Question

Calculate the integral:

Answer 1

i hope its a good one

Answer 2

\[\int\limits_{0}^{\infty}|9e ^{-(1/4)t}-6e ^{-(1/2)t}|dt\]

Answer 3

It is, it is.... in another topic i asked about the ≤

Answer 4

first, break it down into the sum of two integrals.

Answer 5

(or difference, in this case)

Answer 6

then factor out the constants.

Answer 7

focus on the generic: e^(-x) type

Answer 8

\[\int e^{ax}dx=\frac{1}{a}e^{ax}+c\]

Answer 9

i know that but with these sign | |

Answer 10

\[\frac{d}{dx}(e^{-kx})=-k\ e^{-kx}\]
so all your missing to match is the "-k" part

Answer 11

eww.. absolute value integrals

Answer 12

it might help to know that |a| = a, and |-a| = a

Answer 13

id have to read up on the abs value stuff again

Answer 14

http://tutorial.math.lamar.edu/Classes/CalcI/ComputingDefiniteIntegrals.aspx

this helps

Answer 15

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

Answer 16

and of course using the imporper integration techniques for this one might be useful as well

Answer 17

but guys.. i have a question...

do i have to take the abs value of my integral? or is that the last thing i have to do?

Answer 18

\[[-36e ^{-(1/4)t}+12e ^{-(1/2)t}]\]

or will it be

[+36e ^{-(1/4)t}+12e ^{-(1/2)t}]

Answer 19

\[9e^{−(1/4)t} − 6e^{−(1/2)t}\]
take the first derivative to find any critical points:
\[\frac{-9}{4}e^{−(1/4)t} + \frac{6}{2}e^{−(1/2)t}=0\]
after some algebra which I messed up on but the wolf gives: we get:

t = 8 ln(2) - 4ln(3)

but better yet, graphing this thing tells us there is no cusp in the interval so we just regard this as the + side

http://www.wolframalpha.com/input/?i=%7C9e%5E%28-t%2F4%29-6e%5E%28-t%2F2%29%7C

Answer 20

in other words, ignore the | | since they are pointless across the interval [3,inf)

Answer 21

where did I get 3 from? doh!!

still, the cusp, or the bad part, is not in your interval so just integrate it without the | |

Answer 22

but amisre64? what do you mean...

there are 2 ways to calculate this, but what is the right one?

Take the abs. first and then fill in the boundaries or fill in the boundaries first and then take the abs.

I will get the answer 24 or 48 whats the right one?

Answer 23

24 is the right answer

an absolute value function tends to have what is called a cusp; that is the pointy part that has no derivative.

if the cusp was in our interval we would integrate it in 2 sets; but since the cusp is not in our interval we can go ahead and ignore the | | notation

Answer 24

if 24 is the right answer...

i have to fill in the boundaries and then take the abs. value of it

Answer 25

then id go that route :) since wolfram gives us 24 as an answer

Answer 26

http://www.wolframalpha.com/input/?i=integrate+%7C9e%5E%28-t%2F4%29-6e%5E%28-t%2F2%29%7C+from+0+to+infinity

Answer 27

hmm... oke.. weird why is my teacher saying that i have to take the +36 instead of -36... then he just gave me the wrong answer

Answer 28

but thank you for your explanation amistre64 :)