How to Find the Average Value of a Function Over a Given Interval?

KingKai
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Homework Statement



Determine the average value of the following function over the interval [0,6]

f(t) = (t2-1)e-0.5t

Homework Equations



1/(b-a) ∫ f(x) = {f(x)}

The Attempt at a Solution



Substitution?

let u = (t2-1)

du/dt = 2t

du/2t = dt

1/(6) ∫ u e-0.5t (du/2t)And here, I reach a roadblock.
 
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so to get the average of the function you integrate the function and then divide by the interval length.
 
Try a different technique.
 
KingKai said:

Homework Statement



Determine the average value of the following function over the interval [0,6]

f(t) = (t2-1)e-0.5t

Homework Equations



1/(b-a) ∫ f(x) = {f(x)}

The Attempt at a Solution



Substitution?

let u = (t2-1)

du/dt = 2t

du/2t = dt

1/(6) ∫ u e-0.5t (du/2t)

And here, I reach a roadblock.
Use integration by parts, twice.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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