Question

find the positive value of k for the definite integral intergral 0 to pi/2 mod cos x - kx dx is minimum

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx}$

$\bf{\mapsto\,\,\,We\,\,\,know\,,}\\\\\displaystyle\bf{\left|\int^{b}_{a}\,f(x)\,dx\right|\le\int^{b}_{a}\,|f(x)|\,dx}$

So,

$\displaystyle\tt{\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx\ge\left|\int^{\frac{\pi}{2}}_{0}\,\left(cos(x)-k\,x\right)\,dx\right|}$

$\displaystyle\tt{\implies\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx\ge\left|\int^{\frac{\pi}{2}}_{0}\,cos(x)\,dx-\int^{\frac{\pi}{2}}_{0}\,k\,x\,dx\right|}$

$\displaystyle\tt{\implies\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx\ge\left|[sin(x)]^{\frac{\pi}{2}}_{0}-\dfrac{k}{2}\left[x^2\right]^{\frac{\pi}{2}}_{0}\right|}$

$\displaystyle\tt{\implies\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx\ge\left|1-\dfrac{k}{2}\cdot\dfrac{\pi^2}{4}\right|}$

$\displaystyle\tt{\implies\int^{\frac{\pi}{2}}_{0}\,\left|cos(x)-k\,x\right|\,dx\ge\left|1-\dfrac{\pi^2\,k}{8}\right|}$

We also know a minimum of a modulus function is at the point where it is zero

So,

$\sf{Required\,\,\,value\,\,\,of\,\,\,k\,=\dfrac{8}{\pi^2}}$