{\frac{5}{2}}x^{5/2} = \frac{8}{5}x^{5/2}.$

Now, we evaluate from $0$ to $0.25$:

$\text{Area} = \left[\frac{8}{5}x^{5/2}\right]_0^{0.25} = \frac{8}{5}\left(0.25^{5/2} - 0^{5/2}\right) = \frac{8}{5}\left(\frac{1}{32}\right) = \frac{8}{160} = \frac{1}{20} = 2^{-1} \times 5^{-1}.$

Thus, $m = -1$ and $n = -1$.

The integral \int_{6.3}^{6.4} 2x^3 , dx is not a suitable approximation because the function \sin x - \sin x \cos 2x is periodic and its behavior over the interval from $6.3$ to $6.4$ may introduce significant changes not captured by a constant polynomial approximation like $2x^3$. The approximation is valid only for small values of $x$.

The integral \int_{6.3}^{6.4} (\sin x - \sin x \cos 2x) , dx can be approximated by identifying a and b such that the behavior of \sin x - \sin x \cos 2x closely resembles that of $2x^3$ within the chosen limits. We can use $a = 6.3$ and $b = 6.4$ to align the approximation with the evaluated limits and provide a suitable comparison between both integrals. Therefore, one might choose values driven by periodicity or particular characteristics of the interval.