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
To find the average value of a function on an interval, you need to perform the following steps:
1. Find the definite integral of the function over the interval.
2. Divide the result from step 1 by the length of the interval.
Let's solve the problem using these steps:
1. First, we want to find the definite integral of the function f(x) = x^2 + bx + 1 over the interval [-1, 1].
The integral of the polynomial function f(x) = x^2 + bx + 1 can be calculated term by term.
∫ (x^2 + bx + 1) dx = ∫x^2 dx + ∫bx dx + ∫1 dx
= (x^3/3 + bx^2/2 + x) + C, where C is the constant of integration.
2. Now, let's evaluate the definite integral over the interval [-1, 1]:
∫[-1,1] (x^2 + bx + 1) dx
= [(x^3/3 + bx^2/2 + x)]|[-1,1]
= [(1/3 + b/2 + 1) - (-1/3 + b/2 - 1)]
= [(1/3 + b/2 + 1) + (1/3 - b/2 + 1)]
= (2/3 + b + 1 - b/2 + 2/3 - b/2 + 1)
Simplifying this expression:
= (4/3 - b/2 + b + 2)
3. Next, we need to calculate the length of the interval [-1, 1]:
Length of interval = 1 - (-1) = 2
4. Finally, we divide the value from step 2 by the length of the interval from step 3 to find the average value of the function:
Average value of f(x) over the interval [-1, 1] = (4/3 - b/2 + b + 2)/2 = (8/3 - b/2 + 2b + 4)/2.
Hence, the average value of the function f(x) = x^2 + bx + 1 on the interval [-1, 1] is (8/3 - b/2 + 2b + 4)/2.