Question

Find the average value of f(x)=√25−x² over the interval [0,5]. Write the exact answer.

Answer 1

To find the average value of f(x)=√25−x² over the interval [0,5], integrate the function over the interval and divide it by the width of the interval.

Step-by-step explanation:

To find the average value of the function f(x) = √(25-x²) over the interval [0,5], we need to find the definite integral of the function over that interval and divide it by the width of the interval. The average value can be calculated as:

average value = (1/5) * ∫[0,5] √(25-x²) dx

Using the property that the integral of the square root of a function is equal to the integral of the function divided by 2, we can simplify the calculation to:

average value = (1/5) * (1/2) * ∫[0,5] (25-x²) dx

Now we can evaluate the definite integral to find the average value.

Answer 2

The average value of f(x) = √(25 - x²) over the interval [0,5] is π/4.

Step-by-step explanation:

To find the average value of f(x) = √(25 - x²) over the interval [0,5], we need to calculate the definite integral of f(x) over that interval and divide it by the length of the interval.
First, let's find the indefinite integral of f(x):
∫(√(25 - x²)) dx

We can rewrite this integral using trigonometric substitution. Let x = 5sin(θ), dx = 5cos(θ) dθ

∫(√(25 - (5sin(θ))²)) (5cos(θ)) dθ

The integral becomes:

∫(5cos(θ)√(25 - 25sin²(θ))) dθ

We can simplify it further as:

∫(5cos(θ)√(25cos²(θ))) dθ

Now the integral is simplified and we can calculate it:

5∫cos²(θ) dθ

We know that the integral of cos²(θ) is θ/2 + (1/4)sin(2θ).

Substituting our angle back:

5(θ/2 + (1/4)sin(2θ))

We need to find the limits of integration based on the original interval [0,5]. When x = 0, θ = 0, and when x = 5, θ = π/2.
Plugging in these values:

5((π/2)/2 + (1/4)sin(2(π/2))) - 5(0/2 + (1/4)sin(2(0)))

5((π/4) + (1/4)sin(π)) - 5(0 + 0)

5((π/4) + (1/4)0) - 0

5(π/4) = π/4