On applying the mean value theorem on integral, then find out the average value of f(x) = 3 - 3/2xor on [0, 2].Correct answer is '1.5'. Can you explain this answer?

Question

Answer

Understanding the Mean Value Theorem for Integrals
The Mean Value Theorem for Integrals states that if a function is continuous on a closed interval [a, b], then there exists at least one point c in (a, b) such that:
f(c) = (1 / (b - a)) * ∫[a to b] f(x) dx
This theorem helps in finding the average value of a function over a given interval.
Finding the Average Value of f(x)
To find the average value of the function f(x) = 3 - (3/2)x on the interval [0, 2], we follow these steps:
- Step 1: Verify Continuity
The function f(x) is a linear function, hence it is continuous on [0, 2].
- Step 2: Evaluate the Integral
We need to calculate the integral of f(x) from 0 to 2:
∫[0 to 2] (3 - (3/2)x) dx
- Step 3: Calculate the Integral
- The integral can be computed as follows:
- ∫(3) dx from 0 to 2 = [3x] from 0 to 2 = 3(2) - 3(0) = 6
- ∫(- (3/2)x) dx from 0 to 2 = [-(3/4)x^2] from 0 to 2 = - (3/4)(4) + 0 = -3
Combining these results gives:
6 - 3 = 3
- Step 4: Average Value Calculation
Using the Mean Value Theorem formula:
Average value = (1 / (b - a)) * ∫[a to b] f(x) dx
= (1 / (2 - 0)) * 3 = 1.5
Conclusion
Thus, the average value of the function f(x) = 3 - (3/2)x over the interval [0, 2] is 1.5.

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