Evaluate the definite integral of (x-5)^21 dx from 0 to 10

Question

hartnn · Answer

put x-5 = u
du=... ?

anonymous · Answer

1? because isn't du just derivative of x-5?

hartnn · Answer

not exactly 1 
when 
x-5 = u 
1-0 = du/dx
so, du = dx 
got this ?

anonymous · Answer

Yes and then would it be (u)^21 du?

hartnn · Answer

yes!
 now don't forget to change the limits
 when x= 0, u=... ?
when x = 10, u =.... ?

anonymous · Answer

-5? and 5?

hartnn · Answer

yes! 
so, you have $\int \limits_{-5}^5u^{21}du$
tight ?
any ideas how to solve this ?

anonymous · Answer

could you put the 5 and -5 into x-5? and take antiderivative of u^21? and put 5 and -5 in and subtract?

hartnn · Answer

can you think of an easier way ? 
i mean look at the limits 
-5 to +5
sounds like -a to +a 
do you recollect any property of definite integral ??

anonymous · Answer

i dont remember one that had a and -a in it

hartnn · Answer

ok, let me tell ya, 
$\huge \int \limits_{-a}^a f(x)dx = 0$
IF f(x) is odd function. 
so, here IS u^21 an ODD function ?
do you know what is an odd function ?

anonymous · Answer

yes

hartnn · Answer

yes for what ?

anonymous · Answer

is it odd because you can get f(x)=-f(x)?

hartnn · Answer

yes, its odd, so it makes the integral = 0 
simple, right ?

anonymous · Answer

so it's just a basic property?

hartnn · Answer

yup.

anonymous · Answer

that's much simpler than I had thought.

anonymous · Answer

Thank you for explaining this to me!

hartnn · Answer

welcome ^_^