Why does it equal to zero? I know it has something to do with sine being an odd function, but why does the integral of an odd function equal 0 ? (if it does)
If it exists it must be zero as the integrand is an odd function.
This is because if the integral of odd function $\displaystyle f(x)$ exits it is:
$\displaystyle \displaystyle I=\lim_{L\to \infty} \int_{-L}^L f(t)\ dt = \lim_{L\to \infty} \left[\int_{-L}^0 f(t)\ dt + \int_{0}^L f(t)\ dt \right] $
.... $\displaystyle \displaystyle =\lim_{L\to \infty} \left[\int_{0}^L f(-t)\ dt + \int_{0}^L f(t)\ dt \right]=\lim_{L\to \infty} \left[\int_{0}^L -f(t)\ dt + \int_{0}^L f(t)\ dt \right]=0$
Alternatively draw a picture and it will be obvious.
CB
When you have e negative integral from minus infinity to 0, the minus sign in front of the integral changes it to being from 0 to minus infinity, correct? Then how come it appears as being from 0 to positive infinity? Does that sign change aswell?
Also, i don't think i know how to draw the graph of that function.
And, how did you write the reply using math symbols (integral symbol, limit with subscript, etc) directly onto the reply?
It is not at all clear what you are asking here! If f(x) is an odd function, then f(-x)= -f(x). I think that, by "the minus sing in front of the integral" you are referring to the minus sign you get from f(-x)= -f(x). But to get that, you have to change from "f(x)" to "f(-x)".
$\displaystyle \int_{-a}^a f(x)dx= \int_{-a}^0 f(x)dx+ \int_0^a f(x) dx$.
Now, make the change of variable u= -x in the first integral. Since u= -x, du= -dx and the lower limit is -(-a)= a:
$\displaystyle \int_{-a}^a f(x)dx= -\int_a^0 f(-u)du= \int_0^a f(-u)du= -\int_0^a f(u)du$.
If f(x) has anti-derivative F(x) (remember Captain Black said "if it exists ...") then $\displaystyle -\int_0^a f(u)du= -F(a)+ F(0)$.
Of course $\displaystyle \int_0^a f(x)dx= F(a)- F(0)$ so the two parts, $\displaystyle \int_{-a}^0 f(x)dx$ and $\displaystyle \int_0^a f(x)dx$ cancel.
As far as drawing the graph is concerned, I don't think Captain Black intended for you to draw a detailed graph (although that wouldn't be difficult using a graphing calculator). But you should be able to see that since f(-x)= -f(x), any region where f(x)> 0 to the right of x= 0 is mirrored by a region to the left where f(x)< 0 and those areas cancel each other.Also, i don't think i know how to draw the graph of that function.
Go back to the home page and look under "Math Resources" "Latex Help".And, how did you write the reply using math symbols (integral symbol, limit with subscript, etc) directly onto the reply?
Well you said that $\displaystyle -\int_0^a f(u)du= -F(a)+ F(0)$.
and that $\displaystyle \int_0^a f(x)dx= F(a)- F(0)$
But isn't F(a) from the first integral different from the F(a) in the second, since u is different than x, wouldnt that make the two F(a)'s to be different, thus not cancelling? (u=-x, which means its different than x, correct?) Basicly what im saying is, the function f(u) is different from f(x), since u is not equal to x, which would mean their integrations would be different aswell, wouldn't it?
No, it isn't. $\displaystyle \int_a^b f(x)dx= \int_a^b f(u)du= \int_a^b f(t)dt= \int_a^b f(y)dy= \cdot\cdot\cdot$
The variable inside the integand is a "dummy" variable. It does not matter what you call it.
No, the functions in f(x), f(u), f(t), f(y), etc. are all the same function- f.(u=-x, which means its different than x, correct?) Basicly what im saying is, the function f(u) is different from f(x)
No, you need to review your "function notation"., since u is not equal to x, which would mean their integrations would be different aswell, wouldn't it?