# Question from my test

• Dec 14th 2009, 03:50 PM
Latszer
Question from my test
I had a problem on my final today and was just wondering if I got it right.

if $\displaystyle \int_0^1 f(t)dt = 7$ then what does $\displaystyle \int_2^2.5 f(5-2t)dt$

I think that is the problem.
• Dec 14th 2009, 03:51 PM
Latszer
The second integral should be from 2 to 2.5 of f(5-2t) dt, sorry.
• Dec 14th 2009, 04:44 PM
oblixps
use a u-substitution. let u = 5-2t so du = -2 dt

you will get (-1/4)integral from 2 to 2.5 of f(u)du. now we need to change the limits on integration to be in terms of u since as of now they are in terms of t. remember the substitution we made u = 5-2t. substitute 2 and 2.5 for t and you will get u=1 and u=0 respectively.

you will now have (-1/4)integral from 1 to 0 of f(u)du. now this looks pretty similar to the integral that was given. if we switch the limits of integration we switch the sign of the integral. (this is a property of definite integrals).

our integral is now: (1/4) [integral from 0 to 1 of f(u)du], notice that inside the brackets is exactly the integral you mentioned above that equals 7. even though at the top it was f(t)dt, in the integral we have now, f(u)du is the same thing since its just a dummy variable. we can substitute that integral for 7 so the final answer should be 7/4.
• Dec 14th 2009, 04:50 PM
treetheta
Quote:

Originally Posted by oblixps
use a u-substitution. let u = 5-2t so du = -2 dt

you will get (-1/4)integral from 2 to 2.5 of f(u)du. now we need to change the limits on integration to be in terms of u since as of now they are in terms of t. remember the substitution we made u = 5-2t. substitute 2 and 2.5 for t and you will get u=1 and u=0 respectively.

you will now have (-1/4)integral from 1 to 0 of f(u)du. now this looks pretty similar to the integral that was given. if we switch the limits of integration we switch the sign of the integral. (this is a property of definite integrals).

our integral is now: (1/4) [integral from 0 to 1 of f(u)du], notice that inside the brackets is exactly the integral you mentioned above that equals 7. even though at the top it was f(t)dt, in the integral we have now, f(u)du is the same thing since its just a dummy variable. we can substitute that integral for 7 so the final answer should be 7/4.

7/2 i think the .5 was supposed the appart of the integral.