Question

integration with limits. question attached

Answer 1

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

Answer 2

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

Answer 3

That's the answer?

Answer 4

http://openstudy.com/code-of-conduct

"Don’t post only answers - guide the asker to a solution."

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

forget the limits at first; they are used later

Answer 9

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.

Answer 10

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

Answer 11

?

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

just to put things back on track :)

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

3^-5 is just a number

e = 2.71828181845905 .... is just a number

e^-5 is still just a number

Answer 21

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

Answer 22

yes :)

Answer 23

Yep!

Answer 24

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*

Answer 25

now we use the limits

Answer 26

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

Answer 27

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

Answer 28

okay I'm a bit lost

Answer 29

the latex is showing up wierd lately

Answer 30

@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.

Answer 31

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

Answer 32

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.

Answer 33

yes, but its important that you relate this to the general rules so that you know how they are being applied in specific cases

Answer 34

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.

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

so now I am plugging in 100?

Answer 39

what are your limits?

Answer 40

100 and 80

Answer 41

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

Answer 42

F(100)=-0.0009079986 ?

Answer 43

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

Answer 44

\[\Large e^{-5}\frac{e^{-0.05(100)}}{-0.05}-\Large e^{-5}\frac{e^{-0.05(80}}{-0.05}\]
\[\Large e^{-5}\frac{e^{-5}}{-0.05}-\Large e^{-5}\frac{e^{-4}}{-0.05}\]
\[\Large \frac{e^{-10}}{-0.05}-\frac{e^{-9}}{-0.05}\]
\[\Large -\frac{e^{-10}}{0.05}+\frac{e^{-9}}{0.05}\]
\[\Large \frac{e^{-9}-e^{-10}}{0.05}\]

Answer 45

@amistre64 I got that too.

Answer 46

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

Answer 47

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

Answer 48

yes

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

my previous wrong answer haha

Answer 53

thank you!

Answer 54

lol ... oh :)

Answer 55

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 :)

Answer 56

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

Answer 57

thats as "all the way through" as you can get .... maybe do something with the decimal in the denom , but other than that

Answer 58

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

Answer 59

100/5 = 20

20(e^-9 - e^-10)

Answer 60

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?

Answer 61

the -.05 in the denominator

Answer 62

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

Answer 63

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.

Answer 64

i have moments of lucidity ;)

Answer 65

:-)

Answer 66

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