# 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?