integrate 2e^(t-2)x
you sure?
yes
whats the variable to integrate by? x or t?
dx
k.... the t-2 would act like a constant
(2e^(t-e)x)(1/(t-2)?
take the 2 to the outside since its a constant
then multiply by (t-2)/(t-2) and take the bottom one out since its acting like a constant as well
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)
... +c lol
why are you multiplying top and bottom by t-2
\[2(xe ^{t-2} - e ^{t-2})+c\]
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..
(t-2)/(t-2) = 1 so the value of the function stays the same; only thing that changes is the way it looks
wait so that "x" was "dx"?
how do you integrate this int (t-2) e^(t-2)x
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?
integrating wrt x
t-2 = say...u u e^ux ints up to e^ux
oh you have to use u-sub for this problem?
you can if it helps to see whats going on, but yeah....
u sub is just a way to clean up the function to see it better; gets the distracting stuff out the way
so putting it in that form is the easiest way to solve?
integration is not an 'easy' thing. it is an art more than a science
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
.... +C
so you're integrating 2e^ux?
yes, but recall that i defined u = t-2; and the t-2 is acting like a constant
yeah. i think i need to play around with similar problems and practice
thanks!
yw :)
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