Question

Question part points submissions used use a power series to approximate the definite integral, i, to six decimal places. 0.3 ln(1 + x4) dx 0

Answer 1

$\ln(1+x)=\displaystyle-\sum_{n=1}^\infty\frac{(-x)^n}n$

$\implies\ln(1+x^4)=\displaystyle-\sum_{n=1}^\infty\frac{(-x^4)^n}n$

$\displaystyle-\int_0^{0.3}\sum_{n=1}^\infty\frac{(-x^4)^n}n\,\mathrm dx$
$=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\int_0^{0.3}x^{4n}\,\mathrm dx$
$=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\dfrac{x^{4n+1}}{4n+1}\bigg|_0^{0.3}$
$=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}(0.3)^{4n+1}}{n(4n+1)}$
$\approx0.000485$