Question

integration of e(-y^2) in the limit of 0 to 1

Answer 1

hyy...

The series is quite simple, just integrate 1−x2+x^4/2−x^6/3!+−…1−x^2+x4/2!−x^6/3!+−… term by term to get 1−1/3+1/10−1/42+−…1−1/3+1/10−1/42+−….

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hope it helps

Answer 2

$\begin{gathered}\huge{\bold{\underline{\underline{\:\:\:\: hello\: friend: - }}}} \\ \\\: \: \large{\bold{\underline{\underline{here \: is \: your \: answer: - } }}} \\ \\ \\ \sf \qquad \implies \: The \: series is \: quite \: simple, \: just \: integrate \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\sf \qquad \implies \: 1-x²+x^{4/2}-x^{6/3}+-- \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\sf \qquad \implies \: 1-x²+x^{4/2}-x^{6/3}-\: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\sf \qquad \implies \: term \: by \: term \: to \: get \: 1-1/3+1/10-1/42+- \\ \\ \large{\bold{\underline{\underline{i \: hope \: this \: helps \: you \: }}}} \\ \\ \\ \\\end{gathered}$