Question

how can I find the anti-derivative of 1.2^(k)+1

Answer 1

Do you know how to take the derivative of \(\Large\bf\sf 1.2^k+1\) ?

Answer 2

nope...

Answer 3

but I was looking online, and I think I found the answer- (1.2^k)/(ln1.2) + K

Answer 4

You basically do (a^x)/(lna)

Answer 5

Yes,\[\Large\bf\sf (a^k)'\quad=\quad \ln a(a^k)\]When we integrate we divide by the coefficient instead,\[\Large\bf\sf \int\limits (a^k)dk\quad=\quad \frac{(a^k)}{\ln a}\]

Answer 6

And then for the other term, you just integrate 1 --> k.

Answer 7

Ya it's not too bad, just gotta remember your rules for differentiating exponentials :)

Answer 8

yeah- I just haven't worked with exponentials before, and they just sorta sprung them at us in the assignment randomly... Do you know much about applications of the definite integral?

Answer 9

Like real life applications? +_+

Answer 10

kind of... I have a question that I'm having issues with..

Answer 11

A certain computer algorithm used to solve very complicated differential equations uses an iterative method. That is, the algorithm solves the problem the first time very approximately,and then uses that first solution to help it solve the problem a second time just a little bit better, and then uses that second solution to help it solve the problem a third time just a little bit better, and so on. Unfortunately, each iteration takes a progressively longer amount of time. In fact, the amount of time it takes to process the k-th iteration is given by T(k)=1.2^k+1 seconds. Use a definite integral to approximate the time(in hours) it will take the computer algorithm to run through 60 iterations. (Note the T(k) is the amount of time it takes to process just the k-th iteration.)

Answer 12

Can I just use \[\int\limits_{0}^{60} 1.2^{k}+1 dk\]

Answer 13

and then divide by 3600 to find hours? or no

Answer 14

thinkingggg +_+

Answer 15

join the club :/

Answer 16

The amount of time it takes to complete the first iteration is: \(\Large\bf\sf T(1)\)
And time for the second iteration: \(\Large\bf\sf T(2)\).

So to complete all 60 iterations we would sum up all of those times,\[\Large\bf\sf \sum_{k=1}^{60}T(k)\]

Answer 17

And they want us to use the indefinite integral to approximate this?
Mmm my brain.

Answer 18

@Zarkon fix it!! XD

Answer 19

So we want a Riemann sum? The question asks us to use definite integrals...

Answer 20

Well the `definition` of the integral uses the Riemann sum :)
It's just the sum of a bunch of rectangles.

Answer 21

So they're closely related.

Answer 22

yes, as n approaches infinity, so if we just use the definite integral of the function from 0-60, would we get the same thing???

Answer 23

I have issues with application problems...

Answer 24

If it makes you feel better there is a part B ;D

Answer 25

Ya I dunno, I'm a little confused on this also :3
Definite Integrals give us area.
And so we're including all of the iterations between that we don't want.
Like between \(\Large\bf\sf T(1)\text{ and } \Large\bf\sf T(2)\) there is \(\Large\bf\sf T(1.5)\) and all those goofy numbers that we don't want.

I dunno, maybe it's a good approximation though +_+
My brain just isn't working.

What do we have for part b?

Answer 26

The maximum error in the computer's solution after k iterations is given by Error=2k^-2. Approximately how long (in hours) will it take the computer to process enough iterations to reduce the maximum error to below .0001?

Answer 27

Hmm that part seems pretty straight forward.
They just want a k value that makes the max error `less than` .0001.\[\Large\bf\sf \text{max error}=2k^{-2}\]\[\Large\bf\sf .0001\lt 2k^{-2}\]And then solve for k :o

Answer 28

That's what I figured. And then when you have k, plug it into the function that fives time (rounding to the next greater integer)

Answer 29

Yes* lol :)

Answer 30

I still can't make sense of part a, since we're only adding the integer values I don't see it approximates to the area under the curve.. very weird.

Answer 31

ok, so about 142 iterations for B?

Answer 32

Mmm yah that sounds about right!

Answer 33

I don't really know what to do about A. I thought the same you did.. but a Riemann sum is not a definite integral, and we can't really do it with the function in an integral... just stuck.