How do you evaluate the definite integral #intte^(-t)dt# from #[0,6]#?

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Answer 1 · 2016-11-22T20:50:57

The answer is #=1-7/e^6#

#intuv'=uv-intu'v#

For our integral #intte^(-t)dt#

Let #u=t# , #=>#, #u'=1#

and #v'=e^(-t)#, #=>#, #v=-e^(-t)#

Therefore,

#intte^(-t)dt=-te^(-t)-int-e^(-t)dt#

#=-te^(-t)-e^(-t)+C#

#int_0^6te^(-t)dt= [-e^(-t)(t+1)] _0^6#

#=-7/e^6+1#