A Riemann Sum clarification

Question

I have to find the area under the curve $f(x) = x^2$ in the segment $[-2, 1]$ using Riemann Sum. So here is what I did, but it's wrong and I need some clarification about (see the questions in the end).

So I chose $\Delta x_k = \frac{3}{n}$ and $\xi_k = x_k = k \Delta x_k = \frac{3k}{n}$

So

$$S_n = \sum_{k = 0}^{n-1} f(\xi_k)\Delta x_k = \sum_{k = 0}^{n-1} \left(\frac{3k}{n}\right)^2 \frac{3}{n} = \sum_{k = 0}^{n-1} \frac{27k^2}{n^3} = \frac{27}{n^3} \sum_{k = 0}^{n-1} k^2 = \frac{27}{n^3}\frac{1}{6}(2n^3-3n^2+n)$$

Now $$\lim_{n\to +\infty} S_n = \frac{27}{3}$$

Whereas $$\int_{-2}^1 x^2\text{d}x = \frac{7}{3}$$

• Where am I wrong?

• Also: here I chose $\xi_k = x_k = k\Delta x_k$ but what would it be if I chose $\xi_k = x_{k+1}$. How would I use this? What would $\xi_k$ be?

I have some confusions about

Answer 1

Where am I wrong?

In your sum, you have $\xi_k$ going from $0$ to $3$, not from $-2$ to $1$, so you computed $\displaystyle\int_0^3x^2dx=\dfrac{27}3.$

Try instead $\; \displaystyle\sum_{k = 0}^{n-1} \left(-2+\dfrac{3k}{n}\right)^2\; \dfrac{3}{n} $, and you should get $\displaystyle\int_{-2}^1x^2dx=\color{red}3$.