Help with average value of a function

credd741
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i have a question :cry:
how would you find a number b such that the average value of
f(x)= 1+10x-3x^2 on the interval [0,b] is equal to 4.

i have tried to set this equal to 4 but everytime i try to solve for b i get some number but it is always wrong. please help!:cry::cry::cry::cry:
 
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Welcome to the PF. How do you calculate the average of a function over an interval? What is the relevant equation?
 
the relevant equation is the f(x)=1+10x-3x^2
 
In order to find the average value of a function over a specified interval [0,b] you must integrate the function on that interval and divide by the interval length.

Can you formulate an equation which states these facts in terms of b and solve for b?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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