Using the definition of the integral to find an exact value

[andromeda_sky]

Okay, so i'm trying to find the value of the integral from -1 to 3 of (x² )-2x+1dx. Using the fundamental theorem of calculus, I found the value I want to be 16/3. I checked this with a graphing calculator.

Unfortunately, I can't make the limit as x->Infinity of the sum of (deltax)(f(xi)) come out as 16/3.

So far i've found delta x, which is (b-a)/n to be 4/n, and xi, which is a+i(deltax), to be (4i/n)-1.

I then plugged xi into the original integrand, x² -2x+1 and got an answer of((-1+4i/n)² )-2(-1+(4i/n)+1 (simplified to 4-16i/n+(16i² /n² )), which I then multiplied by my deltax value 4/n to get (16/n)-(64i/n² )+(64i² /n³ ).

Using the rules of summation, I separated the variable i and replaced it with the appropriate functions of n. (i=n(n-1)/2, i² =n(n-1)(n-2)/6), and then multiplied the separated values back in to get, after a ton of simplification, ((-128n³ +285n² +128n)/6n³ )). The limit as x->Infinity of this function seems to be -128/6 by dominating terms. This is wrong. Where did I mess up?

 

[Ishuda]

First of all, the sum you have for the i² is incorrect. That makes what you did following incorrect. Using your notation, i² =n(n-1)(2n-1)/6 not n(n-1)(n-2)/6. That would leave your i² summation off by about a factor of 2.

Next, rather than go through the 'ton of simplification', can you argue something like
\(\displaystyle lim_{n \to \infty} \frac{1}{n} \Sigma^{i=n-1} 1 \to 1\)
and something similar for the other sums and therefore the total sum goes to
64 \(\displaystyle \frac{1}{n^3}\Sigma i^2 - 64 \frac{1}{n^2}\Sigma i + 16 \frac{1}{n}\Sigma 1 \to 64/3 - 64/2 + 16 \)= 16/3