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More integral issues

  1. Nov 27, 2005 #1
    Here's the question:


    b b
    ∫ f(x)dx = a + 2b, then ∫ (f(x) + 5)dx = ?
    a a

    I'm thinking myself into circles... I want to say I need to take the derivative of a+2b to then find out what equals f(x) and then just take the integral of that +5... but its just not working out.
  2. jcsd
  3. Nov 27, 2005 #2


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    If I am reading correctly what you wrote then you are merely adding 5(b-a) to the original integral.
  4. Nov 28, 2005 #3


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    [tex]\int_a^b (f(x)+ 5)dx= \int_a^b f(x)dx+ 5\int_a^b dx[/tex]
  5. Nov 28, 2005 #4
    So would it be just a+2b+5?
  6. Nov 28, 2005 #5

    Physics Monkey

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    Nope, what does [tex] \int^b_a dx [/tex] equal?
  7. Nov 28, 2005 #6
    Ok, so then I just do 5(b-a)?
  8. Nov 28, 2005 #7


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    You have to add that, yes :smile:
  9. Nov 28, 2005 #8
    Yeah, that's what I meant to say.

    So in this problem:

    If f(x)=g(x)+7 from 3 to 5, then the integral from 3 to 5 of [f(x)+g(x)]dx is?

    Can I just use the same method and get

    2 ∫ g(x)dx+7
  10. Nov 28, 2005 #9


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    Almost, don't forget that the 7 was in the integrand!

    [tex]\int\limits_3^5 {f\left( x \right) + g\left( x \right)dx} = \int\limits_3^5 {g\left( x \right) + 7 + g\left( x \right)dx} = 2\int\limits_3^5 {g\left( x \right)dx} + 7\int\limits_3^5 {dx} [/tex]
  11. Nov 28, 2005 #10
    how do i figure dx in this case? do i use g(x) or f(x)?
  12. Nov 28, 2005 #11


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    You either use f(x) and substitute g(x) by f(x)-7 or you use g(x), and substitute f(x) by g(x)+7.
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