OpenStudy (anonymous):

Find the integral of t^3(t-4)^(1/2)dt

4 years ago
OpenStudy (zarkon):

u=t-4

4 years ago
OpenStudy (anonymous):

So du=dt

4 years ago
OpenStudy (zarkon):

yes

4 years ago
OpenStudy (anonymous):

Ok then after I substituted some of the things, I got (u+4)^3u^(1/2)du but then I got stuck there...

4 years ago
OpenStudy (zarkon):

expand\((u+4)^3\)

4 years ago
zepdrix (zepdrix):

\[\large \int\limits (u+4)^3\sqrt u \; du\]Just so it's easier to read D:

4 years ago
OpenStudy (anonymous):

Ok so did you get u^3+12u^2+48u +64?

4 years ago
OpenStudy (zarkon):

that is correct...now distribute the \(\sqrt{x}\)

4 years ago
OpenStudy (zarkon):

\[\sqrt{u}\]

4 years ago
OpenStudy (zarkon):

then integrate term by term

4 years ago
OpenStudy (anonymous):

I got 2u^(9/2)/9+4u^(7/2)/7+96u^(5/2)/5+64u(3/2)/3+c, and all I need to do is change the u's to (t-4)?

4 years ago
OpenStudy (anonymous):

Wait the second term should be 24u^(7/2)/7 and the last should be 128 u^(3/2)/3

4 years ago
OpenStudy (zarkon):

correct

4 years ago
OpenStudy (anonymous):

Thank you :)

4 years ago
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