A vanishing cubic term does not make an integral exact · source reference

Printed question

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

(a) Given that $f(x)=e^{\sin3x}$, find the Maclaurin expansion of $f(x)$ in ascending powers of $x$, up to and including the term $x^2$. Show that this expansion is independent of $x^3$. [4]

(b) Use your expansion from part (a) and integration to find an approximate expression for $\int\frac{e^{\sin3x}}{x^2}\,dx$. Hence obtain an approximate value for $\int_{0.1}^{0.2}(\frac2x)^2e^{\sin3x}\,dx$, correct to 4 decimal places. [3]

(c) Use your calculator to find $\int_{0.1}^{0.2}(\frac2x)^2e^{\sin3x}\,dx$, correct to 4 decimal places. [1]

(d) A student compares the answers to parts (b) and (c) and concludes that the approximation is accurate. Give a reason to support his conclusion. [1]

[2023/DHS/Prelim/P1/6]

Printed working

$\sin3x=3x-(3x)^3/3!+\cdots=3x-9x^3/2+\cdots$.

$e^{\sin3x}=1+(3x-9x^3/2+\cdots)+\frac12(3x-9x^3/2+\cdots)^2+\frac16(3x-9x^3/2+\cdots)^3+\cdots=1+3x-9x^3/2+9x^2/2+27x^3/6+\cdots=1+3x+9x^2/2+0x^3$.

Printed differentiation alternative. $y'=3\cos3x\,y$; $y''=3\cos3x\,y'-9\sin3x\,y$; $y'''=3\cos3x\,y''-9\sin3x\,y'-9\sin3x\,y'-27\cos3x\,y$. At 0: $y=1$, $y'=3$, $y''=9$, $y'''=0$. Thus $1+3x+9x^2/2+0x^3+\cdots$.

$\int e^{\sin3x}/x^2\,dx\approx\int(x^{-2}+3x^{-1}+9/2)\,dx=-x^{-1}+3\ln|x|+9x/2+C$.

$4[-x^{-1}+3\ln x+9x/2]_{0.1}^{0.2}=30.1178$ (4 d.p.). Calculator value $29.9995$ (4 d.p.). Printed reason: values 0.1–0.2 are close to 0, so powers $x^4$ and higher can be neglected. Alternative: percentage error $|30.1178-29.9995|/29.9995\times100=0.3943\%$, which is small.

Teacher note 1

From MF27 replace $x$ by $3x$ in $\sin x=x-x^3/3!+\cdots$ to obtain the expansion for $\sin3x$. For $e^{\sin3x}$, replace $x$ by $\sin3x=3x-9x^3/2+\cdots$ in $e^x=1+x+x^2/2!+x^3/3!+\cdots$.

Mathematical review

(a) $\sin3x=3x-9x^3/2+O(x^5)$. In $e^u=1+u+u^2/2+u^3/6+\cdots$, the cubic coefficient is $-9/2+27/6=0$. Hence $f(x)=1+3x+9x^2/2+O(x^4)$.

(b) Divide by $x^2$ and integrate: $\int f(x)/x^2\,dx\approx-1/x+3\ln|x|+9x/2+C$. Multiply by 4 for the definite integral: $4[-1/x+3\ln x+9x/2]_{0.1}^{0.2}=30.1178$.

(c) Numerical quadrature gives $29.9995$ (4 d.p.). (d) Relative error is about $0.3943\%$. Both endpoints are close to zero and the first omitted numerator term is quartic, supporting this accuracy for the specified interval.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 10. Working page(s): 10, 11. The two Series source files are byte-identical; their shared visible content is expressed once and mapped to both page sets. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.