Asked by Ishaan
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]
Answers
Answered by
anon
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
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
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