find $\int_{-1}^1(1-2f(1-2x))dx$

Question

I am given $$\int_0^1f(x)dx=1$$ and I need to find $$\int_{-1}^1(1-2f(1-2x))dx=\int_{-1}^11dx+\int_{-1}^1-2f(1-2x)dx$$ what I was trying to do is: $$t=1-2x$$ $$x=\frac{1-t}{2}$$ $$\int_{x=-1}^{x=1}-2f(1-2x)dx = \int_{t=3}^{t=-1}-2f(t)\frac{dx}{dt}\cdot dt= \int_{t=3}^{t=-1}-2f(t)(-\frac{1}{2})\cdot dt= -\int_{t=-1}^{t=3}f(t)dt$$

but the bounds don't work out. am I missing something?
If the bounds of the desired integral and given integral were switched, I could use my method, so maybe there's a mistake in the problem.

Answer 1

It does look like you would need some additional information. For example, consider the functions $f_1, f_2:\Bbb R\to\Bbb R$ defined as $$f_1(x) = 1;\quad f_2(x) = \begin{cases}1 & x \in [0, 1]\\0 & x\not\in[0, 1]\end{cases}$$

Clearly, both $f_1$ and $f_2$ satisfy the original (integral) condition for $f$. However, they give different values for the desired integral.

Answer 2

Or, maybe the problem should have stated we are given $$ \int_{-1}^{1} f(x) \, dx = 1$$ and asked for $$ \int_0^1 (1- 2f(1-2x)) \, dx ?$$

Now your $t-$substitution works!