hobz Posted February 3, 2009 Share Posted February 3, 2009 I was wondering why/how the following statement is true. [math] \int_a^b f(x)\quad dx = \frac{b-a}{2} \int_{-1}^1 f \left( \frac{b-a}{2} x + \frac{a+b}{2} \right) [/math] I have come so far as to insert 1 and -1 instead of the x, and subtracting lower limit from upper limit. But that gives me [math] \frac{b-a}{2} F(b) - F(a) [/math] which has a strange constant in front of it. I have tried to interpret this constant as some sort of a mean-value of interval. I don't know why it's there or what is means. Can somebody shed some light on this matter? Thanks. Link to comment Share on other sites More sharing options...

D H Posted February 3, 2009 Share Posted February 3, 2009 I was wondering why/how the following statement is true. [math] \int_a^b f(x)\quad dx = \frac{b-a}{2} \int_{-1}^1 f \left( \frac{b-a}{2} x + \frac{a+b}{2} \right) \; dx [/math] Note: I added a 'dx' to the integral on the right-hand side. To see why this must be true, consider the simple u-substitution [math]u = \frac {2x}{b-a} - \frac{b+a}{b-a}[/math] The inverse of which is [math]x = \frac {b-a} 2 \,u + \frac{b+a}2[/math] With this substitution, the integration limits map to -1 and 1. Inside the integral, [math]f(x)\to f\left(\frac {b-a} 2 \,u + \frac{b+a}2\right)[/math] and [math]dx \to \frac {b-a} 2 \, du[/math] Putting this all together, [math]\int_a^b f(x)\, dx = \frac {b-a} 2 \int_{-1}^1 f\left(\frac {b-a} 2 \,u + \frac{b+a}2\right) du[/math] That the right-hand side uses du is irrelevant; u is just a dummy variable here. Changing the us to xs will not change the value of the integral at all. To see that the right-hand side still evaluates to F(b)-F(a), note that for any function g(x), if [math]\int g(x) dx = G(x)[/math] then [math]\int g(kx+a) = \frac 1 k G(kx+a)[/math] if the integral exists. (Proof: Differentiate.) With this, the right-hand side of the identity in question evaluates to [math]\frac {b-a} 2 \frac{2}{b-a} \left.F\left(\frac {b-a} 2 \,u + \frac{b+a}2\right)\right|_{-1}^1 = F(b) - F(a)[/math] Link to comment Share on other sites More sharing options...

hobz Posted February 4, 2009 Author Share Posted February 4, 2009 Thank you for answering. I see your point, and understand that then constant is there to "kill" [math]du = \frac{2}{b-a}[/math]. However, I fail to see why substitution is needed. What is wrong with the logic I use? Link to comment Share on other sites More sharing options...

D H Posted February 4, 2009 Share Posted February 4, 2009 Since you didn't show how you arrived at your conclusion, it's a bit hard to tell where you went wrong. Read my post again. I worked the problem forwards by showing how to derive the identity and backwards by showing that the identity leads to F(b)-F(a). Link to comment Share on other sites More sharing options...

hobz Posted February 4, 2009 Author Share Posted February 4, 2009 I insert the limits on the right hand side: [math] \frac{b-a}{2} \int_{-1}^1 f \left( \frac{b-a}{2} x + \frac{b+a}{2} \right) dx [/math] [math] \frac{b-a}{2} \left( F \left( \frac{b-a}{2} \cdot 1 + \frac{b+a}{2} \right) - F \left( \frac{b-a}{2} \cdot -1 + \frac{b+a}{2} \right) \right) [/math] [math] \frac{b-a}{2} \left( F \left( \frac{b}{2}-\frac{a}{2} + \frac{a}{2} + \frac{b} {2} \right) - F \left( -\frac{b}{2}+\frac{a}{2} + \frac{a}{2} + \frac{b} {2} \right) \right) [/math] giving [math] \frac{b-a}{2} \left( F(b) - F(a) \right) [/math] which is not equal to the left hand side. Link to comment Share on other sites More sharing options...

D H Posted February 4, 2009 Share Posted February 4, 2009 Your mistake here is that your second equation does not follow from the first. I'll repeat the pertenant part of post #2. For any function g(x), if [math]\int g(x) dx = G(x)[/math] then, if the integral exists, [math]\int g(kx+c) = \frac 1 k G(kx+c)[/math] (Proof: Differentiate.) Link to comment Share on other sites More sharing options...

hobz Posted February 4, 2009 Author Share Posted February 4, 2009 Ahh, but of course. Thanks for spelling it out. I have apparently developed a nasty habbit of ignoring constants. Link to comment Share on other sites More sharing options...

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