How Do Inverse Functions Relate When Integrated?

[MathewsMD]

If f and g are inverse functions and f ' is continuous, prove that:

[from a to b] ∫ f(x) dx = b f(b) - a (a) - [from f(a) to f(b)] ∫ g(y) dy

Hint: Use part (a) and make the substitution y = f(x)

I have been trying to rearrange the equations and have looked online at answers, and still don't quit understand how to solve this question.

From this link (http://math.berkeley.edu/~bgillesp/m1b-s12/files/worksheets/ws1-solutions.pdf) it says: First use integration by parts, and then in the remaining integral use integration by substitution with u = f(x). Keep in mind the defining relation x = g(f(x)) = g(u), and also recall the derivative of an inverse function, g′(x) = 1/f′(g(x)), which is obtained by implicit differentiation on f(g(x)) = x.

I tried to solve the question using this method, but have not really gotten anywhere. Also, from here (http://answers.yahoo.com/question/index?qid=20090908094826AAQg3Ky) the answer states: ∫(a to b) f(x) dx = xf(x) [for x = a to b] - ∫(a to b) xf '(x) dx
= (b f(b) - a f(a)) - ∫(a to b) xf '(x) dx.

As for the second term, substitute y = f(x).
So, x = g(y) and dy = f'(x) dx.
Bounds: x = a ==> y = f(a) and x = b ==> y = f(b).

Thus, we get
∫(a to b) f(x) dx = (b f(b) - a f(a)) - ∫(f(a) to f(b)) g(y) dy.

which I don't entirely understand since it looks like he just went in a circle and ended up with the original equation without showing anything. I'm just a little confused on how original functions and their inverses relate to each other by this relation when integrated, and any help would be greatly appreciated! Thank you!

 

[Curious3141]

If f and g are inverse functions and f ' is continuous, prove that:

[from a to b] ∫ f(x) dx = b f(b) - a (a) - [from f(a) to f(b)] ∫ g(y) dy

Hint: Use part (a) and make the substitution y = f(x)

I have been trying to rearrange the equations and have looked online at answers, and still don't quit understand how to solve this question.

From this link (http://math.berkeley.edu/~bgillesp/m1b-s12/files/worksheets/ws1-solutions.pdf) it says: First use integration by parts, and then in the remaining integral use integration by substitution with u = f(x). Keep in mind the defining relation x = g(f(x)) = g(u), and also recall the derivative of an inverse function, g′(x) = 1/f′(g(x)), which is obtained by implicit differentiation on f(g(x)) = x.

I tried to solve the question using this method, but have not really gotten anywhere. Also, from here (http://answers.yahoo.com/question/index?qid=20090908094826AAQg3Ky) the answer states: ∫(a to b) f(x) dx = xf(x) [for x = a to b] - ∫(a to b) xf '(x) dx
= (b f(b) - a f(a)) - ∫(a to b) xf '(x) dx.

As for the second term, substitute y = f(x).
So, x = g(y) and dy = f'(x) dx.
Bounds: x = a ==> y = f(a) and x = b ==> y = f(b).

Thus, we get
∫(a to b) f(x) dx = (b f(b) - a f(a)) - ∫(f(a) to f(b)) g(y) dy.

which I don't entirely understand since it looks like he just went in a circle and ended up with the original equation without showing anything. I'm just a little confused on how original functions and their inverses relate to each other by this relation when integrated, and any help would be greatly appreciated! Thank you!

Your post is very difficult to read without LaTex. But here's how I would approach it.

Let $I = \int_a^bf(x)dx$.

Let $y = f(x)$. So $I = \int_a^b ydx$.

The next step is to use $x = g(y)$ to change the element of integration to $y$ (i.e. integral (something) dy). Can you make this substitution? You'll need chain rule. Remember to change the bounds accordingly.

The integration of parts comes later. But first make the sub.

 

[MathewsMD]

Your post is very difficult to read without LaTex. But here's how I would approach it.

Let $I = \int_a^bf(x)dx$.

Let $y = f(x)$. So $I = \int_a^b ydx$.

The next step is to use $x = g(y)$ to change the element of integration to $y$ (i.e. integral (something) dy). Can you make this substitution? You'll need chain rule. Remember to change the bounds accordingly.

The integration of parts comes later. But first make the sub.

Hmmm..okay

I did:

$f(x)=y$

$\int_{f(a)}^{f(b)}\frac {y}{f'(x)}dy = bf(b) - af(a) - \int_{f(a)}^{f(b)}g(y)dy$ [1]

$\int_{f(a)}^{f(b)}g(y)dy = yg(y)\mid_a^b - \int_{f(a)}^{f(b)}yg'(y)dy$ [2]

Plugging [2] into [1] and simplifying

$\int_{f(a)}^{f(b)}\frac {y}{f'(x)}dy = \int_{f(a)}^{f(b)}yg'(y)dy$

I am now a little confused and not sure if I should continue or stop. Any guidance would be great!

 

[Curious3141]

Hmmm..okay

I did:

$f(x)=y$

$\int_{f(a)}^{f(b)}\frac {y}{f'(x)}dy = bf(b) - af(a) - \int_{f(a)}^{f(b)}g(y)dy$ [1]

EDIT: Now I see where you're coming from. You did the sub ($dx = g'(y)dy$), but I wanted you to leave it as $I = \int_{f(a)}^{f(b)} yg'(y)dy$, which can then be integrated by parts. Instead, you've expressed it as $\int_{f(a)}^{f(b)}\frac{y}{f'(x)}dy$, which is not incorrect (because $g'(y) = \frac{1}{f'(x)}$), but also not useful, because there's no point in leaving a function of $x$ in the integrand when you're integrating wrt $y$.

I found the rest of your derivation confusing - didn't check every step (and can't tell if it's even correct), but it looks circular (going nowhere).

Why don't you just apply integration parts to $\int_{f(a)}^{f(b)} yg'(y)dy$ after doing the sub as I advised?

 

[MathewsMD]

Okay, maybe I'm misunderstanding something but by substituting f(x) = y and g(y) = x into the equation, I find:

$\int^b_a ydx = I = xy l_a^b - \int_a^b xdy$

From here, I keep trying to simplify but don't see how any relationships that could be applied yield $I$

 

[vela]

Hmmm..okay

I did:

$f(x)=y$

$\int_{f(a)}^{f(b)}\frac {y}{f'(x)}dy = bf(b) - af(a) - \int_{f(a)}^{f(b)}g(y)dy$ [1]

How did you come up with this equation? You appear to be assuming what you're trying to prove.

It looks like what you did was differentiate $y=f(x)$ to get $dy = f'(x)\,dx$ so that $dx = \frac{dy}{f'(x)}$. So changing variables from $x$ to $y$ gives you
$$\int_a^b f(x)\,dx = \int_{f(a)}^{f(b)} y\frac{dy}{f'(x)}.$$ The problem here is you still have that function in terms of $x$ when you want everything in terms of $y$. Try using $x = g(y)$ instead to relate $dx$ and $dy$. (Or use what Curious3141 said about how the derivatives of inverses are related.) You can show then that
$$\int_a^b f(x)\,dx = \int_{f(a)}^{f(b)} yg'(y)\,dy.$$ It's most straightforward from here if you use integration by parts on the integral, as Curious3141 suggested.

$\int_{f(a)}^{f(b)}g(y)dy = yg(y)\mid_a^b - \int_{f(a)}^{f(b)}yg'(y)dy$ [2]

Here, you apparently integrated by parts. The limits on first term on the righthand side are incorrect. Everything is in terms of $y$ so the limits should be values of $y = f(x)$. They should still be $f(a)$ and $f(b)$.

Plugging [2] into [1] and simplifying

$\int_{f(a)}^{f(b)}\frac {y}{f'(x)}dy = \int_{f(a)}^{f(b)}yg'(y)dy$

I am now a little confused and not sure if I should continue or stop. Any guidance would be great!

You're making the proof difficult because you're manipulating both sides and trying to meet in the middle somewhere. It's easiest if you start with one side of the equation and work steadily until you end up with the other side.