I know from integration that the answer is -4. However, I am messing something up somewhere while working through the Riemann sums. Going cross-eyed trying to find my mistake. I included the pertinent steps and skipped the details. I do have those; just didn't type them in. Can anybody help me out?

$$\int_{-1}^{1}\left(3x^{2}-3\right)\mathrm{d}x$$ $$\lim_{n\to \infty}\sum_{i=1}^{n}\left(3\left(\frac{2i}{n}\right)^{2}-3\right)\left(\frac{2}{n}\right)$$ $$\left(\frac{24}{n^{3}}\right)\left(\frac{2n^{3}+3n^{2}+n}{6}\right)-6$$ $$8-6=2$$


When rewriting a definite integral as a Riemann sum, it is rewritten as so:

$$\int_a^b f(x)\, dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i)\Delta x$$

Where the following are given:

$$\Delta x = {b - a\over n}$$ $$x_i = a + i\Delta x $$

You've forgotten to add $a$ in $x_i = a + i\Delta x$. It should be:

$$\lim_{n\to \infty} \sum_{i=1}^n \left(3\left(-1 + {2i\over n}\right)^2 -3\right)\left(\frac 2n\right)$$

In actuality, the Riemann sum you had was the definite integral:

$$\int_0^2 \left(3x^2 - 3\right) dx = 2$$

Because you forgot to account for $a$ for $x_i$—thus it was $0$, the lower bound was treated as $0$.

  • $\begingroup$ I just came back to edit for that. I had tried putting the -1 in front of the squared term, then combining with -3 for a final -4 constant. That zeroed out my area. I completely get it now!!! Thanks! $\endgroup$ – user147485 Mar 15 '18 at 4:21

You want to find the Riemann Sum for $$\int_{-1}^{1}\left(3x^{2}-3\right)\mathrm{d}x$$

Your limits are from $-1$ to $1$.

Thus your partition points are $$-1+2i/n , \text { for } 1\le i\le n$$

Thus you should have $$\lim_{n\to \infty}\sum_{i=1}^{n}\left(3\left(-1+\frac{2i}{n}\right)^{2}-3\right)\left(\frac{2}{n}\right)$$ instead of $$\lim_{n\to \infty}\sum_{i=1}^{n}\left(3\left(\frac{2i}{n}\right)^{2}-3\right)\left(\frac{2}{n}\right)$$


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