Prove \int _a ^b f(x) \: dx with Continuous f

[DivGradCurl]

Problem:

(a) If $$f$$ is one-to-one and $$f^{\prime}$$ is continuous, prove that

$$\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy$$

(b) In the case where $$f$$ is a positive function and $$b > a > 0$$, draw a diagram to give a geometric interpretation of part (a).

My work:

(a) $$\int _a ^b f(x) \: dx$$

Integrating by parts gives

$$u = f(x) \Rightarrow \frac{du}{dx} = f ^{\prime} (x) \Rightarrow du = f ^{\prime} (x) \: dx$$
$$dv = dx \Rightarrow v = x$$

$$\int _a ^b f(x) \: dx = \left. xf(x) \right] _a ^b - \int _a ^b x f ^{\prime} (x) \: dx$$
$$\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _a ^b x f ^{\prime} (x) \: dx$$

Applying the Substitution Rule gives

$$y = f(x) \Leftrightarrow x = f^{-1} (y) \Rightarrow \frac{dy}{dx} = f ^{\prime} (x) \Rightarrow dx = \frac{dy}{f ^{\prime} (x)}$$

$$y(b) = f(b)$$
$$y(a) = f(a)$$

$$\int _a ^b f(x) \: dx = bf(b) - af(a) - \int _{f(a)} ^{f(b)} f ^{-1} (y) \: dy$$

(b) I'm not sure how I should handle this one. The left-hand side is quite easy to visualize: it corresponds to a generic integral from a to b. The right-hand side does not seem to be that simple, and I need some help.

Any help is highly appreciated.

 

[arildno]

1) Draw orthogonal x-and y-axes on a piece of paper
2) Mark the interval [a,b] on the x-axis, and draw f(x) over it, so that it ranges between f(a) and f(b); mark f(a) and f(b) on the y-axis.
3) What rectangle can you naturally construct whose area is a*f(a)?
4) What rectangle can you naturally construct whose area is b*f(b)?
5) Look and behold, and see if you find an easy geometric interpretation of the equality..

 

[saltydog]

You know Thiago, I don't claim to know much math, no more than 1% in fact, so I'm not surprised your relation is news for me and I'm sure you will follow Arildno so I don't think I'm giving anything away by posting the attached plot for the function $y(x)=0.2 x^2$ from 2 to 4.

 

[arildno]

I met this relation in my first analysis course, and it was, to me, one of those "Wow, math is really cool!"-experiences..

 

[DivGradCurl]

Thank you guys! I can now see what you're talking about. It's pretty straight-forward. I was a bit confused by a "big fat" generic equation. :)

 

[saltydog]

You know Thiago, I don't claim to know much math, no more than 1% in fact, so I'm not surprised your relation is news for me and I'm sure you will follow Arildno so I don't think I'm giving anything away by posting the attached plot for the function $y(x)=0.2 x^2$ from 2 to 4.

Just want to correct my statement above: The function $y(x)=0.2x^2$ is NOT one-to-one and only because it's so in the first quadrant would it qualify for the relation above with a and b in the same quadrant.