Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 72

$\dfrac{2}{k}$

We know that $\sin x$ is a periodic function with period $\dfrac{2\pi}{k}$. Now, $A=\int_0^{\pi/k} \sin kx dx$ This implies that $[(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(1/k)-(\dfrac{-1}{k}\cos 0)$ Thus, $A=\dfrac{2}{k}$