# Simple Calculus question about integrating a function

I'm learning Calculus but don't have the tools to figure out a questions that I'm wondering about. If you could help that would be appreciated.

Suppose that you have a non-constant function $f(x)$ on the interval $[a,b]$. Define a function $F = \int_a^bf(x)dx$ Now suppose that you take the function $f(x)$ and rotate it about $x=\frac {a+b}2$ by $\theta$ from $0$ to $\frac {\pi}2$. We then project this back into $\mathbb{R}^2$ and measure $F$ along the rotation. We know that when $\theta = 0$, $F=\int_a^bf(x)dx$ and when $\theta = \frac {\pi}2$ then $F = 0$. So my question is, is F a linear function, and if it isn't, then how do we define it and graph it? Does the degree of f(x) change anything?

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Your definition of $F$ doesn't make sense. $\int_a^b f(x)\ dx$ is a number, independent of $x$. – Javier Mar 4 '13 at 20:18
Sorry is should be more like $F(\theta)$ So, I changed it to just F. – AlexHeuman Mar 4 '13 at 20:19
But how do you define $F$ in terms of $\theta$? How do you integrate a rotated function? Maybe a picture would help. – Javier Mar 4 '13 at 20:20
Sorry, I haven't learned this stuff at all yet so I have trouble figuring out the notation. What I imagine is a 2-D polynomial function between the interval [a,b]. Then I imaging rotating that function about the y-axis into the z dimension by $\theta$. Then I project that as to what it would look like in 2-D. I imagine |b-a| becoming smaller and approaching 0 as $\theta$ approaches $\frac {\pi}2$ – AlexHeuman Mar 4 '13 at 20:23
I think I understand now. You mean project the rotated graph of $f$ onto the $xy$ plane, and then integrate this new function? – Javier Mar 4 '13 at 20:26

Based on the conversations in the comments, I believe this is what you mean.

Say you have a curve which is the graph of a real function. If we rotate this by $\theta$ about the line $x = \frac{a+b}{2}$, that is, the center of the interval, and then project that onto the $xy$ plane, the graph of $f$ will end up compressed by a factor $\cos \theta$.

To understand this, try to think of the following. Suppose you're in the $xy$ plane, looking in the positive $y$ direction, with positive $z$ being up, looking at the graph of $f$. Since it's in the same plane as you, it just looks like a horizontal straight line to you. Now rotate it by $\theta$ about its midpoint. It will still look like a straight line, but now it's rotated so it's no longer aligned with the $xy$ plane. If you project it back down again, the distance of any point to the center (that is, $\frac{a+b}{2}$) is $\cos \theta$ times what it was before. Try drawing some pictures to get a feel for it.

If we define $x_0 = \frac{a+b}{2}$ for convenience, we will get a new function $g_\theta$ (which is this "compressed" $f$), defined on $[x_0-\frac{b-a}{2}\cos \theta, x_0+\frac{b-a}{2}\cos \theta]$. This are just $a$ and $b$ modified so their distance to $x_0$ is scaled by $\cos \theta$. At any point $x$, $g_\theta(x)$ is $f$ evaluated at the point we get if we take $x$ and undo the shrinking we did before, when took $a$ and $b$ and changed them to be closer to $x_0$. In symbols, $g_\theta(x) = f(x_0+\frac{x-x_0}{\cos\theta})$.

We want $F(\theta)$ to be defined as the integral of this $g_\theta$, so $F(\theta) = \int_{x_0-\frac{b-a}{2}\cos \theta}^{x_0+\frac{b-a}{2}\cos \theta}f(x_0+\frac{x-x_0}{\cos\theta})\ dx$. The substitution $u = x_0+\frac{x-x_0}{\cos\theta}$ lets us rewrite this as $F(\theta) = \cos \theta \int_a^bf(u)\ du$. So it turns out that $F(\theta)$ is proportional to $\cos\theta$.

I understand this might be hard to grasp. I cannot stress enough the importance of drawing some pictures to convince yourself of everything I've said. I would post some, but I don't really know enough about drawing math in a computer to do it.

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I think you know what I mean, but I don't quite understand your math. Can you recommend some reading for me? – AlexHeuman Mar 4 '13 at 21:01
I'll do some editing. – Javier Mar 4 '13 at 21:02
I appreciate it. – AlexHeuman Mar 4 '13 at 21:03
Thanks Javier for the edits. It is perfectly clear to me now. Don't know why I didn't see that before. I was looking at it from the x-y plane and not the x-z plane. – AlexHeuman Mar 4 '13 at 21:21

What would make sense is to define $F(x)=\int_a^x f(t)dt$. As long as $f$ is reasonable, this will be a fine function. Rotating $f$ is hard to make sense of, because it may well not even be a function. Probably the most reasonable try is that if $(x,f(x))$ is a point on the curve, the rotated point is $(x \cos \theta - f(x) \sin \theta,f(x) \cos \theta + x \sin \theta)$. This will give you a rotated graph of $f$, but if the tangent of $f$ is high enough and $\theta$ is high enough, it won't be a function any more-it will fail the vertical line test. Imagine $f(x)=x^2$ on $(0,1)$ and $\theta = \frac \pi 2$. Even if it is a function, it is hard to relate the integral of the rotated function with the original function. There is no reason for $F$ to be linear unless $f$ is constant.

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I believe that $f$ is supposed to be rotated about the line $x=\frac{b-a}{2}, z=0$ (it's probably supposed to be $\frac{a+b}{2}), projected onto the$xy$plane, and then integrated. So it will still be a function, it will be like a compressed version of$f$. – Javier Mar 4 '13 at 20:42 Hey there, that's not exactly what I meant. Perhaps I have a better way of putting it. Let S be a plane that is the ordinate of f from [a,b]. As we rotate S at its midpoint$x= \frac {a+b}2 $and project it back onto$\mathbb{R}^2$then F is the integral of the projection measure by$\theta$from 0 to$\frac {\pi}2$– AlexHeuman Mar 4 '13 at 20:45 Yes, it's supposed to be a+b/2, sorry dumb mistake. – AlexHeuman Mar 4 '13 at 20:46 I'm sorry, but that's not really clearer. What does it mean for a plane to be the ordinate of$f$? – Javier Mar 4 '13 at 20:47 meaning$ S{(x,y)|a \le x \le b, 0 \le y \le f(x)}\$ – AlexHeuman Mar 4 '13 at 20:48