I have the integral from 1 to 5 of [h(t-2) + h(t)]lnt dt

I broke this integral into integral from 1 to 5 of [h(t-2)ln t] + integral from 1 to 5 of [h(t) lnt]

but now I don't know how to actually use the step function to find the answer..

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- August 31st 2010, 02:12 AMCookieCIntegrals using step functions
I have the integral from 1 to 5 of [h(t-2) + h(t)]lnt dt

I broke this integral into integral from 1 to 5 of [h(t-2)ln t] + integral from 1 to 5 of [h(t) lnt]

but now I don't know how to actually use the step function to find the answer.. - August 31st 2010, 03:25 AMmr fantastic
- August 31st 2010, 04:21 AMCookieC
h(t - a) = 1 if t>= a and 0 if t<a ? but uumm how do i know what t is..? :S

- August 31st 2010, 04:23 AMCookieC
Wait is it because its bewteen 1 to 5?? so if its in between there we can say it is equal to 1 and if it is not then it is equal to 0?

- August 31st 2010, 04:59 AMCookieC
Actually I don't understand it..

I have h(t-2) = 1 when t>=2 and 0 when t < 2

With the integral from 1 to 5 of h(t-2)lnt do i do integral from 1 to 2 of lnt x 0 is equal to zero then plus integral from 2 to 5 of lnt x 1? - August 31st 2010, 07:27 PMmr fantastic