1.10: Past-paper question 9

4PM1/1R/June/2023 — Question 9 · 14 marks

(a) Expand (1+2x)13\displaystyle (1 + 2x)^{-\frac{1}{3}} in ascending powers of x\displaystyle x up to and including the term in x3\displaystyle x^3 expressing each coefficient as a fraction in its lowest terms.
(3)
(b) Find the range of values of x\displaystyle x for which your expansion is valid.
(1)
f(x)=2+kx2(1+2x)13f(x) = \frac{2 + kx^2}{(1 + 2x)^{\frac{1}{3}}}
(c) Obtain a series expansion of f(x)\displaystyle f(x) in ascending powers of x\displaystyle x up to and including the term in x3\displaystyle x^3
Give your coefficients in terms of k\displaystyle k where appropriate.
(3)
The coefficient of x3\displaystyle x^3 in the series expansion of f(x)\displaystyle f(x) is 83\displaystyle -\frac{8}{3}
(d) Find the exact value of k\displaystyle k
(2)
(e) Hence, using algebraic integration, estimate the value of
0.10.2f(x)dx\int_{0.1}^{0.2} f(x) \, \mathrm{d}x
Give your answer to 4 decimal places.
(5)