How do you integrate #int troot3(t-4)dt#?

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Answer 1 · 2017-01-29T23:47:30

#3/7(t-4)^(7/3)+3(t-4)^(4/3)+C#

#I=inttroot3(t-4)color(white).dt#

Apply the substitution #u=t-4#. This also implies that #t=u+4# and #du=dt#. Then:

#I=int(u+4)root3ucolor(white).du#

We can write #root3u# with a fractional exponent and then distribute:

#I=int(u+4)u^(1/3)color(white).du#

#I=intu^(4/3)color(white).du+4intu^(1/3)color(white).du#

Integrate both using the rule #intu^ncolor(white).du=u^(n+1)/(n+1)+C#:

#I=u^(7/3)/(7/3)+4(u^(4/3)/(4/3))+C#

#I=3/7u^(7/3)+3u^(4/3)+C#

Since #u=t-4#:

#I=3/7(t-4)^(7/3)+3(t-4)^(4/3)+C#