Minimizing Integral

Calculus Level 4

∫ 0 π ∣ sin ⁡ x − A x ∣ d x \large \int_0^\pi | \sin x - Ax | \, dx

Find the value of A A such that the value of the integral above is minimized. If the value of A A can be expressed as P π Q sin ⁡ ( π R ) , \dfrac{\sqrt P}{\pi^Q} \sin \left ( \dfrac{\pi}{\sqrt R} \right) , submit your answer as P + Q + R P+Q+R .


The answer is 5.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

Brian Lie
Mar 13, 2018

I ( t ) = ∫ 0 π ∣ sin ⁡ x − A x ∣ d x sin ⁡ t = A t , t ∈ ( π 2 , π ) = ∫ 0 t ( sin ⁡ x − A x ) d x + ∫ t π ( A x − sin ⁡ x ) d x = A π 2 2 − A t 2 − 2 cos ⁡ t A = sin ⁡ t t = π 2 sin ⁡ t 2 t − t sin ⁡ t − 2 cos ⁡ t I ′ ( t ) = ( 2 t − π ) ( 2 t + π ) ( sin ⁡ t − t cos ⁡ t ) 2 t 2 \begin{aligned} I(t)&=\int_0^\pi|\sin x-Ax|dx&\small\color{#3D99F6}\sin t=At,t\in\left(\frac\pi 2,\pi\right) \\&=\int_0^t(\sin x-Ax)dx+\int_t^\pi(Ax-\sin x)dx \\&=\frac{A\pi^2}2-At^2-2\cos t&\small\color{#3D99F6}A=\frac{\sin t}t \\&=\frac{\pi^2\sin t}{2t}-t\sin t-2\cos t \\\\I'(t)&=\frac{(\sqrt 2t-\pi)(\sqrt 2t+\pi)(\sin t-t\cos t)}{2t^2} \end{aligned}

Therefore, I ( t ) I(t) has a minimum value when I ′ ( t ) = 0 I'(t)=0 , that is,

t = π 2 , A = 2 π sin ⁡ ( π 2 ) , \begin{aligned} t&=\frac\pi{\sqrt 2},\\ A&=\frac{\sqrt 2}\pi\sin\left(\frac\pi{\sqrt 2}\right), \end{aligned}

making the answer 2 + 1 + 2 = 5 2+1+2=\boxed 5 .

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