How do you find the definite integral for: #tln(t) dt# for the intervals #[1, 10]#?

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Answer 1 · 2016-07-13T18:17:30

#= 25 ( ln 100 - 1)#

#int_0^10 \ t ln t \ dt#

#= int_0^10 \ d/dt(t^2/2) ln t \ dt#

and by IBP: #int u v' = uv - int u' v#

#= t^2 / 2 ln t |_0^10 - int_0^10 \ t^2/2 d/dt(ln t) \ dt#

#= t^2 / 2 ln t |_0^10 - int_0^10 \ t^2/2 1/t \ dt#

#= t^2 / 2 ln t |_0^10 - int_0^10 \ t/2 \ dt#

#= t^2 / 2 ln t - t^2 /4 |_0^10 qquad triangle#

clearly #t^2 / 2 ln t |_{t = 0} # is a problem and we need to know

#lim_{t to 0^+} t^2 / 2 ln t#

#= 1/2 lim_{t to 0^+} ln t / (1/t^2)# which is now indeterminate so L'Hopital'sRule

#= 1/2 lim_{t to 0^+} (1/t) / (-2/t^3)#

#= -1/4 lim_{t to 0^+} t^2 = 0#

so #triangle# becomes

#= 10^2 / 2 ln 10 - 10^2 /4#

#= 10^2 / 4 * 2 ln 10 - 10^2 /4#

#= 10^2 / 4 ( ln 100 - 1)#

#= 25 ( ln 100 - 1)#