# Integral Help

• Oct 8th 2012, 10:08 PM
ineedhelplz
Integral Help
Attachment 25121

Why is it that f(x) is treated the same as f(x/5)?

I know the answer is 40a, I just don't understand why the factor isn't taken into account.

Thanks
• Oct 8th 2012, 10:21 PM
chiro
Re: Integral Help
Hey ineedhelplz.

The factor is taken into account within the limits of the new integral. Let z*5 = x. 5*dz = dx/5. If the old limits for z were [0,a] then when z = 0, x = 0 and when z = a, x = 5a. This is just an integral substitution that retains the value of the integral, but compensates for the change in variable by changing the limits.
• Oct 8th 2012, 10:31 PM
ineedhelplz
Re: Integral Help
Where is the z coming from?

Sorry I'm a bit confused
• Oct 8th 2012, 10:38 PM
chiro
Re: Integral Help
It's just a dummy variable. where we define 5*z = x or z = x/5 and look at the integral in relation to x/5 instead of x. So if you think of the first integral in terms of z = x and the next integral in terms of x = 5*z you go from a normal f(z) integral to a f(5*z) or f(x/5) which is what is in the integral on the RHS.\

You can go from f(x) to f(x/5) but I've just introduced a dummy variable so that you don't confuse the two x's as being the same.

The integration substitution formula is that if we have an integral of Integral [a,b] f(g(x))g'(x)dx then we make the substitution u = g(x) and this gives us the new integral [g(a),g(b)] f(u)du and both integrals have the same value.
• Oct 8th 2012, 10:45 PM
MarkFL
Re: Integral Help

$2\int_0^{5a}f\left(\frac{x}{5} \right)+3\,dx=2\int_0^{5a}f\left(\frac{x}{5} \right)\,dx+6(5a-0)$

We may make a substitution:

$u=\frac{x}{5}\,\therefore\,dx=5\,du$ and we have:

$2\int_0^{5a}f\left(\frac{x}{5} \right)+3\,dx=10\int_0^{a}f(u)\,du+30a=10a+30a=40a$

As mentioned, the variable of integration is a "dummy" variable, as it gets integrated out in the evaluation of the definite integral.
• Oct 8th 2012, 10:49 PM
ineedhelplz
Re: Integral Help
Thanks very much!