Kkutie7
  • Kkutie7
I've forgotten how to set this problem up:
Mathematics
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SOLVED
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chestercat
  • chestercat
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Kkutie7
  • Kkutie7
If \[\int\limits_{1}^{3}f(w)dw=7\] find the value of \[\int\limits_{1}^{2}f(5-2x)dx\]
anonymous
  • anonymous
Substitute \[5-2x=w\]
random231
  • random231
by f(omega) it means a periodic function right?

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Kkutie7
  • Kkutie7
uh.... I don't think so
random231
  • random231
nishant is right!
Kkutie7
  • Kkutie7
@Nishant_Garg that doesn't work
Kkutie7
  • Kkutie7
Would work if the limits were the same
random231
  • random231
that does! 5-2x=w then -2dx=dw you need to change the limits then
anonymous
  • anonymous
\[5-2x=w\]\[dx=-\frac{1}{2}dw\] Limits become 3 to 1 |dw:1449810288528:dw|
anonymous
  • anonymous
After that you can use the fact that \[\int\limits_{a}^{b} f(x) dx=-\int\limits_{b}^{a}f(x)dx\]
Kkutie7
  • Kkutie7
right...
random231
  • random231
^^^^^
Kkutie7
  • Kkutie7
So \[\frac{7}{2}\]

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