- Thread starter
- #1

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 ?

- Thread starter Erfan
- Start date

- Thread starter
- #1

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 ?

- Admin
- #2

- Jan 26, 2012

- 4,197

$$\langle f \rangle=\frac{1}{b-a} \int_{a}^{b}f(x) \, dx.$$

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?