anonymous
  • anonymous
integrate 2e^(t-2)x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
you sure?
anonymous
  • anonymous
yes
amistre64
  • amistre64
whats the variable to integrate by? x or t?

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More answers

anonymous
  • anonymous
dx
amistre64
  • amistre64
k.... the t-2 would act like a constant
anonymous
  • anonymous
(2e^(t-e)x)(1/(t-2)?
amistre64
  • amistre64
take the 2 to the outside since its a constant
amistre64
  • amistre64
then multiply by (t-2)/(t-2) and take the bottom one out since its acting like a constant as well
amistre64
  • amistre64
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)
amistre64
  • amistre64
... +c lol
anonymous
  • anonymous
why are you multiplying top and bottom by t-2
anonymous
  • anonymous
\[2(xe ^{t-2} - e ^{t-2})+c\]
amistre64
  • amistre64
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..
amistre64
  • amistre64
(t-2)/(t-2) = 1 so the value of the function stays the same; only thing that changes is the way it looks
anonymous
  • anonymous
wait so that "x" was "dx"?
anonymous
  • anonymous
how do you integrate this int (t-2) e^(t-2)x
amistre64
  • amistre64
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?
anonymous
  • anonymous
integrating wrt x
amistre64
  • amistre64
t-2 = say...u u e^ux ints up to e^ux
anonymous
  • anonymous
oh you have to use u-sub for this problem?
amistre64
  • amistre64
you can if it helps to see whats going on, but yeah....
amistre64
  • amistre64
u sub is just a way to clean up the function to see it better; gets the distracting stuff out the way
anonymous
  • anonymous
so putting it in that form is the easiest way to solve?
amistre64
  • amistre64
integration is not an 'easy' thing. it is an art more than a science
amistre64
  • amistre64
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
amistre64
  • amistre64
.... +C
anonymous
  • anonymous
so you're integrating 2e^ux?
amistre64
  • amistre64
yes, but recall that i defined u = t-2; and the t-2 is acting like a constant
anonymous
  • anonymous
yeah. i think i need to play around with similar problems and practice
anonymous
  • anonymous
thanks!
amistre64
  • amistre64
yw :)

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