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Math Help - Simple integration problem

  1. #1
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    Simple integration problem

    Okay so I know how to integrate really well, but I came across a problem that left me confused.

    x[0,1] f(x) = (1-x)^9 dx

    I was told that when integrating this problem to put a negative sign in front of the integrand, but not why. Why would you do this?
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  2. #2
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    Quote Originally Posted by radioheadfan View Post
    Okay so I know how to integrate really well, but I came across a problem that left me confused.

    x[0,1] f(x) = (1-x)^9 dx

    I was told that when integrating this problem to put a negative sign in front of the integrand, but not why. Why would you do this?
    Try the substitution u= 1- x. Then (1- x)^9 becomes u^9 which is easy to integrate. But you also have to replace "dx". With u= 1- x, du/dx= -1 so du= -dx and dx= -du. Your integral becomes -\int u^9(-du)= -\int u^9 du.

    Notice that I didn't put the limits of integration here. You could integrate -u^9, to get -1/10 u^{10}, of course, change that to -1/10 (1-x)^9 and evaluate that at x=0 and x=1.

    But I prefer to change the limits of integration as I substitute. Those are x= 0 and x= 1 and they have to be changed to u also. When x= 0, u= 1 and when x= 1, u= 0.
    \int_0^1 (1-x)^9 dx= -\int_1^0 u^9 du
    That "-" in "1- x" has reversed the order of the limits of integration- but \int_a^b f(x)dx= -\int_b^a f(x)dx so you really have
    \int_0^1 (1-x)^9 dx= -\int_1^0 u^9 du= \int_0^1 u^9 du
    and the negative has disappeared!

    Another way of looking at it is to remember that integration is the opposite of differentiation. Whatever function you integrated to, you would have f(u)= f(1- x). And you would use the "chain rule" to differentiate: the derivative of f(1-x) is the derivative of f times the derivative of 1- x, which is -1.

    With any substitution for a decreasing function, like 1- x, the derivative will be negative. That's why you need the "-" in the anti-derivative.
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  3. #3
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    oh wow I get it.
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