integration with limits. question attached

- anonymous

integration with limits. question attached

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- anonymous

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- anonymous

i think we should be able to let u equal -0.05x-5

- amistre64

\[a^{n+m}=a^na^m\]

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- anonymous

That's the answer?

- amistre64

http://openstudy.com/code-of-conduct
"Donâ€™t post only answers - guide the asker to a solution."

- anonymous

@shaynaderby, you can split this up into two parts. That's what @amistre64 hinted at.

- anonymous

okay well that makes sense. I just have no idea what do to with limits, they get plugged in, right?

- anonymous

Okay I can walk you through this. Try splitting it up.

- amistre64

forget the limits at first; they are used later

- 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

e^(-0.05)-e^(-5)

- anonymous

?

- amistre64

e^(-5) is a constant; no need to integrate it

- anonymous

well I know integrating e^x is always e^x

- 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

\[\int e^{-0.05x-5}dx=\int e^{-0.05x}e^{-5}dx\to\ e^{-5}\int e^{-0.05x}dx\]

- amistre64

just to put things back on track :)

- anonymous

what exactly about e^-5 makes it a constant, for future reference

- anonymous

e^-5 is just a number. Just like the number 4.

- amistre64

\[\int e^{kx}dx=\frac{e^{kx}}{k}\]
might e a rule in the back of a book

- amistre64

3^-5 is just a number
e = 2.71828181845905 .... is just a number
e^-5 is still just a number

- anonymous

so the integral of e^(-0.05x) is e^(-0.05x)/-0.05 ?

- amistre64

yes :)

- anonymous

Yep!

- 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

now we use the limits

- anonymous

but where are we at this point now that we integrated it?

- amistre64

\[\int_{a}^{b} f(x)dx=F(b)-F(a)\]

- anonymous

okay I'm a bit lost

- amistre64

the latex is showing up wierd lately

- 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

\[\int f(x)dx = F(x)+c\]
\[\int_{a}^{b} f(x)dx = F(b)-F(a)\]

- 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

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

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

so what does my function look like now though, before I do that

- amistre64

you found F(x): \(\Large e^{-5}\frac{e^{-0.05x}}{-0.05}\)

- anonymous

okay great that's what I wrote down actually when I tried it myself

- anonymous

so now I am plugging in 100?

- amistre64

what are your limits?

- anonymous

100 and 80

- amistre64

then find F(100) and subtract F(80)

- anonymous

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- anonymous

F(100)=-0.0009079986 ?

- anonymous

ohhh I see where I may have screwed up, I never put e^(-5) on top

- 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

@amistre64 I got that too.

- amistre64

good, that means my latex coding dint mess up my keepiong track of signs and such :)

- anonymous

@shaynaderby did your class just start doing integrals where they gave upper and lower limits?

- anonymous

yes

- anonymous

everything is covered way too fast for me, it's business calculus, so stuff like this helps

- amistre64

what is the big red x in the pic you posted?

- 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

my previous wrong answer haha

- anonymous

thank you!

- amistre64

lol ... oh :)

- 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

should I solve all the way through? This program is webassign

- 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

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

100/5 = 20
20(e^-9 - e^-10)

- 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

the -.05 in the denominator

- amistre64

that - sign can be pulled out as -1/-1 and used where we need it

- 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

i have moments of lucidity ;)

- anonymous

:-)

- anonymous

okay that makes sense. I have another problem if you guys are up for helping me with it

- anonymous

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