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integration: estimate s(4), given v(t) = 40(1 - e^(-2t)) and
- Thread starter Seimuna
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Re: integration
Seimuna said: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?
Re: integration
Hello, Seimuna!
Well, you began correctly ... sort of . . .
\(\displaystyle \text{Integrate: }\;40\int\left(1 - e^{-2t}\right)\,dt \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) + C\)
\(\displaystyle \text{Since }s(0) \,=\,0\text{, we have: }\;40\left(0 + \tfrac{1}{2}e^0\right) + C \:=\:0 \quad\Rightarrow\quad C \:=\:-20\)
. . \(\displaystyle \text{Hence, the function is: }\;s(t) \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) - 20\)
\(\displaystyle \text{Therefore: }\;s(4) \;=\;40\left(4 + \tfrac{1}{2}e^{-8}\right) - 20 \;=\;140.0067093 \;\approx\;\boxed{140}\)
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 . . .
\(\displaystyle \text{Integrate: }\;40\int\left(1 - e^{-2t}\right)\,dt \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) + C\)
\(\displaystyle \text{Since }s(0) \,=\,0\text{, we have: }\;40\left(0 + \tfrac{1}{2}e^0\right) + C \:=\:0 \quad\Rightarrow\quad C \:=\:-20\)
. . \(\displaystyle \text{Hence, the function is: }\;s(t) \;=\;40\left(t + \tfrac{1}{2}e^{-2t}\right) - 20\)
\(\displaystyle \text{Therefore: }\;s(4) \;=\;40\left(4 + \tfrac{1}{2}e^{-8}\right) - 20 \;=\;140.0067093 \;\approx\;\boxed{140}\)