Question to integral calculation

Here is an explanation of the calculus involved...

\(y=f(x)\\ \frac{dy}{dx}=f'(x)\\ \int_{x_1}^{x_2} \frac{dy}{dx}dx\\=\left[y\right]_{x_1}^{x_2}\\=\left[f(x)\right]_{x_1}^{x_2}\\ =f(x_2)-f(x_1) \)

So if you graph the differential curve against x and then find the area under that curve between x1 and x2 (you do this by finiding the definite integral of it between x1 and x2)    you will always get   f(x2)-f(x1)

I understand this bit perfectly I just don't comprehend why it works this way.  Maybe a pictorial explanation might help. ://  idk.

Hi Amnesia, welcome to the forum.

I like your question - it is making me think.

Maybe another mathematican would like to weigh in here as I would also like to see this discussed.  :))

Here is something for you to think about amnesia :))

It is not very technical but

You are graphing  dy/dx against  x from x1 to x2

The integral

\(\int_{x_1}^{x_2}\;\frac{dy}{dx}\;dx\)

is the sum of an infinite number of rectangles from x1 to x2

remember that d stands for difference and The integral sign is a stylized S which stands for sum (that is my interpretation anyway and it makes total sense)

So with the integral we have (from x1 to x2)

\(\frac{\text{difference in ys}}{\text{difference in xs}}*\text{difference in xs}\\ =\text{difference in ys because the x'es cancel out}\\ =y_2-y_1\)

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This is why definite integration gives an area.

\(\int_{x_1}^{x_2}\;y\;dx\\ \text{This is all the infinitely thin y heights multiplied by all the infinitely thin x widths }\\\text{over the domain }x_1 \;to \;x_1\\ and\\ \text{y *x is the area of a rectangle. Add all these skinny areas together and you have the area under the curve.}\)

In all this discussion i have used y instead of f(x) because it is usually easier to interprete, but they are the same thing.

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Anyone is welcome to comment of course but do not criticize my lack of technical talk unless you truly believe it to be totally incorrect.

This is how I see it in a fairly non-mathematically precise way.  :)

Hi Amnesia,

I felt like drawing a picture.  I might do another if I get up to it.   I have a second slightly different pic already in my mind  :)

Take a look, see if you can see the relevance :))   Ask questions if you want to.

I am just looking at the the function   y=0.75x+3   and its derivative   y'=0.75  I have graphed both thes lines.  See if you can work out the rest of what I have done.

Let's take Melody's picture idea a little further.  Imagine you have the following function, f(x), which is comprised of a sequence of straight lines (this is an arbitrary function I made up off the top of my head - it isn't specially chosen):

The gradient, df(x)/dx, of this function is easily plotted:

The integral of df(x)/dx with respect to x from, say x = 1 to x = 7, is just the area under the curve. It is easy to calculate because df(x)/dx is just constructed from a number of constants (i.e. horizontal straight lines).  We can see that the area under df(x)/dx from x = 1 to x = 7 is 9.  

Using the plot of f(x) we can also see this is the same as the difference between f(7) and f(1). i.e. f(7) - f(1) = 9

You can change the values of the gradients of  the various sections of f(x) as much as you like and you will always find the area under the corresponding graph of df(x)/dx will equal f(x2) - f(x1).  Try it!

Now a general curve can be thought of as an infinite number of (infinitely small) straight line segments, so the same result applies to "curvy" curves as well!

. 

Much appreciated  Alan,    

It is very interesting how calculus works and if amnesia can sort out these concepts she will have a good head start on her class mates I think. :))

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So amnesia, be a sponge and soak up as many concepts as you possibly can :)

I have another demonstration graph in mind but I might not get up to constructing it.   ://

I think I might understood it. Well its still a bit cloudy, but when understanding something new its always cloudy at first and gets better the more I invest into the topic. I might have to look a few more times at your explanations before I fully get the context.Thanks to both you Melody and Alan for all those greate and comprehensive Explanations :) Great stuff to learn from

You are very welcome amnesia :)

You can do a lot of mathematics, including calculus, with limited understanding but the more underlying concepts that you can grasp the easier it all is.  Plus it is MUCH easier to remember, and utilize, and build on concepts that you fully understand. 

You are right, it will be fuzzy for a while. :)