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Suppose $\int^{-1.5}_{-6}f(x)$dx = 1, $\int^{-4.5}_{-6}f(x)$dx =9, $\int^{-1.5}_{-3}f(x)$dx = 9. I solved $\int^{-3}_{-4.5}f(x)$dx = -13. However, how do I get $\int^{-4.5}_{-3}(1(f(x))-9)$dx? I know that whatever integral I got from -4.5 to -3 has to be negated. What rule(s) do I need to solve this equation?

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2 Answers

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You essentially have a system of equations.

  • $\int^{-1.5}_{-6}f(x)$dx = 1 reduces to F(-1.5) - F(-6) = 1, where F'(x) = f(x).
  • $\int^{-4.5}_{-6}f(x)$dx = 9 reduces to F(-4.5) - F(-6) = 9

You can use those two equations to deduce that F(-1.5) - F(-4.5) = -8

Keep at it can you can find what F(-4.5) - F(-3) is.

$\int^{-4.5}_{-3}(f(x)-9)$dx evaluates to F(-4.5) - F(-3) + 9(-4.5) - 9(-3).

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Rule#1: $\int_a^bf(x)dx + \int_b^cf(x)dx = \int_a^cf(x)dx$

(Just from Newton's definition as $F(b) - F(a) + F(c) - F(b) = F(c) - F(a)$)

You have:

  • F(-1.5) - F(-6) = 1
  • F(-4.5) - F(-6) = 9
  • F(-1.5) - F(-3) = 9

I have calculated, that for $\int_{-4.5}^{-3}$: $F(-3) - F(-4.5) = (F(-1.5) - F(-6)) - (F(-1.5) - F(-3)) - (F(-4.5) - F(-6)) = 1 - 9 - 9 = -17$

Rule#2(reversion of integration bounds makes changes the sign): $\int_a^bf(x)dx = -\int_b^af(x)dx$ $\int_{-3}^{-4.5} = -\int_{-4.5}^{-3} = 17$

Rule#3(integral is linear operator): $\int_a^b(pf(x) + qg(x))dx = p\int_a^bf(x)dx + q\int_a^bg(x)dx$

You want to calculate: $\int_{-3}^{-4.5}(f(x)-9)dx$

$\int_{-3}^{-4.5}(f(x)-9)dx = \int_{-3}^{-4.5}f(x)dx - 9\int_{-3}^{-4.5}1dx$

$\int_{-3}^{-4.5}1dx = [x]_{-3}^{-4.5} = -4.5 - (-3) = -1.5$

$\int_{-3}^{-4.5}(f(x)-9)dx = 17 - 9 (-1.5) = 30.5$

(I hope a haven't made any mistake).

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The process is correct, but the integral from -4.5 to -3 is 13, not 17. –  cuabanana Apr 28 '13 at 0:01
    
Where did you get 13? –  V-X Apr 28 '13 at 8:17
    
I got it from the rule that when you reverse the limits of integration, you negate the original answer. BTW, the integral from -4.5 to -3 was given as -13. I negated because the problem reversed the limits of integration. –  cuabanana Apr 30 '13 at 0:54
    
number -13 doesn't correspond to the other known finite integrals. Please, check, if the question is correct. –  V-X Apr 30 '13 at 8:43
    
//except when there are delta "functions" at -3 or -4.5 and the whole question has no sense. –  V-X Apr 30 '13 at 8:44
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