Riemann's integral problem

[Ashford]

hey guys,
I was reading a calculus text book and I'm on the section of Riemann's integral, now the example my text uses is, to find the area under the curve f(x)=x²+1, now the book explains the solution like this



It's later simplified like this:





NOTE: The curve occurs between x = 0 and x = 3

Now the part that I don't understand is the part where the book tells you that the sum of the first n squares is equal to:



Is this a general formula that the author knows from previous experience or did he get this by simplifying any of the latter formulas? The reason I want to know this is because the formula is then written like this:



is replaced with n since if you add 1 n times you get n

So if it the author didn't get this from simplifying any of the latter formulas, what would have happened if we were looking at a different function (eg. f(x)=x³-2x)

 

[Ratchit]

Ashford,

A simple web search finds Math Forum - Ask Dr. Math .

Ratch

 

[Ashford]

Ok this just confirm that the author knows this from previous knowledge, so the question now is what would have happened if we were dealing with a different function, one that didn't work quite the same way, like "f(x)=x³-2x" because I dont know anyway of writing in algebraic notation and if it can't be done then how can the equation be solve

 

[MrAl]

Hello there,


Sometimes you just have to look these summations up in a math handbook. For the i^2 example you might find that in a calculus textbook, and for the i^3 example that would be:
sum[i=1 to n](i^3)=(n*(n+1)/2)^2

Here's a little more on the subject:

https://mathrefresher.blogspot.com/2006/09/summation-formula.html

Be sure to check any calculated solution VERY carefully before assuming it is valid.

 

[Ashford]

Thanks a lot, this really helps.

 

[MrAl]

Hi again,


Sure, you're welcome. It's been quite a while since i did these problems back in college, but i used the finite differences method and came up with the generator equation for your other question about sum(x^3+2*x) and it came out to:
y=a*x^4+b*x^3+c*x^2+d*x+e

where:
a=1/4
b=1/2
c=5/4
d=1
e=0

and after factoring it comes out to:
y=(x*(x+1)*(x^2+x+4))/4

I checked this over a few values but you may want to check it out further.

Good luck with it...