Calculus
posted by Ishaan on .
If b is a positive constant, find the average value of the function f defined by f(x) = x2 + bx + 1 on the interval [1,1]

f(x)= x^2 + bx + 1 on [1,1]
First, integrate
 = integration symbol
x^2 + bx + 1 = 1/3 x^3 + 1/2 bx^2 + x + C
Evaluate from 1 to 1
1/3 x^3 + 1/2 bx^2 + x + C
1/3 (1)^3 + 1/2 b(1)^2 + 1  (1/3 (1)^3 + 1/2 b(1)^2  1
1/3 + 1/2 b + 1  (1/3 + 1/2 b  1)
1/2 b + 4/3  (1/2 b  4/3) = 8/3
Average Value
V = 1/(b  a) * f(x)
V = 1/(1  (1)) * 8/3
V = 1/2 * 8/3
V = 8/6 = 4/3