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March 28, 2015

March 28, 2015

Posted by **Anon** on Sunday, March 16, 2014 at 6:14pm.

- AP Calculus -
**Steve**, Sunday, March 16, 2014 at 11:46pm∫[a,b] x^2 dx

= 1/3 x^3 [a,b]

= 1/3 (b^3 - a^3)

Consider only integer values. Then that will give you

a^2 + (a+1)^2 + ... + (b-1)^2 + b^2

Now, we know that

n

∑ k^2 = n(n+1)(2n+1)/6

k=1

So, we want sum from a-b = sum from 1-b - sum from 1-a, or

b(b+1)(2b+1)/6 - a(a+1)(2a+1)/6

= 1/6 (2b^3+3b^2+b - 2a^3-3a^2-a)

= 1/3 (b^3-a^3) plus some lower-order junk

So, our integral exact value appears to agree with the integer approximation.

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