Calculus
posted by Robin .
A) Find the average value of
f(x)=x^3x+1 on the interval (0,2)
B) Find c so that f(c)equals the average value

avg value is area divided by width, so
1/2 ∫[0,2] x^3x+1 dx
= 2
So, since the area is 4, and the width is 2, a rectangle of height 2 will have the same area.
We need to solve
f(x) = 2
x^3x+1 = 2
x^3x1 = 0
x = 1.32
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