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Mean value of second derivative of a function

Erfan

New member
Jul 19, 2013
9
The curve C is defined parametrically by x = 4t - t^2 and y = 1 - e^-t where 0≤t<2 .
Show that the mean value of d^2 y / dx^2 with respect to x over the interval 0≤x≤7/4 is (4e^(-1/2) - 3)/ 21 .

I've figured out d^2 y/dx^2 as ((t-1)e^-t)/(4(2-t)^3) . Any idea how to do the the other part ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
I'm not sure you need to compute $y''(x)$. Why not use the FTC? You know that the average value $\langle f \rangle$ of a function $f$ on the interval $[a,b]$ is given by
$$\langle f \rangle=\frac{1}{b-a} \int_{a}^{b}f(x) \, dx.$$
So the average value of $y''(x)$ on the interval $[0,7/4]$ would be
$$\langle y''(x) \rangle=\frac{1}{7/4-0} \int_{0}^{7/4}y''(x) \, dx
=\frac{4}{7} [y'(x)] \big|_{0}^{7/4}.$$
So now you need to compute $y'(x)$, and evalute at the two endpoints. Does that make sense?