If f is even then then left and right integrals are equal

1. Jun 4, 2015

Euklidian-Space

1. The problem statement, all variables and given/known data
If $f$ is an even function then $$\int_{-a}^{0} f = \int_{0}^{a} f$$

2. Relevant equations

3. The attempt at a solution
My attempt was trying to show the upper sum of both integrals were equal.

Take a partition of [-a,0] call it $P_{1}$, and $P_{2}$ for [0.a].

if we can show that $U(f,P_{1}) = U(f,P_{2})$ I think that might do it. However, i am not sure how to get there

2. Jun 4, 2015

fourier jr

I think it would be helpful to know that f is even if f(x) = f(-x)

3. Jun 5, 2015

momoko

Just do a substitution, say, y=-x. Then the result follows.

4. Jun 5, 2015

Euklidian-Space

is that a substitution on the limits or the variable?

5. Jun 5, 2015

Staff: Mentor

Yes (to both). If you change the variable, the limits of integration change as well.

6. Jun 5, 2015

Euklidian-Space

ok i keep getting a negative in there and I can't figure out how to get rid of it. here is my working

$$\int_{0}^{a} f(x) dx = \int_{0}^{a} f(-x) dx$$

let $y = -x$ $\rightarrow$ $dy = -dx$. So we now have $-\int_{0}^{-a} f(y) dy = \int_{-a}^{0} f(y) dy$

now if i sub -x back in for y I get a negative infront again since du = -dx. Any thoughts?

7. Jun 5, 2015

Staff: Mentor

Yes. When you reverse the order of the integration limits, the sign of the integral changes.

IOW, $\int_a^b f(x)dx = -\int_b^a f(x)dx$

8. Jun 5, 2015

momoko

Try to start from LHS and end with RHS.

9. Jun 5, 2015

SammyS

Staff Emeritus
He (or she) is not starting with what needs to be proved.

10. Jun 6, 2015

Noctisdark

Let f be any odd function, then lim x -> -a (f(-a)-f(x))/(-a-x) = (-f(a) - f(x))/(-a-x) = (f(a) + f(x))/(a+x) = (f(a) - f(-x))/(a-(-x)) = lim y -> a (f(a) - f(y))-(a-y)this proves that f'(-a) = f'(a), now since your function is even, it must be a derivative of and odd function , Good luck ;)
Edit, y = -x , just for the fancy notation !

11. Jun 6, 2015

Euklidian-Space

Yeah but isn't that not what we want. Maybe i am not seeing something but it seems we wouldnt want the negative out in front?

12. Jun 6, 2015

vela

Staff Emeritus
Try doing the substitution only on one side of the equation you started with.

13. Jun 6, 2015

Noctisdark

lim y -> a (f(a) - f(y))-(a-y) this is meant to be lim y-> a (f(a)-f(y))/(a-y)

14. Jun 6, 2015

Euklidian-Space

but i have been working on just one side. i do not see a point where i stop working on the side that I "started" with

15. Jun 6, 2015

vela

Staff Emeritus
Oh, I misunderstood what you wrote. You got
$$\int_0^a f(x)\,dx = \int_0^a f(-x)\,dx = -\int_0^{-a} f(y)\,dy = \int_{-a}^0 f(y)\,dy.$$ Do you see that you've shown what you setting out to prove?

Last edited: Jun 7, 2015
16. Jun 7, 2015

Euklidian-Space

But the variables are different no? When you sub -x back in for y wouldnt you get $$-\int_{-a}^{0} f(-x) dx?$$

17. Jun 7, 2015

vela

Staff Emeritus
Remember the variable of integration is a dummy variable.

18. Jun 7, 2015

Euklidian-Space

I am sorry if I am being difficult vela, but I just dont understand why showing

$$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(y) dy$$
shows that
$$\int_{0}^{a}f(x)dx = \int_{-a}^{0} f(x) dx$$

if anything i think i have shown...

$$\int_{0}^{a} f(x) dx = -\int_{-a}^{0} f(x) dx$$

Last edited: Jun 7, 2015
19. Jun 7, 2015

SammyS

Staff Emeritus
You may want to use the Edit feature to fix your typo !

20. Jun 7, 2015

Euklidian-Space

No thats what i meant...

Since dy = -dx we get that negative out in front done we?