1. ## Equivalent Integrals

Given the definite integral $\int_{\sqrt{2}}^{2} \frac{f(x)}{2\pi}dx$, which of the following integrals are equivalent and justify your answer. (There could be more then one.)

A. $\int_{0}^{2\pi} f(\frac{x}{\pi})dx$
B. $\int_{\frac{1}{4}}^{\frac{1}{2}}f(2sin(x\pi))cos(x \pi)dx$
C. $-\frac{1}{2\pi}\int_{1}^{2}\frac{f(\frac{1}{x})}{x^ 2}dx$

I need some help on this one...

2. Originally Posted by ebonyscythe
Given the definite integral $\int_{\sqrt{2}}^{2} \frac{f(x)}{2\pi}dx$, which of the following integrals are equivalent and justify your answer. (There could be more then one.)

A. $\int_{0}^{2\pi} f(\frac{x}{\pi})dx$
B. $\int_{\frac{1}{4}}^{\frac{1}{2}}f(2sin(x\pi))cos(x \pi)dx$
C. $-\frac{1}{2\pi}\int_{1}^{2}\frac{f(\frac{1}{x})}{x^ 2}dx$

I need some help on this one...
I'd suggest that you work backward: Take each answer and perform a substitution that will bring the argument of the function f back to x.

For example, look at the first answer:
$\int_{0}^{2\pi} f(\frac{x}{\pi})dx$

Let $u = \frac{x}{\pi} \implies du = \frac{dx}{\pi}$

So
$\int_{0}^{2\pi} f(\frac{x}{\pi})dx = \int_0^{2}f(u) \cdot \pi ~ du$

Change the "dummy" variable u to x and we get that
$\int_{0}^{2\pi} f(\frac{x}{\pi})dx = \pi \int_0^{2}f(x)~ dx$
which is not the same as the integral you are looking for.

-Dan