Question

integrate 2e^(t-2)x

Answer 1

you sure?

Answer 2

yes

Answer 3

whats the variable to integrate by? x or t?

Answer 4

dx

Answer 5

k....

the t-2 would act like a constant

Answer 6

(2e^(t-e)x)(1/(t-2)?

Answer 7

take the 2 to the outside since its a constant

Answer 8

then multiply by (t-2)/(t-2) and take the bottom one out since its acting like a constant as well

Answer 9

and we int((t-2) e^(t-2)x) which = e^(t-2)x

bring over the constants to get:

2 e^(t-2)x
---------
(t-2)

Answer 10

... +c lol

Answer 11

why are you multiplying top and bottom by t-2

Answer 12

\[2(xe ^{t-2} - e ^{t-2})+c\]

Answer 13

because I know how to int (t-2) e^(t-2)x; so I need to use a useful form of '1' to help me out..

Answer 14

(t-2)/(t-2) = 1 so the value of the function stays the same; only thing that changes is the way it looks

Answer 15

wait so that "x" was "dx"?

Answer 16

how do you integrate this int (t-2) e^(t-2)x

Answer 17

and the way it looks is important to me... becasue I can int (t-2) e^(t-2)x :) the rest is just constants that can be dragged outside right?

Answer 18

integrating wrt x

Answer 19

t-2 = say...u

u e^ux ints up to e^ux

Answer 20

oh you have to use u-sub for this problem?

Answer 21

you can if it helps to see whats going on, but yeah....

Answer 22

u sub is just a way to clean up the function to see it better; gets the distracting stuff out the way

Answer 23

so putting it in that form is the easiest way to solve?

Answer 24

integration is not an 'easy' thing. it is an art more than a science

Answer 25

t-2 = u

{S} 2 e^ux dx

2 {S} e^ux dx

2 {S} (u/u) e^ux dx

2/u {S} u e^ux dx

(2/u) e^ux

Answer 26

.... +C

Answer 27

so you're integrating 2e^ux?

Answer 28

yes, but recall that i defined u = t-2; and the t-2 is acting like a constant

Answer 29

yeah. i think i need to play around with similar problems and practice

Answer 30

thanks!

Answer 31

yw :)