Change in integration interval

[hobz]

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.

[D H]

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]

[hobz]

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?

[D H]

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).

[hobz]

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.

[D H]

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.)

[hobz]

Ahh, but of course. Thanks for spelling it out.

I have apparently developed a nasty habbit of ignoring constants.