Find the indefinite integral
(integral sign) 3te^2tdt
13 answers
use a calculator, that's what calculus is all about. duh!
Excuse you that DUH is not necessary and if I could use the calculator I would have used it but my teacher did not teach me to how to use the calculator for this so please mind your business
do t e^(2t) dt
and multiply by 3 later
by parts
u = t
du = dt
dv = e^(2t) dt
v = (1/2) e^(2t)
u v = (1/2) t e^(2t)
v du = (1/2)e^(2t) dt
integral v du = e^(2t)
u v - integral v du = (1/2) t e^(2t) -e^(2t)
and multiply by 3 later
by parts
u = t
du = dt
dv = e^(2t) dt
v = (1/2) e^(2t)
u v = (1/2) t e^(2t)
v du = (1/2)e^(2t) dt
integral v du = e^(2t)
u v - integral v du = (1/2) t e^(2t) -e^(2t)
then multiply by 3
integral(u dv) = uv - integral(v du)
let u = 3t
du/dt = 3 , so du = 3 dt
let dv = e^(2t)dt
v = (1/2)e^(2t)
so integral(3t(e^(2t)))dt = 3t(1/2)e^(2t) - integral((1/2)e^(2t)(3 dt)
= (3/2)te^(2t) - (3/4)e^(2t)
I don't know how much further you want to simplify this, but I differentiated my last answer and it works
let u = 3t
du/dt = 3 , so du = 3 dt
let dv = e^(2t)dt
v = (1/2)e^(2t)
so integral(3t(e^(2t)))dt = 3t(1/2)e^(2t) - integral((1/2)e^(2t)(3 dt)
= (3/2)te^(2t) - (3/4)e^(2t)
I don't know how much further you want to simplify this, but I differentiated my last answer and it works
Let 3t = u and e^2t dt = dv
du = 3 dt v = (1/2) e^2t
The integral is
uv - INTEGRAL v du
= (3/2)t e^2t - INTEGRAL (3/2)e^2t
= (3/2)t e^2t - (3/4)e^2t
du = 3 dt v = (1/2) e^2t
The integral is
uv - INTEGRAL v du
= (3/2)t e^2t - INTEGRAL (3/2)e^2t
= (3/2)t e^2t - (3/4)e^2t
Thank You Damon, Reiny and Drwls for your help
but why does v=(1/2)e^2t
all three of us had chosen
let dv = e^(2t)dt
or dv/dt = e^(2t)
wouldn't you have to integrate that to get v ?
v = (1/2)e^(2t)
I hope you recognized that we used a method called integration by parts
in choosing the "u" and "dv"
let u be something that you can differentiate, and
let dv be the part that you can integrate,
then hope for the best
let dv = e^(2t)dt
or dv/dt = e^(2t)
wouldn't you have to integrate that to get v ?
v = (1/2)e^(2t)
I hope you recognized that we used a method called integration by parts
in choosing the "u" and "dv"
let u be something that you can differentiate, and
let dv be the part that you can integrate,
then hope for the best
because d/dt of e^2t = 2 e^2t
so you need the (1/2) to get one of them instead of 2
so you need the (1/2) to get one of them instead of 2
oooo okay Thank You
yeah you better say thank you and if i were to mind my own business then why did you post a question where everyone can see dufus?
Ummm because I wasn't looking for help from you and of course I'm going to say Thank You because i have manners unlike you you don't have any and you don't address people that's looking for help form you like that so Goodbye and Have a nice day
1 answer