Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 62

$\dfrac{2}{k}$

Since, the function $\sin x$ is a periodical function having period $\dfrac{2\pi}{k}$. Thus, we have $A=\int_0^{\pi/k} \sin kx dx$ $\implies [(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(\dfrac{1}{k})-(\dfrac{-1}{k})(\cos 0)$ $\implies A=\dfrac{2}{k}$