Question

can some one teach me integral and derivative?

Answer 1

How deep do you want to know?

Answer 2

um, 1st and 2nd derivatives, and defined and undefined integral for polynomials, trigonometric functions, exponential and logarithmic functions.

Answer 3

Do you want to know how integrals and derivatives are rigorously defined? (It is a bit deep, but it is essential to a true understanding.)

Answer 4

well, I geuss what ever you can teach me is good, because i don't know alot..

Answer 5

guess*

Answer 6

Okay. Integrals were first rigorously defined by Bernard Riemann. Later mathematicians modified his definition and, currently, the Lebesgue definition of the integral is considered the best. Nonetheless, the Riemann integral gives you a good idea of what an integral really is.

An integral, in its most intuitive definition, is defined as the area underneath a one-variable function. Integrals are also defined for multivariable functions. For example, an integral of a two variable function is the volume underneath the surface formed by the two variable function.

It is best that you use pictures to fully understand what I am talking about. Here is a picture for the integral of a one-variable function. (Ignore the a and b, I will explain that later. Focus on the S.): http://upload.wikimedia.org/wikipedia/commons/thumb/f/f2/Integral_as_region_under_curve.svg/682px-Integral_as_region_under_curve.svg.png

Here is an example of what the integral of a two variable function is:
http://en.wikipedia.org/wiki/File:Volume_under_surface.png

Note that the integral of a two variable function is considered a "double integral" because you must integrate twice since there are two variables.

Do you follow so far?

Answer 7

okay, i'll look at then :)

Answer 8

Do you understand what I've said so far? I was going to explain the actual definition next.

Answer 9

if a function is of this form \[f(x)=x^n+b=x^n+bx^0\qquad\qquad n,b\in\mathbb R\] the first derivative with respect to x of the function is\[\frac{\text d}{\text dx}f(x)=f^{\prime}(x)=\dot f(x)=\text D_xf(x)\]
\[=nx^{n-1}\qquad=nx^{n-1}+0bx^{-1}\]

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the indefinite integral of that function with respect to x is \[F(x)=\int f(x)\text dx=\int (x^n+b)\cdot\text dx=\int(x^n+bx^0)\cdot\text d x\]
\[=\frac{x^{n+1}}{n+1}+bx+c\qquad\qquad n,b,c\in \mathbb R\qquad n\neq -1\]

Answer 10

oh my word, so sorry.. I was away but yes I do :) sorry @Limitless

Answer 11

but what do you mean by the area underneath a one-variable or multi-variable function? which area?

Answer 12

sorry nvm, i saw the area.. didnt see the pictures! so a 2 variable function is in a 2D form?

Answer 13

What do you mean by 2D form? If you mean, a 2 variable function is a surface, then yes.

Answer 14

well the first picture showed the area under the graph in a different way compared to the second one with the 2-variable function.

Answer 15

The second picture shows the volume produced by a 2-variable function. It's an extension of the concept; it's not the exact same thing.

Answer 16

oh okay :) then I understand so far

Answer 17

Alright.

There are two types of integrals. There are definite integrals and indefinite integrals. Definite integrals are integrals which are taken over an interval of the real line. That is, they look like this \(\int_{a}^{b}f(x)dx\) and represent the area covered by the function on the interval \([a,b]\). Indefinite integrals are integrals which are not taken over a particular interval, hence the naming 'indefinite'. They are represented by \(\int f(x)dx\).

The indefinite integral is crucial to understanding the definite integral. There's just one issue, though...

Not all indefinite integrals can be expressed in terms of elementary functions (i.e. things which we find nice and easy to work with). So, we simply put that aside and continue on to understanding definite integrals.

(I use \(f(z)\) instead of \(f(x)\) only because these results apply to complex functions as well.)

The first fundamental theorem of calculus states the following:
If \(f(z)\) is a continuous function, then \(\frac{d}{dz}\int f(z)dz=f(z).\) This tells us that integration and derivation are inverses of one another, like how addition and subtraction are inverses. This theorem also guarantees the existence of indefinite integrals for continuous functions.

The second fundamental theorem of calculus is heavily related to this and states the following:
If \(F(z)=\int f(z)dz\), then \(\int_{a}^{b}f(z)dz=F(b)-F(a).\) This means that if the indefinite integral of a function does exist, we can calculate the definite integral in terms of this function created by the indefinite integral. It is pretty mind-blowing when you think about it because there is no apparent reason that definite and indefinite integration are related.

Do you feel as though you have a good understanding of the theory behind integration now?

Answer 18

yes, I understand :)

Answer 19

Alright. Wanna learn about derivatives now?

Answer 20

but I am going for lunch now.. so i'll be back in 20mins. yes please, and thank so much for the help!=)

Answer 21

A derivative is an operation on a function which tells us the rate of change of a function. For a single variable real function, it is denoted \(\frac{d}{dx}f(x)\) typically. It is defined in terms of a limit of the difference quotient. i.e., \[\frac{d}{dx}f(x)=\lim_{\Delta x \to 0}\underbrace{\frac{f(x+\Delta x)-f(x)}{\Delta x}}_{\textrm{difference quotient}}.\]
Note that we are assuming the derivative exists.

Derivatives have various notations. Some popular ones include \(f'(x)\), \(\frac{df}{dx}\), and \(\frac{df(x)}{dx}.\) They all mean the same thing. The first one is accredited to Joseph Louis Lagrange and the second two are credited to Gottfried Willhelm von Leibniz.

The second derivative of a function is denoted in various different forms as \(f''(x)\), \(\frac{d^2}{dx^2}f(x),\) and \(\frac{d^2f}{dx^2}.\) Once again, they all mean the same thing. The second derivative, assuming it exists, is defined \[f''(x)=\lim_{\Delta x \to 0}\frac{f'(x+\Delta x)-f'(x)}{\Delta x}.\]

Interestingly enough, there are two equivalent definitions for the first and second derivative. They are, respectively, \[
\begin{align}
f'(x)&=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x-\Delta x)}{2\Delta x}\\
\text{and } f''(x)&=\lim_{\Delta x \to 0}\frac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2}.
\end{align}\]

What else would you like to know? The majority of the things with respect to "[. . .]polynomials, trigonometric functions, exponential and logarithmic functions" are trivial computations.

Answer 22

i think that is enough for the exam, I just didnt want to be lost if they gave a question about it.

Answer 23

What is your exam over, precisely? Typically trivial calculations are part of an exam. If so, you should memorize the "basic" integrals and derivatives and the respective techniques

Answer 24

I am doing a Medicine entrance exam in 13 days :/ So with what you have taught me I can calculate surfaces?

Answer 25

Computation requires a bit more. What I have shown you is the theory behind the computations. You may want to learn more specific things in Calculus.
Look for more info here http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeIntro.aspx
and here http://tutorial.math.lamar.edu/Classes/CalcI/ComputingIndefiniteIntegrals.aspx
:)

Answer 26

BTW, feel free to ask any question on any of the things you are studying.

Answer 27

okay, thank you! :)

Answer 28

You're welcome. :D

Answer 29

So I balanced some equations yesterday but I am not sure if they are correct, do you mind checking them because I don't have answers at the back of the book..

Answer 30

Post them as a new question. This will make the information more easily shifted through by users on OS.

Answer 31

You should probably tag me in it, incase I'm not on.

Answer 32

okay :)