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].
I got help and realized that one way to turn this into a telescoping sum is to realize that in an interval [c,d], $$\Delta x_k=d-c$$ and $$x^*_k=\frac{ c+ d} {2}$$ Thus obtaining $$2x^*_k\Delta x_k = d^2-c^2$$
My question is whether this alone is enough to solve the problem in letter (a).
In letter (b) if we consider the first interval as [a,c], then it would be c^2-a^2. Doing this up to one last interval, say [d,b], we would have b^2-d^2. So the answer would be c^2-d^2? I'm so confused