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Mathematics 96 Online
OpenStudy (inkyvoyd):

Note: This is NOT a question. This is a tutorial. How to do basic integration? See comment below for a tutorial on the more basic concepts of integration. I will be covering starting concepts, and I will be providing an explanation for u-substitution as well as some examples. Proofs shall follow if I have time.

OpenStudy (inkyvoyd):

Now, this tutorial is assuming that you already understand the more important parts of differential calculus. Without this knowledge you will not be able to follow. You should know: What a derivative is. Basic rules for taking the derivatives of a function; look at @lgbasallote 's guide. Understanding of the notation of derivative. A basic (not conceptual) knowledge of what a differential is.

OpenStudy (inkyvoyd):

We shall first work with finding the "anti-derivative" of a function, or, a function that has a derivative of a given function. Let's start with an easy example; \[f(x)=3x^2-2x+5\] -> This is our function that we seek to find the anti-derivative of; we will denote the anti-derivative by F(x) Then, we seek to find (by our definition) the function whose derivative is \[3x^2-2x+5\] You experience with finding the derivative of a function should tell you that you will get something along the lines of \[F(x)=ax^3-bx^2+cx\] In fact, why don't we start from here? \[F(x)=ax^3-bx^2+cx\] \[df(x)/dx =3ax^2-2bx+c\] Now, let's match our original function and the derivative of our guesswork. \[f(x)=3x^2-2x+5\] \[df(x)/dx =3ax^2-2bx+c\] it seems that a=1, b=1, and c=5 Let's substitute these values and try to double check. \[F(x)=(1)x^3-(1)x^2+(5)x\] taking the derivative, we have \[f(x)=3x^2-2x+5\] It appears that our guess is by our definition, the anti-derivative of f(x).

OpenStudy (inkyvoyd):

One with experience in integral calculus, would, however, point out a flaw in our reasoning. There is actually more than one solution to finding the anti-derivative of f(x). If you recall, the derivative of any constant is 0. If we added any arbitrary constant (1,2,3, e,pi, sqrt(2),any real constant, actually), and took the derivative, the result would be the same. i.e. \[F(x)=x^3-x^2+5x+1\] -> solution \[F(x)=x^3-x^2+5x-2\] -> solution \[F(x)=x^3-x^2+5x+e\] -> solution We recognize this by adding a "C" to the end of our solution, because any constant added to our result will still give the same, correct answer. In Tex, that would be \[F(x)=x^3-x^2+5x+C\]

OpenStudy (inkyvoyd):

One may ask, "Why is this"? The answer can be explained by a simple graph. Let's say we \[y=x\] Then anti-derivative of this function is \[y=x^2/2+C\] (Check this one :D ) Let's try graphing these. |dw:1335709361433:dw|

OpenStudy (inkyvoyd):

|dw:1335709424970:dw| My horrible graphing skills aside, you will instantly realize that I forgot to "take into account the C". In fact, I have. Why don't we choose C for a positive value and a negative value as well?

OpenStudy (inkyvoyd):

|dw:1335709488606:dw|

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