Question

A particle moves along a straight line with velocity v(t) = te^-t meters per second after t seconds. How far will it travel during the first 75 seconds?

Answer 1

\[ v(t) = te^{-t} \]
\[\frac{dx}{dt}=te^{-t}\]
\[\int\limits_{}^{}1dx=\int\limits_{}^{} te^{-t}dt\]
\[x=\int\limits\limits_{}^{} te^{-t}dt\]

Answer 2

umm

Answer 3

integration by parts

Answer 4

do the x= part using IBP?

Answer 5

\[x=\int\limits_{0}^{75} e^{-t} t \, dt\]
\[x\text{}=\int\limits\limits_{0}^{75} e^{-t} \, dt-e^{-t} t\]
\[\large x=[-e^{-t} t-e^{-t}]_0^{75} \]

Answer 6

ok, just going by your last post.. i don't get the second step.

Answer 7

i dont get the -et^-t part

Answer 8

isn't that the integral?... not sure why you stuck it on as a subtraction

Answer 9

Its from integration by parts,
\[x=\int\limits\limits_{0}^{75} e^{-t} t \, dt\]
let,
u=t dv=e^-t
du=dt v=-e^-t
\[x=-te^{-t}-\int\limits_{}^{}-e^{-t}dt\]
Rearranging it
\[x\text{}=\int\limits\limits\limits_{0}^{75} e^{-t} \, dt-e^{-t} t\]

Answer 10

ok, it's starting to make sense

Answer 11

let me look it over for a bit