Question:
Consider the function f(x) = 2x^2 - 3x + 1 on the interval [0, 4].
a) Find the average value of f(x) on the interval [0, 4].
b) Is there a value c in the interval [0, 4] such that f(c) is equal to the average value found in part (a)? Justify your answer.
Answer:
a) To find the average value of a function f(x) on an interval [a, b], we need to evaluate the definite integral of f(x) from a to b, and then divide the result by the length of the interval (b - a).
In this case, our function is f(x) = 2x^2 - 3x + 1 on the interval [0, 4].
The average value, denoted by "Avg" of f(x) on [0, 4] can be calculated as:
Avg = (1/(4 - 0)) ∫[from 0 to 4] (2x^2 - 3x + 1) dx.
To simplify the integral, we can compute each term separately:
∫ 2x^2 dx = (2/3) x^3 + C ∫ -3x dx = (-3/2) x^2 + C ∫ 1 dx = x + C
Plugging the integral results back into the initial equation:
Avg = (1/(4 - 0)) [ (2/3) x^3 - (3/2) x^2 + x ] |_0^4
Substituting the upper limit (4) into each term:
Avg = (1/4) [ (2/3)(4^3) - (3/2)(4^2) + (4) ] - (1/4) [ (2/3)(0^3) - (3/2)(0^2) + (0) ]
Evaluated expression:
Avg = (1/4) [ (2/3)(64) - (3/2)(16) + 4 ] - (1/4) [ 0 - 0 + 0 ]
Avg = (1/4) [ (128/3) - (48/2) + 4 ]
Avg = (1/4) [ (128/3) - (96/3) + 12/3 ]
Avg = (1/4) [ (32/3) ]
Simplifying further:
Avg = 8/3
Therefore, the average value of f(x) on the interval [0, 4] is 8/3.
b) To determine if there exists a value c in the interval [0, 4] such that f(c) is equal to the average value found in part (a), we need to check if f(c) = 8/3 has any solutions.
Let's solve the equation 2c^2 - 3c + 1 = 8/3:
2c^2 - 3c + 1 - 8/3 = 0
Multiplying by 3 to get rid of the fractions:
6c^2 - 9c + 3 - 8 = 0
6c^2 - 9c - 5 = 0
Using the quadratic formula:
c = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 6, b = -9, and c = -5.
c = (-(-9) ± √((-9)^2 - 4(6)(-5))) / (2(6))
c = (9 ± √(81 + 120)) / 12
c = (9 ± √201) / 12
As √201 is not a rational number, it is clear that there are no values of c in the interval [0, 4] such that f(c) = 8/3.
Therefore, there is no value c in the interval [0, 4] for which f(c) is equal to the average value of f(x) on the interval [0, 4].