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## Homework Statement

I need to use the right endpoint formula to find the area under the curve f(x) = 4-x

^{2}over the interval [1,2]

I integrated this problem and got 5/3. But doing it the long way got me 11/3. Can someone where I made my mistake on applying the right endpoint formula, thanks!

## Homework Equations

right endpoint formula = [tex] f(\frac{b-a}{n}*k) [/tex] and [tex]

\sum_{k=1}^n K^2 = \frac{n(n+1)(2n+1)}{6}[/tex]

## The Attempt at a Solution

Before I use the right endpoint formula I tried integration and got:

[tex]\int_{1}^{2} 4-x^2 dx = \left[ 4x-\frac{x^3}{3}\right]_{1}^{2} = \frac{5}{3}[/tex]

here is what I got using the right endpoint formula:

[tex] \lim_{n\rightarrow\infty} \sum_{k=1}^n f(\frac{k}{n})*\frac{1}{n}

= \left[4-(\frac{k^2}{n^2})\right]*\frac{1}{n} [/tex]

[tex] = \frac{4}{n} - \frac{K^2}{n^3} [/tex]

[tex] \lim_{n\rightarrow\infty} 4 - \frac{1}{n^3}(\frac{(n(n+1)(2n+1)}{6} )[/tex]

[tex] = 4 - \frac{1}{3} = \frac{11}{3} [/tex]