anonymous
  • anonymous
integration with limits. question attached
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
anonymous
  • anonymous
i think we should be able to let u equal -0.05x-5
amistre64
  • amistre64
\[a^{n+m}=a^na^m\]

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anonymous
  • anonymous
That's the answer?
amistre64
  • amistre64
http://openstudy.com/code-of-conduct "Don’t post only answers - guide the asker to a solution."
anonymous
  • anonymous
@shaynaderby, you can split this up into two parts. That's what @amistre64 hinted at.
anonymous
  • anonymous
okay well that makes sense. I just have no idea what do to with limits, they get plugged in, right?
anonymous
  • anonymous
Okay I can walk you through this. Try splitting it up.
amistre64
  • amistre64
forget the limits at first; they are used later
anonymous
  • anonymous
Yep, forget the limits at this point. Just split the e^(-0.05x-5) into two parts. Hint: each part will have "e" in it.
anonymous
  • anonymous
e^(-0.05)-e^(-5)
anonymous
  • anonymous
?
amistre64
  • amistre64
e^(-5) is a constant; no need to integrate it
anonymous
  • anonymous
well I know integrating e^x is always e^x
anonymous
  • anonymous
Okay well if you have e^3x, then the integral of that is (1/3)*e^(3x). Try taking the derivative of (1/3)*e^(3x) and see if you come up with e^3x.
amistre64
  • amistre64
\[\int e^{-0.05x-5}dx=\int e^{-0.05x}e^{-5}dx\to\ e^{-5}\int e^{-0.05x}dx\]
amistre64
  • amistre64
just to put things back on track :)
anonymous
  • anonymous
what exactly about e^-5 makes it a constant, for future reference
anonymous
  • anonymous
e^-5 is just a number. Just like the number 4.
amistre64
  • amistre64
\[\int e^{kx}dx=\frac{e^{kx}}{k}\] might e a rule in the back of a book
amistre64
  • amistre64
3^-5 is just a number e = 2.71828181845905 .... is just a number e^-5 is still just a number
anonymous
  • anonymous
so the integral of e^(-0.05x) is e^(-0.05x)/-0.05 ?
amistre64
  • amistre64
yes :)
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
yay! as you guys can probably tell at this point, I have a fantastic professor and have a great capacity for paying attention in class *sarcasm*
amistre64
  • amistre64
now we use the limits
anonymous
  • anonymous
but where are we at this point now that we integrated it?
amistre64
  • amistre64
\[\int_{a}^{b} f(x)dx=F(b)-F(a)\]
anonymous
  • anonymous
okay I'm a bit lost
amistre64
  • amistre64
the latex is showing up wierd lately
anonymous
  • anonymous
@shaynaderby as a sidenote, here's why the derivative of e^x is e^x. You first write down e^x, and then multiply it by the derivative of what's in that exponent. The derivative of x is 1, so you end up having 1*e^x. So if you have e^4x, then the derivative ends up being 4e^4x.
amistre64
  • amistre64
\[\int f(x)dx = F(x)+c\] \[\int_{a}^{b} f(x)dx = F(b)-F(a)\]
anonymous
  • anonymous
The whole functions and rules things confuses me. Can you guys just keep sort of telling me what to do step by step? That's helping me more.
amistre64
  • amistre64
yes, but its important that you relate this to the general rules so that you know how they are being applied in specific cases
anonymous
  • anonymous
Okay so you've established that your integral ends up being (e^-5)*(e^(-0.05x)/-0.05). Plug the upper limit in to x minus the lower limit that's plugged into x.
anonymous
  • anonymous
so what does my function look like now though, before I do that
amistre64
  • amistre64
you found F(x): \(\Large e^{-5}\frac{e^{-0.05x}}{-0.05}\)
anonymous
  • anonymous
okay great that's what I wrote down actually when I tried it myself
anonymous
  • anonymous
so now I am plugging in 100?
amistre64
  • amistre64
what are your limits?
anonymous
  • anonymous
100 and 80
amistre64
  • amistre64
then find F(100) and subtract F(80)
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
F(100)=-0.0009079986 ?
anonymous
  • anonymous
ohhh I see where I may have screwed up, I never put e^(-5) on top
amistre64
  • amistre64
\[\Large e^{-5}\frac{e^{-0.05(100)}}{-0.05}-\Large e^{-5}\frac{e^{-0.05(80}}{-0.05}\] \[\Large e^{-5}\frac{e^{-5}}{-0.05}-\Large e^{-5}\frac{e^{-4}}{-0.05}\] \[\Large \frac{e^{-10}}{-0.05}-\frac{e^{-9}}{-0.05}\] \[\Large -\frac{e^{-10}}{0.05}+\frac{e^{-9}}{0.05}\] \[\Large \frac{e^{-9}-e^{-10}}{0.05}\]
anonymous
  • anonymous
@amistre64 I got that too.
amistre64
  • amistre64
good, that means my latex coding dint mess up my keepiong track of signs and such :)
anonymous
  • anonymous
@shaynaderby did your class just start doing integrals where they gave upper and lower limits?
anonymous
  • anonymous
yes
anonymous
  • anonymous
everything is covered way too fast for me, it's business calculus, so stuff like this helps
amistre64
  • amistre64
what is the big red x in the pic you posted?
anonymous
  • anonymous
Well I always pass Paul's notes on. I recommend to anyone in calc and hopefully you'll get some help out of them :-) http://tutorial.math.lamar.edu/Classes/CalcI/ComputingDefiniteIntegrals.aspx
anonymous
  • anonymous
my previous wrong answer haha
anonymous
  • anonymous
thank you!
amistre64
  • amistre64
lol ... oh :)
amistre64
  • amistre64
plug in the fraction we got at the end and see if it likes it .... i can never tell what a program will accept for an answer. MathLab hates me :)
anonymous
  • anonymous
should I solve all the way through? This program is webassign
amistre64
  • amistre64
thats as "all the way through" as you can get .... maybe do something with the decimal in the denom , but other than that
amistre64
  • amistre64
if youre thinking of changed the es to an approximation make sure that is what the program is asking for then, otherwise exact is the best route
amistre64
  • amistre64
100/5 = 20 20(e^-9 - e^-10)
anonymous
  • anonymous
nope, it worked. Thank you guys so much, that really did help me to understand the process. I do have a few more questions about this particular problem though. Where did all of the sign changes come in, like when we went from subtracting one fraction from the other and then the first one became positive and we were adding them, then we put them together on top being subtracted?
amistre64
  • amistre64
the -.05 in the denominator
amistre64
  • amistre64
that - sign can be pulled out as -1/-1 and used where we need it
anonymous
  • anonymous
0.05 = 5/100 = 1/20 So if you have x/(1/20) = x*(20/1) = 20x Same concept as when he re-wrote the answer.
amistre64
  • amistre64
i have moments of lucidity ;)
anonymous
  • anonymous
:-)
anonymous
  • anonymous
okay that makes sense. I have another problem if you guys are up for helping me with it
anonymous
  • anonymous

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