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Let M be the average value of f(x)=x² on [0,10]. Find a value of c in [0,10] such that f(c)=M c=

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Final answer:

To find a value of c in the interval [0,10] such that f(c) is equal to the average value of f(x)=x² on the interval [0,10], we calculate the average value using the formula M = (1/(b-a)) * ∫(a to b) f(x) dx. The average value is 100/3. Then, solving the equation x² = (100/3) gives us the value of c as √(100/3).

Step-by-step explanation:

To find a value of c in the interval [0,10] such that f(c) is equal to the average value of f(x)=x² on the interval [0,10], we need to calculate the average value first.

The average value, M, of f(x)=x² on the interval [0,10] is given by the formula:

M = (1/(b-a)) * ∫(a to b) f(x) dx

In this case, a = 0 and b = 10. So, we have:

M = (1/(10-0)) * ∫(0 to 10) x² dx = (1/10) * [x³/3] from 0 to 10 = (1/10) * [(10³/3) - (0³/3)] = (1/10) * (1000/3) = 100/3.

To find a value of c such that f(c) = M, we need to solve the equation x² = (100/3) for x.

Taking the square root of both sides, we get:

x = ±√(100/3)

Since c must be in the interval [0,10], we can disregard the negative solution. Therefore, c ≈ √(100/3).

User Joe Day
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