I am trying to solve this problem:
Use a calculator or computer to make a table of values of left and right Riemann sums $L_n$ and $R_n$ for the integral $\int_0^1 e^{-x^2} dx$ with n = 5, 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral $\int_{-1}^2 e^{-x^2}dx$ Explain?
So using this calculator here, I have determined these values:
$$ \begin{array}{c|lcr} n & \text{left} & \text{right} \\ \hline 5 & 1.07746 & 0.68479 \\ 10 & 0.980007 & 0.783670 \\ 50 & 0.901705 & 0.862437 \\ 100 & 0.89189 & 0.872262 \\ \end{array} $$
So the area of the integral must between $0.872262$ and $0.89189$ right? Also we can say something similar if the integral was $\int_{-1}^2 e^{-x^2} dx$ right? The lower and the upper bound will be higher since $e^{-1^2}$ is a positive value and is above $0$ so the integral will be = the area under the curve which is positive right? Does this sound right?
What you can say is that the correct value is between the upper and lower Riemann sums. Your conclusion about the first integral is correct, but it depends on the fact that the function is monotonic in the interval of interest. In this case the function is decreasing, so the left Riemann sum is the upper sum and the right one is the lower sum. When the function is not monotonic that does not apply. It could be that the function has a narrow but very tall spike between the points you evaluate it at and the integral could be enormous.
As an example, consider computing $\int_{-1}^1e^{-x^2}dx$. By symmetry the left and right sums will be the same. If you use two intervals, the left sum is $f(-1)+f(0)=\frac 1e+1 \approx 1.3679$ and the right sum is $f(0)+f(1)=1+\frac 1e$. The correct value is about $1.49365$. The upper sum is $2f(0)=2$ and the lower sum is $f(-1)+f(1)=\frac 2e\approx 0.7358$. The correct integral is nicely between the upper and lower sums but not between the left and right sums.