I can't find my dumb mistake.
I'm figuring the definite integral from first principles of $2x+3$ with limits $x=1$ to $x=4$. No big deal! But for some reason I can't find where my arithmetic went screwy. (Maybe because it's 2:46am @_@).
so
$\delta x=\frac{3}{n}$ and $x_i^*=\frac{3i}{n}$
where $x_i^*$ is the right end point of each rectangle under the curve.
So the sum of the areas of the $n$ rectangles is
$\Sigma_{i=1}^n f(\delta xi)\delta x$
$=\Sigma_{i=1}^n f(\frac{3i}{n})\frac{3}{n}$
$=\Sigma_{i=1}^n (2(\frac{3i}{n})+3)\frac{3}{n}$
$=\frac{3}{n}\Sigma_{i=1}^n (2(\frac{3i}{n})+3)$
$=\frac{3}{n}\Sigma_{i=1}^n ((\frac{6i}{n})+3)$
$=\frac{3}{n} (\frac{6}{n}\Sigma_{i=1}^ni+ 3\Sigma_{i=1}^n1)$
$=\frac{3}{n} (\frac{6}{n}\frac{n(n+1)}{2}+ 3n)$
$=\frac{18}{n}\frac{(n+1)}{2}+ 9$
$=\frac{9(n+1)}{n}+ 9$
$\lim_{n\to\infty} \frac{9(n+1)}{n}+ 9 = 18$
But the correct answer is 24.