Consider the Riemann sum $$\sum_{k=1}^n 2x^∗_k ∆x_k$$ of the integral of f(x) = 2x in an interval [a, b].
(a) Show that if $$x^∗_k$$ is the midpoint of the k−th subinterval, then the Riemann sum is telescopic.
(b) Use part (a) to evaluate the definite integral of f(x) = 2x in [a, b].
For letter (a) I was thinking about considering $$x^*_k=a+(k-\frac{1}{2})\Delta x_k$$ But for some reason, I couldn't cancel the terms. What am I doing wrong?
(a) Since $\Delta x_k = x_k-x_{k-1}$ and $x_k^* = \frac{1}{2}(x_k+x_{k-1})$, one has $2x_k^*\Delta x_k = x_k^2-x_{k-1}^2$, hence $$ \sum_{k=1}^n 2x_k^*\Delta x_k = \sum_{k=1}^n x_k^2-x_{k-1}^2 = (x_1^2-x_0^2) + (x_2^2-x_1^2) + \ldots + (x_n^2-x_{n-1}^2) = x_n^2 - x_0^2 $$
(b) Given that $x_0 = a$ and $x_n = b$, one finds in the end : $$ \int_a^b 2x \,\mathrm{d}x = \sum_{k=1}^n 2x_k^*\Delta x_k = x_n^2 - x_0^2 = b^2 - a^2 $$