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

Answer 1

Isaiah Allen

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

We use integration by parts

#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#

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Answer 2

Evelyn Agard

To evaluate the definite integral ∫₀⁶ e^(-t) dt, we use the antiderivative of e^(-t), which is -e^(-t). Then, we substitute the upper and lower limits of integration into this antiderivative and subtract the result evaluated at the lower limit from the result evaluated at the upper limit.

So, ∫₀⁶ e^(-t) dt = -e^(-6) - (-e^(-0)) = -e^(-6) + e^0 = -e^(-6) + 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.