Question

show that if k is a positive constant, then the area between the x-axis and one arch of the curve y=sin kx is 2/k? need help :(

Answer 1

\[\sin(kx)=0=> x=0, \pi, ...\]
\[\int\limits_{0}^{\pi}\sin(kx)dx=\frac{-1}{k}\cos(kx)|_{0}^{\pi}=\frac{-1}{k}(\cos(k \pi)-\cos(k*0))\]
if k is odd then,
\[=\frac{-1}{k}(-1-1)=\frac{-1}{k}(-2)=\frac{2}{k}\]
if k is even then,
\[=\frac{-1}{k}(1-1)=0\]