mahts help
heya guys, i have a problem, maybe, some of you can help me, i have to make an excelsheet in maths for some calculations
i should also draw a graphic
it is called Integral - "Obersumme und Untersumme"
it has something to do with f(x)=x²
maybe someone of you can help me, if more details are needed, just tell me, thx for help in advance
shy1
i guess it depends what x is like most maths
so...if f(x)=x² then f must = x so u have a straight line gg
hmm. well from what i can see its basically just a way to plot the integral of x^2. unless its a graphical method to calculate integrals. (which is not all that complicated).
but before i can go further, i need u to translate the ubersumme and untersumme for me. infact it mite just be better to post a precise translation of the entire question/problem.
ps: tyr, its more likely he has to plot the integral of f(x), where the function is x^2. even if it was integral of f(x) = x^2, the answer would be 2x not x.
Lets see if I remember the mathlessons. I am not sure this is right, but this is how I remember it
The integral of x^2 is not 2x. 2x is the derivate of x^2, integral is the other way around. x^2 is the integral of 2x + C (any constant to find out how much above the y=0 line the curve is).
since the derivate of Ax^y + C -> A*yx^y-1 (the constant is removed) then you have to find out what the function to calculate the derivate from would look like to get x^2
I think like this
x^2 = Ax^y where A=1 and y=2
So first you have to add 1 to y which gives Ax^y+1 = x^3
Then you have to figure out what number to multiply with A to get 1 which is 1/3
So f(x) = x^2 gives the integral A*1/3*x^3 + C (You remove the C when you derivate so you have to add it to the integral)
x^3 + C
3
Lets see if I remember the mathlessons. I am not sure this is right, but this is how I remember it
The integral of x^2 is not 2x. 2x is the derivate of x^2, integral is the other way around. x^2 is the integral of 2x + C (any constant to find out how much above the y=0 line the curve is).
since the derivate of Ax^y + C -> A*yx^y-1 (the constant is removed) then you have to find out what the function to calculate the derivate from would look like to get x^2
I think like this
x^2 = Ax^y where A=1 and y=2
So first you have to add 1 to y which gives Ax^y+1 = x^3
Then you have to figure out what number to multiply with A to get 1 which is 1/3So f(x) = x^2 gives the integral A*1/3*x^3 + C (You remove the C when you derivate so you have to add it to the integral)
x^3 + C
3
I was gonna say that ... Just 1 thing, make sure to plot x^3/3 + C and not f(x) = x^2. Basically an integral in the area under the curve and the derivate is the tangent point of 1 curve... so if u r going to plot it make sure to put some shade under the curve, only put shade until the shade touches the axis ( x,y) ...peace...
nicely argued. except u havent touched upon the fact that its a fixed integral with limits, and ur supposed to plot the function not the area. unless otherwise mentioned.
as i said. if u have the problem statement, have it translated and post it. its not that hard a problem. but without knowing the exact statement, ur probly gonna interpret it wrong and fuck it up. .
its not that hard a problem. but without knowing the exact statement, ur probly gonna interpret it wrong and fuck it up. .
hehe yeah . And ur gonna blame Q3 people for making u fail on ur test ...peace...
Lets see if I remember the mathlessons. I am not sure this is right, but this is how I remember it
The integral of x^2 is not 2x. 2x is the derivate of x^2, integral is the other way around. x^2 is the integral of 2x + C (any constant to find out how much above the y=0 line the curve is).
since the derivate of Ax^y + C -> A*yx^y-1 (the constant is removed) then you have to find out what the function to calculate the derivate from would look like to get x^2
I think like this
x^2 = Ax^y where A=1 and y=2
So first you have to add 1 to y which gives Ax^y+1 = x^3
Then you have to figure out what number to multiply with A to get 1 which is 1/3So f(x) = x^2 gives the integral A*1/3*x^3 + C (You remove the C when you derivate so you have to add it to the integral)
x^3 + C
3
this is right.
i just finished doing integrals and derivates
thank god its over. yuck. :pukeleft:
omfg my maths was so much easier go away nerds
hmm, thx for helpt, i think the obersumme and untersumme is the sum over and under the line which rises from left bottom to right top, damn, i am so bad in maths....
...