Answer
$\int_a^{b} cdx=c(b-a)$
Work Step by Step
Here, we have $\triangle x=\dfrac{b-a}{n}$ and $c_k=a+\dfrac{k(b-a)}{n}$
Consider the RIEMANN SUM:
$\Sigma_{k=1}^n f(c_k) \triangle x=\Sigma_{k=1}^n \dfrac{c(b-a)}{n}=c(b-a)$
Now, $\lim\limits_{n \to \infty} \Sigma_{k=1}^n f(c_k) \triangle x=\lim\limits_{n \to \infty}c(b-a)$
Thus, $\int_a^{b} cdx=c(b-a)$
