Answer: (i) The mean value of \(\dfrac{dy}{dx}\) on \(0\le x\le 2\) is \(\dfrac{y(2)-y(0)}{2-0}\).

Since \(y=e^{-2x}\), \(y(0)=1\) and \(y(2)=e^{-4}\). Hence the mean value is \(\dfrac{e^{-4}-1}{2}\approx -0.491\).

(ii) The centroid of the region is \((\bar{x},\bar{y})\) where \(\bar{x}=0.463\) and \(\bar{y}=0.255\) (3 s.f.).

(i) The mean value of a derivative over an interval is

\(\dfrac{y(2)-y(0)}{2-0}\).

For \(y=e^{-2x}\),

\(y(0)=e^0=1\), and \(y(2)=e^{-4}\).

So

\(\text{mean value of }\dfrac{dy}{dx}=\dfrac{e^{-4}-1}{2}\approx -0.490842...\).

Therefore, correct to 3 significant figures, the mean value is \(-0.491\).

(ii) Let the required centroid be \((\bar{x},\bar{y})\). For a region under \(y=f(x)\) from \(x=0\) to \(x=2\),

\(\bar{x}=\dfrac{\int_0^2 x f(x)\,dx}{\int_0^2 f(x)\,dx}\),

and

\(\bar{y}=\dfrac{\frac12\int_0^2 (f(x))^2\,dx}{\int_0^2 f(x)\,dx}\).

Here \(f(x)=e^{-2x}\), so

\(\bar{x}=\dfrac{\int_0^2 x e^{-2x}\,dx}{\int_0^2 e^{-2x}\,dx}\), \(\bar{y}=\dfrac{\frac12\int_0^2 e^{-4x}\,dx}{\int_0^2 e^{-2x}\,dx}\).

First,

\(\int_0^2 e^{-2x}\,dx=\left[-\frac12 e^{-2x}\right]_0^2=\frac12(1-e^{-4}).\)

Next, integrate by parts for \(\int_0^2 x e^{-2x}\,dx\):

Take \(u=x\), \(dv=e^{-2x}dx\). Then \(du=dx\), \(v=-\frac12 e^{-2x}\).

So

\(\int_0^2 x e^{-2x}\,dx=\left[-\frac{x e^{-2x}}{2}\right]_0^2+\frac12\int_0^2 e^{-2x}\,dx\).

Hence

\(\int_0^2 x e^{-2x}\,dx=\left[-\frac{x e^{-2x}}{2}-\frac{e^{-2x}}{4}\right]_0^2=\frac14-\frac54 e^{-4}.\)

Therefore

\(\bar{x}=\dfrac{\frac14-\frac54 e^{-4}}{\frac12(1-e^{-4})}\approx 0.463013...\), so \(\bar{x}=0.463\) to 3 s.f.

For \(\bar{y}\),

\(\frac12\int_0^2 e^{-4x}\,dx=\frac12\left[-\frac14 e^{-4x}\right]_0^2=\frac18(1-e^{-8}).\)

Thus

\(\bar{y}=\dfrac{\frac18(1-e^{-8})}{\frac12(1-e^{-4})}=\frac14\cdot\dfrac{1-e^{-8}}{1-e^{-4}}\approx 0.254994...\),

so \(\bar{y}=0.255\) to 3 s.f.

Therefore the centroid is \((0.463,\,0.255)\).