Question

suppose ∫f(t)dt (from 0 to 1)=3. calculate the following:
∫f(2t)dt (from 0 to 0.5)
∫f(1-t)dt (from 0 to 1)
∫f(3-2t)dt (from 1 to 1.5)

Answer 1

I'm not sure how to explain this, it's a bit of a trick question. If you look closely you can see that those three integrals are all really the same as \(\int_0^1f(t)dt\). First they tranform inside f, only to undo it with the integration boundaries.

Answer 2

So for the first one, 2t for t between 0 to 0.5 is really the same as t for t between 0 and 1.

Answer 3

In both cases you get all the points between 0 and 1.

Answer 4

so they all equal 3?

Answer 5

That's right.

Answer 6

can you show me how to do it?

Answer 7

Like I said, I'm having a tough time explaining this, I don't think I can do any better than I already did, sorry.

Answer 8

thanks