Riemann's integral problem

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Ashford

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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)
 
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
 
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.
 
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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...
 
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