{"id":525,"date":"2019-10-28T23:24:47","date_gmt":"2019-10-28T23:24:47","guid":{"rendered":"http:\/\/adrianbell.me\/?p=525"},"modified":"2019-10-28T23:35:27","modified_gmt":"2019-10-28T23:35:27","slug":"integration-revision-2","status":"publish","type":"post","link":"https:\/\/adrianbell.me\/?p=525","title":{"rendered":"Integration Revision - Integration By Parts, Twice"},"content":{"rendered":"

So looking at the first one from a few days ago (Oct 13th):<\/p>\n

\"\int^{\infty}_{0}x^{2}\:\lambda\: e^{-\lambda x}\:dx=\frac{2}{\lambda^{2}}\"<\/p>\n

I was totally lost with this. But running through some old tricks made this seem a lot more approachable. First off, moving the constant out brings a bit more clarity to the integrand,<\/p>\n

\"\int^{\infty}_{0}x^{2}\lambda e^{-\lambda x}\:dx=\lambda\int^{\infty}_{0}x^{2} e^{-\lambda x}\:dx\"<\/p>\n

But given the \"x^{2}\" and the exponential, we'll also probably need to use integration by parts:<\/p>\n

\"\int^{b}_{a} f(x)g^{\prime}(x)dx=[f(x)g(x)]^{b}_{a}-\int^{b}_{a} f^{\prime}(x)g(x)dx\"<\/p>\n

So, as above, letting:
\n\"f(x)=x^{2}\"
\n\"g^{\prime}(x)=e^{-\lambda x}\",<\/p>\n

we have:
\n\"f^{\prime}(x)=2x\"
\n\"g(x)=-\frac{1}{\lambda}e^{-\lambda x}\"<\/p>\n

Then plugging \"f\", \"f^{\prime}\", \"g\" and \"g^{\prime}\" into our lovely integration by parts tool above gives:<\/p>\n

\"\lambda\int^{\infty}_{0}x^{2}\: e^{-\lambda x}\:dx\"<\/p>\n

\"=\lambda\left(\left[-x^{2}\: \frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}-\int^{\infty}_{0}-2x\:\frac{1}{\lambda}\: e^{-\lambda x}\:dx\right)\"<\/p>\n

\"=\lambda\left(0-0+2\frac{1}{\lambda}\int^{\infty}_{0}x\: e^{-\lambda x}\:dx\right)\"<\/p>\n

\"=2\int^{\infty}_{0}x\: e^{-\lambda x}\:dx\"<\/p>\n

Which looks very familiar with what we've started with, except \"x^{2}\" is now just \"x\"! If we can reduce \"x\" by another power, we'll end up with just \"1\" which will surely give us a much easier integral to solve.<\/p>\n

So applying integration by parts again, but letting:
\n\"f(x)=x\"
\n\"g^{\prime}(x)=e^{-\lambda x}\",<\/p>\n

we have:
\n\"f^{\prime}(x)=1\"
\n\"g(x)=-\frac{1}{\lambda}e^{-\lambda x}\"<\/p>\n

Then:<\/p>\n

\"2\int^{\infty}_{0}x\: e^{-\lambda x}\:dx\"<\/p>\n

\"=2\left(\left[-x\: \frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}-\int^{\infty}_{0}-\frac{1}{\lambda}\: e^{-\lambda x}\:dx\right)\"<\/p>\n

\"=2\left(0-0+\frac{1}{\lambda}\int^{\infty}_{0} e^{-\lambda x}\:dx\right)\"<\/p>\n

\"=\frac{2}{\lambda}\int^{\infty}_{0} e^{-\lambda x}\:dx\"<\/p>\n

\"=\frac{2}{\lambda}\left[-\frac{1}{\lambda}\:e^{-\lambda x}\right]^{\infty}_{0}\"<\/p>\n

\"=\frac{2}{\lambda}\left(0-\left(-\frac{1}{\lambda}\right)\right)\"<\/p>\n

\"=\frac{2}{\lambda}\:\frac{1}{\lambda}\"<\/p>\n

\"=\frac{2}{\lambda^{2}}\"<\/p>\n","protected":false},"excerpt":{"rendered":"

So looking at the first one from a few days ago (Oct 13th): I was totally lost with this. But running through some old tricks made this seem a lot more approachable. First off, moving the constant out brings a bit more clarity to the integrand, But given the and the exponential, we'll also probably … Continue reading Integration Revision - Integration By Parts, Twice<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[38],"tags":[],"class_list":["post-525","post","type-post","status-publish","format-standard","hentry","category-calculus"],"_links":{"self":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/525","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=525"}],"version-history":[{"count":41,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/525\/revisions"}],"predecessor-version":[{"id":566,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/525\/revisions\/566"}],"wp:attachment":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=525"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=525"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=525"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}