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\[\int\limits_{0}^{9}f(x)dx=37; \int\limits_{0}^{9}g(x)dx=16\]
Find: \[\int\limits_{0}^{9}[2f(x)+3g(x)]dx\]
  • phi
the integral is "linear" in other words, for a constant "a" \[ \int\limits_{0}^{9} a f(x)dx = a\int\limits_{0}^{9} f(x)dx \]

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so in this case \[2\int\limits_{0}^{9} x.dx\]??
  • phi
and you can separate integrals (assuming they have the same limits \[ \int\limits_{0}^{9} f(x) + g(x) dx= \int\limits_{0}^{9} f(x)dx + \int\limits_{0}^{9} g(x)dx\]
  • phi
not x , f(x) so separate the problem into \[ 2\int\limits_{0}^{9} f(x)dx + 3\int\limits_{0}^{9} g(x)dx \]
yea right! so \[37x(9,0)+16x(9,0)\]
  • phi
I don't know about the x(9,0) part just replace the integral with the number they say it is equal to \[ 2\cancel{(\int\limits_{0}^{9} f(x)dx)}37 + 3\int\limits_{0}^{9} g(x)dx \] and do the same for the other, it is 16
  • phi
I am trying to show you sub in 37 for the result of the first integral
ohh alright, i got it!
so its 2(37) + 3(16) = 122
thanks @phi :)

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