find ∫ p ( x ) d x where: p ( x ) = e − x 2 2 2 π .

Question

find   ∫  p  (  x  )  d  x where:  p  (  x  )  =            e                                    −                          x              2                                2                            2      π        .

Marisa Colon · Accepted Answer

We can integrate the Maclaurin Series term by term as follows.

2 \int_{0}^{1} \frac{1}{\sqrt{2 π}} \sum_{n = 0}^{\infty} \frac{(\frac{- x^{2}}{2})^{n}}{n!}

\sqrt{\frac{2}{π}} \sum_{n = 0}^{\infty} \int_{0}^{1} \frac{(- 1)^{n} x^{2 n}}{2^{n} n!}

\sqrt{\frac{2}{π}} \sum_{n = 0}^{\infty} \frac{(- 1)^{n} x^{2 n + 1}}{2^{n} n! (2 n + 1)} |_{0}^{1}

\sqrt{\frac{2}{π}} \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{2^{n} n! (2 n + 1)} \approx 0.682689492

From this series, we get an approximation of the true value.

Find intp(x)dx where: p(x)=e^(−x^2/2)/sqrt (2π)

Answered question

Answer & Explanation

New Questions in Calculus 2