integration: estimate s(4), given v(t) = 40(1 - e^(-2t)) and

[Seimuna]

estimate s(4).
given,
v(t) = 40(1-e^(-2t)) and s(0)=0

as i know, integrate v(t)?
then ll get s(t)=40[t+(1/2)(e^(-2t))]
but then s(0) not = 0

what i did wrongly?

 

[Deleted member 4993]

Re: integration

estimate s(4).
given,
v(t) = 40(1-e^(-2t)) and s(0)=0

as i noe, integrate v(t) rite? <<< I don't understand what you are talking about

then ll get s(t)=t+(1/2)(e^(-2t)) <<< Where did 40 go?? Where did constant of integration go?

but then s(0) not = 0

what i did wrongly?

 

[Seimuna]

Re: integration

really sorry...i edited on the first post...really sorry...

 

[soroban]

Re: integration

Hello, Seimuna!

$\displaystyle \text{Given: }\:v(t) \:= \:40\left(1-e^{-2t}\right)\quad\text{ and: }s(0)=\,0$

$\displaystyle \text{Estimate }s(4).$

Well, you began correctly ... sort of . . .


$\displaystyle \text{Integrate: }\;40\int\left(1 - e^{-2t}\right)\,dt \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) + C$


$\displaystyle \text{Since }s(0) \,=\,0\text{, we have: }\;40\left(0 + \tfrac{1}{2}e^0\right) + C \:=\:0 \quad\Rightarrow\quad C \:=\:-20$

. . $\displaystyle \text{Hence, the function is: }\;s(t) \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) - 20$


$\displaystyle \text{Therefore: }\;s(4) \;=\;40\left(4 + \tfrac{1}{2}e^{-8}\right) - 20 \;=\;140.0067093 \;\approx\;\boxed{140}$

 

[Seimuna]

Re: integration

thanks...