--- In the question you may use expansions from the List of Formulae (MF27). Find the Maclaurin expa

Question

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In the question you may use expansions from the List of Formulae (MF27).

Find the Maclaurin expansion of $\cos(t^3)$ in ascending powers of $t$, up to and including the term in $t^{12}$. Hence, find the Maclaurin series of $\displaystyle \displaystyle\int_{0}^{x} t^2 \cos(t^3) \, \mathrm{d}t$ up to and including the term in $x^{15}$ given that $x$ is small.[4]
Use your expansion from part (a) to find an approximate value for $\displaystyle \displaystyle\int_{0}^{0.1} t^2 \cos(t^3) \, \mathrm{d}t$, correct to $5$ decimal places.[1]
Find $\displaystyle \displaystyle\int_{0}^{x} t^2 \cos(t^3) \, \mathrm{d}t$ in terms of $x$. Hence evaluate $\displaystyle \displaystyle\int_{0}^{0.1} t^2 \cos(t^3) \, \mathrm{d}t$, correct to $5$ decimal places.[3]
Comparing your answers to parts (b) and (c), and with reference to the value of $x$, comment on the accuracy of your approximations.[1]

Accepted Answer

(a) $\cos(t^3)=1-\frac12t^6+\frac1{24}t^{12}+\cdots$, $\int_0^x t^2\cos(t^3)\,\mathrm{d}t=\frac{x^3}{3}-\frac{x^9}{18}+\frac{x^{15}}{360}+\cdots$. (b) $0.00033$ (5 d.p.). (c) $\frac13\sin(x^3)$; $0.00033$ (5 d.p.). (d) Both agree to 5 d.p.; $x^3=0.001$ makes the neglected terms negligible.

2025 NYJC P1 Q10

2025 NYJC P1 Q10

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