Math Problem Statement
Solution
Let's solve the problem step by step.
Part (a): Find the average value of on the interval .
The formula for the average value of a function on the interval is:
In this case:
- The interval is , so and .
Thus, the average value is:
We will now compute this integral.
The integrals are straightforward:
Now, subtract the two results:
So the average value is:
Part (b): Find a value in the interval such that is equal to the average value.
We need to solve the equation: Substituting :
First, subtract from both sides:
Multiply by to get:
Now, take the square root:
Since must be in the interval , we take the positive root:
Final Answers:
- (a) The average value of on the interval is or approximately 16.67.
- (b) The value such that equals the average value is or approximately 2.89.
Would you like further details on any part of the solution? Here are some follow-up questions for deeper understanding:
- How does the average value formula relate to the definite integral?
- What is the significance of finding a point where equals the average value?
- Can you apply the same process to a different function on the same interval?
- What happens to the average value if the interval is changed?
- What are some common mistakes when calculating definite integrals?
Tip: Always double-check your integral bounds and limits when performing definite integration to avoid small calculation errors!
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Math Problem Analysis
Mathematical Concepts
Integral Calculus
Average Value of a Function
Formulas
Average value of a function: (1 / (b - a)) * integral from a to b of f(x) dx
Integral of x^2: (x^3)/3
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grades 11-12, Early College