In this question, you may use expansions from the List of Formulae (MF26).

1. Use the standard Maclaurin series for $\sin x$ to expand $\sin 3x$ up to $x^3$.

$\sin x = x - \frac{x^3}{3!} + \dots$
$\sin 3x = 3x - \frac{(3x)^3}{3!} + \dots = 3x - \frac{27x^3}{6} + \dots = 3x - \frac{9}{2}x^3 + \dots$

2. Substitute the expansion of $\sin 3x$ into the standard Maclaurin series for $\mathrm{e}^x$, keeping terms up to $x^3$.

$\mathrm{e}^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$
Substitute $u = \sin 3x \approx 3x - \frac{9}{2}x^3$:
$\mathrm{f}(x) = 1 + \left(3x - \frac{9}{2}x^3\right) + \frac{1}{2!}\left(3x - \frac{9}{2}x^3\right)^2 + \frac{1}{3!}\left(3x - \frac{9}{2}x^3\right)^3 + \dots$

3. Expand the polynomial powers and collect terms up to $x^3$.

$\left(3x - \frac{9}{2}x^3\right)^2 = 9x^2 + \dots$
$\left(3x - \frac{9}{2}x^3\right)^3 = 27x^3 + \dots$
$\mathrm{f}(x) = 1 + \left(3x - \frac{9}{2}x^3\right) + \frac{1}{2}\left(9x^2\right) + \frac{1}{6}\left(27x^3\right) + \dots$
$= 1 + 3x + \frac{9}{2}x^2 + \left(-\frac{9}{2} + \frac{27}{6}\right)x^3 + \dots$

4. Determine the final expansion up to $x^2$ and verify the coefficient of $x^3$ is zero.

$\mathrm{f}(x) = 1 + 3x + \frac{9}{2}x^2 + \left(-\frac{9}{2} + \frac{9}{2}\right)x^3 + \dots$
$= 1 + 3x + \frac{9}{2}x^2 + 0x^3 + \dots$
The expansion in ascending powers of $x$ up to the term $x^2$ is $1 + 3x + \frac{9}{2}x^2$.
Since the coefficient of $x^3$ is $0$, it is shown that the expansion is independent of $x^3$.

1. Substitute the Maclaurin expansion of $\mathrm{f}(x)$ from part (a) into the integral to form an approximate integrand.

From part (a), we have the expansion $\mathrm{e}^{\sin 3x} \approx 1 + 3x + \frac{9}{2}x^2$.

2. Integrate the polynomial approximation term-by-term to find the approximate expression for the indefinite integral.

3. Relate the required definite integral to the indefinite integral by manipulating constants.

4. Evaluate the definite integral using the limits $0.1$ and $0.2$.

5. Compute the final approximate value correct to 4 decimal places.

1. Input the function and limits into a graphing calculator and evaluate.

Let $I = \int_{0.1}^{0.2} \left(\frac{2}{x}\right)^2 \mathrm{e}^{\sin 3x} \,\mathrm{d}x$.
Using the definite integral numerical evaluation feature on the graphing calculator,
$I \approx 29.999424...$

2. Round the output to the required 4 decimal places.

$I = 29.9994$ (to 4 d.p.)

1. State the reason relating the limits of integration to the validity of the Maclaurin series approximation.

The Maclaurin series is an expansion about $x=0$. The limits of integration, $x = 0.1$ and $x = 0.2$, are small values that are close to $0$. For such values, the higher powers of $x$ that were omitted from the truncated series expansion are very small and negligible. Consequently, the polynomial closely models the actual function in this interval.