(a) Using the Maclaurin series for $e^x$ and $\sin x$, find the series expansion of $e^x\sin x$ up to and including the term in $x^3$. [5]
(b) Hence estimate $\displaystyle\int_0^{0.5}e^x\sin x\,dx$, giving your answer as a fraction. [2]
- $e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\ldots$ and $\sin x=x-\dfrac{x^3}{6}+\ldots$M1
- Multiplies the series, collecting terms up to $x^3$M1
- Term in $x$: $x$A1
- Term in $x^2$: $x^2$A1
- Term in $x^3$: $\dfrac{x^3}{2}-\dfrac{x^3}{6}=\dfrac{x^3}{3}$A1
- $\int_0^{0.5}\big(x+x^2+\tfrac{x^3}{3}\big)dx=\Big[\tfrac{x^2}{2}+\tfrac{x^3}{3}+\tfrac{x^4}{12}\Big]_0^{0.5}$M1
- $\dfrac{11}{64}$ (accept $0.171875$)A1
- Answer: (a) $x+x^2+\dfrac{x^3}{3}$; (b) $\dfrac{11}{64}$
- Examiner tip: Only products giving powers up to $x^3$ matter; list them systematically so none is missed.