Integral question
@inkyvoyd
I know the substitution formula and how to get U and DU, so on so fourth.
So where we getting stuck? Let u equal the stuff up in the exponent. Ooo this one works out nicely I think!
\[\LARGE\rm \int\limits t^{15}e^{-t^8}dt=\int\limits t^{8+7}e^{-t^8}dt\]\[\Large\rm =\int\limits t^8 e^{-t^8}\left(t^7~dt\right)\]If you break it up like this, it might be easier to plug in your substitution.
\(\Large\rm u=-t^8\) \(\Large\rm du=-8\left(t^7~dt\right)\)Can you kinda see what's going on?
Hmm yes I like how you broke it up.
Well, I thought I was getting it. Just me, I was doing U=t^15 DU=15t^14 DV=e^-t8
So you were trying to jump right into the `integrating by parts` before doing a substitution. See how that runs into trouble? Finding your V would be impossible without borrowing some of those t's from the U.
Confused still senor bus? :U
Oh goodness. It does help to read instructions! Forgive me zepdrix. Thanks for your patience and help! My AmTran is tired lol. Onward!
So the substitution is basically breaking up the 15?
I'll call it something else, not u, so there's no confusion that this has nothing to do with integration by parts. \[\Large\rm \color{orangered}{m=-t^8}\qquad\to\qquad \color{royalblue}{-m=t^8}\]\[\Large\rm \color{green}{-\frac{1}{8}dm=\left(t^7~dt\right)}\]Yah, we're breaking up the 15, and applying a substitution before even assigning any parts.\[\Large\rm =\int\limits\limits \color{orangered}{t^8} e^{\color{royalblue}{-t^8}}\color{green}{\left(t^7~dt\right)}\]
Bahh I got the orange and blue backwards.. my bad :P
Thats fine! The m makes sense!
\[\Large\rm =\frac{1}{8}\int\limits m e^m~dm\]So before applying our parts, we have ummm something like this, yes?
And I'll bet you can handle the parts from there c: Shouldn't be too bad.
I'll try!
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ed8at9flt29
This gives a foundation to work with.
Join our real-time social learning platform and learn together with your friends!