Question

How can you find the absolute difference between the left and right riemann sum on a closed interval? Provide an example with your reasoning.

Answer 1

Suppose that $f(x)$ is a function, which is integrable on the interval $[a,b]$. Suppose that we have points $x_0=a,x_1,x_2,...,x_{n-1},x_n=b$ and $\Delta=x_{i}-x_{i-1}$, $i=1,...,n.$ Then, the left Riemann sum is: $S_l=\sum_{i=0}^{n-1}f(x_i)(x_{i+1}-x_{i})$. The right Riemann sum is: $S_r=\sum_{i=0}^{n-1}f(x_{i+1})(x_{i+1}-x_{i})$. The absolute difference is: $|S_r-S_l|=\sum_{i=0}^{n-1}(f(x_{i+1})-f(x_{i}))(x_{i+1}-x_{i})$. We point out that for small $\Delta$ the difference will approach zero, since both sums will tend to the integral $\int_a^bf(x)dx$ . Thus, the more points we have, the smaller will be the difference between sums. As an example, we take . $a=0,b=1$. $f(x)=x^2$. Consider points: $0,\frac14,\frac24,\frac34,1$. We have: $S_l=\frac{1}{4}(0^2+(\frac14)^2+(\frac24)^2+(\frac34)^2)=\frac{7}{32},\,S_r=\frac{1}{4}((\frac14)^2+(\frac24)^2+(\frac34)^2+1)=\frac{15}{32}.$ The difference is $\frac{8}{32}$. The value of integral is: $\int_0^1x^2dx=\frac13$. For more points the difference between sums will be smaller. Both sums will approach $\frac13.$