Question

Use series to approximate the definite integral I to within the indicated accuracy.

Answer 1

`\displaystyle I=0.0130`

Explanation:

We know the following:

`\displaystyle e^x=\sum^\infty_(n=0)(x^n)/(n!)\\\\e^(-x^2)=\sum^\infty_(n=0)((-1)^nx^(2n))/(n!)\\\\x^3e^(-x^2)=\sum^\infty_(n=0)((-1)^nx^(2n+3))/(n!)`

Now we integrate to get:

`\displaystyle \int^(0.5)_0x^3e^(-x^2)dx=\biggr[\sum^\infty_(n=0)((-1)^nx^(2n+4))/(n!(2n+4))\biggr]^(x=0.5)_(x=0)\\\\\\=\biggr[\sum^\infty_(n=0)((-1)^n(0.5)^(2n+4))/(n!(2n+4))\biggr]-\biggr[\sum^\infty_(n=0)((-1)^n(0)^(2n+4))/(n!(2n+4))\biggr]\\\\\\=\sum^\infty_(n=0)(1)/(2^(2n+4)n!(2n+4))`

Because the series is alternating, we can apply Alternating Series Estimation.

The term with
`n=2` is
`(1)/(4096) < 0.001`, so we use:

`\displaystyle \sum^1_(n=0)(1)/(2^(2n+4)n!(2n+4))=(1)/(64)-(1)/(384)\approx0.0130`

Therefore,
`\displaystyle \int^(0.5)_0x^3e^(-x^2)dx\approx0.0130` with an error less than 0.001