Answer: (i) The mean value of \(\dfrac{dy}{dx}\) on \(0 \leqslant x \leqslant \dfrac{\pi}{2}\) is
\(\dfrac{1}{\frac{\pi}{2}-0}\int_{0}^{\pi/2} \frac{dy}{dx}\,dx = \dfrac{2}{\pi}\bigl(y(\pi/2)-y(0)\bigr).\)
Since \(y=x^{2}\sin x\), we have \(y(\pi/2)=\left(\dfrac{\pi}{2}\right)^{2}\cdot 1\) and \(y(0)=0\), so
\(\dfrac{2}{\pi}\left(\dfrac{\pi^{2}}{4}\right)=\dfrac{\pi}{2}.\)
So the mean value is \(\dfrac{\pi}{2}\).
(ii) First find \(\dfrac{d^{2}y}{dx^{2}}\).
\(\dfrac{dy}{dx}=2x\sin x+x^{2}\cos x\)
so
\(\dfrac{d^{2}y}{dx^{2}}=2\sin x+4x\cos x-x^{2}\sin x.\)
The mean value on \(0 \leqslant x \leqslant \dfrac{\pi}{2}\) is
\(\dfrac{2}{\pi}\int_{0}^{\pi/2}\left(2\sin x+4x\cos x-x^{2}\sin x\right)dx.\)
Using the fact that this is the change in \(\dfrac{dy}{dx}\) across the interval,
\(\dfrac{2}{\pi}\left[2x\sin x+x^{2}\cos x\right]_{0}^{\pi/2} =\dfrac{2}{\pi}\left(\pi\right)=2.\)
So the mean value of \(\dfrac{d^{2}y}{dx^{2}}\) is \(2\).
For a function \(f(x)\), the mean value on \([a,b]\) is \(\dfrac{1}{b-a}\int_a^b f(x)\,dx\).
(i) Here \(f(x)=\dfrac{dy}{dx}\) and \(a=0\), \(b=\dfrac{\pi}{2}\). Hence
\(\text{MV} = \dfrac{2}{\pi}\int_0^{\pi/2} \dfrac{dy}{dx}\,dx = \dfrac{2}{\pi}\bigl[y\bigr]_0^{\pi/2}.\)
Since \(y=x^2\sin x\),
\(y\left(\dfrac{\pi}{2}\right)=\left(\dfrac{\pi}{2}\right)^2\sin\left(\dfrac{\pi}{2}\right)=\dfrac{\pi^2}{4},\quad y(0)=0.\)
Therefore
\(\text{MV} = \dfrac{2}{\pi}\left(\dfrac{\pi^2}{4}-0\right)=\dfrac{\pi}{2}.\)
(ii) Differentiate twice:
\(\dfrac{dy}{dx}=2x\sin x+x^2\cos x,\)
\(\dfrac{d^2y}{dx^2}=2\sin x+2x\cos x+2x\cos x-x^2\sin x=2\sin x+4x\cos x-x^2\sin x.\)
The mean value of \(\dfrac{d^2y}{dx^2}\) on \([0,\pi/2]\) is
\(\text{MV} = \dfrac{2}{\pi}\int_0^{\pi/2} \dfrac{d^2y}{dx^2}\,dx = \dfrac{2}{\pi}\left[\dfrac{dy}{dx}\right]_0^{\pi/2}.\)
Now
\(\dfrac{dy}{dx}\Big|_{x=\pi/2}=2\cdot\dfrac{\pi}{2}\cdot 1+\left(\dfrac{\pi}{2}\right)^2\cdot 0=\pi,\)
and
\(\dfrac{dy}{dx}\Big|_{x=0}=0.\)
So
\(\text{MV}=\dfrac{2}{\pi}(\pi-0)=2.\)
Therefore the mean values are \(\dfrac{\pi}{2}\) and \(2\).
