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(x^(4n+1))/(4n+1)\bigg|_0^(0.3)`

`=\displaystyle\sum_(n=1)^\infty((-1)^(n+1)(0.3)^(4n+1))/(n(4n+1))`

`\approx0.000485`