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

[Euklidian-Space]

Homework Statement

If $f$ is an even function then $$\int_{-a}^{0} f = \int_{0}^{a} f$$

Homework Equations

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

 

[fourier jr]

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

 

[momoko]

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

 

[Euklidian-Space]

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

is that a substitution on the limits or the variable?

 

[Mark44]

is that a substitution on the limits or the variable?

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

 

[Euklidian-Space]

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

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?

 

[Mark44]

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?

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

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

 

[momoko]

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?

Please do not start with what you need to prove.
Try to start from LHS and end with RHS.

 

[SammyS]

Please do not start with what you need to prove.
Try to start from LHS and end with RHS.

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

 

[Noctisdark]

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

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 !

 

[Euklidian-Space]

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

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

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

 

[vela]

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?

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

 

[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 !

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

 

[Euklidian-Space]

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

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

 

[vela]

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?

 

[Euklidian-Space]

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?

But the variables are different no? When you sub -x back in for y wouldn't you get $$-\int_{-a}^{0} f(-x) dx?$$

 

[vela]

Remember the variable of integration is a dummy variable.

 

[Euklidian-Space]

Remember the variable of integration is a dummy variable.

I am sorry if I am being difficult vela, but I just don't 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...

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

 

[SammyS]

I am sorry if I am being difficult vela, but I just don't 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...

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

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

 

[Euklidian-Space]

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

No that's what i meant...

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

 

[SammyS]

No that's what i meant...

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

So $\displaystyle \ \int{-a}^{0} f(y) dy\$ is not a typo ??

What does it mean ?

 

[Euklidian-Space]

So $\displaystyle \ \int{-a}^{0} f(y) dy\$ is not a typo ??

What does it mean ?

oh sorry, didnt see that. it is fixed now.

 

[SammyS]

I am sorry if I am being difficult vela, but I just don't understand why showing

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

Well it's like this:

$\displaystyle \ \int_{-a}^{0} f(y) dy\$

$\displaystyle \ =\int_{-a}^{0} f(t) dt\$

$\displaystyle \ =\int_{-a}^{0} f(u) du\$

...
​

It's even equal to $\displaystyle \ \int_{-a}^{0} f(x) dx\$

That's what vela means by "it's a DUMMY variable."

 

[Euklidian-Space]

Well it's like this:

$\displaystyle \ \int_{-a}^{0} f(y) dy\$

$\displaystyle \ =\int_{-a}^{0} f(t) dt\$

$\displaystyle \ =\int_{-a}^{0} f(u) du\$

...
​

It's even equal to $\displaystyle \ \int_{-a}^{0} f(x) dx\$

That's what vela means by "it's a DUMMY variable."

Again I am sorry man i just don't understand. dy = -dx right? then how does $\int_{-a}^{0} f(y) dy = \int_{-a}^{0} f(x) dx$? when dy = -dx? I do not see how using a "DUMMY variable" justifies dy = -dx to dy = dx

 

[SammyS]

Again I'm sorry man i just don't understand. dy = -dx right? then how does $\int_{-a}^{0} f(y) dy = \int_{-a}^{0} f(x) dx$? when dy = -dx? I do not see how using a "DUMMY variable" justifies dy = -dx to dy = dx

Look at a specific function.

Do the following integrals give different results?
$\displaystyle \ \int_{-2}^{0} (x^2+3) dx\$

$\displaystyle \ \int_{-2}^{0} (u^2+3) du\$