Find the average value of the function on the given interval.

f(x)= √x+2; [1,7]

did you mean it the way you typed it, or did you mean

f(x) = √(x+2) ?

I will assume f(x) = √(x + 2) judging by the values of the interval

first take the definite integral of the function

= ∫(x+2)^(1/2) dx from 1 to 7
= (2/3)(x+2)^(3/2) | from 1 to 7
= (2/3)9^(3/2) - (2/3)(3^(3/2))
= (2/3)(27) - (2/3)(3√3)
= 18 - 2√3

Average value
= definite integral/domain
= (18-2√3)/(7-1)
= 3 - √3/3

check my arithmetic
If you meant your function the way you typed it, follow the same procedure, it does not come out that "nicely".
=

Perfect!!!

To find the average value of a function on a given interval, we need to evaluate the definite integral of the function over that interval and then divide it by the length of the interval.

In this case, we have the function f(x) = √(x+2) and the interval [1,7].

Step 1: Evaluate the definite integral of the function.
To find the definite integral of f(x) over the interval [1,7], we use the following formula:

∫[1,7] f(x) dx

Since f(x) = √(x+2), the integral becomes:

∫[1,7] √(x+2) dx

Integrating this function requires using the u-substitution method. Let's define u as (x+2), so du = dx.

Now we can substitute the values:

∫[1,7] √(x+2) dx = ∫[1,7] √u du

Using the power rule for integration, where ∫√u du = (2/3)u^(3/2), we get:

(2/3) ∫[1,7] u^(1/2) du = (2/3) [(2/3)u^(3/2)] [1,7]

Plugging in the limits of integration:

= (2/3) [(2/3)(7+2)^(3/2) - (2/3)(1+2)^(3/2)]

= (2/3) [(2/3)(9)^(3/2) - (2/3)(3)^(3/2)]

= (2/3) [(2/3)(27) - (2/3)(3)]

= (2/3) [(2/3)(24)]

= (2/3) (16)

= 32/3 or 10.67 (approx)

Step 2: Divide the definite integral by the length of the interval.
The length of the interval [1,7] is found by subtracting the lower limit from the upper limit:

7 - 1 = 6

Now we can divide the definite integral by the length of the interval to get the average value:

Average value = (32/3) / 6 = (32/3) * (1/6)

Simplifying this expression:

= 32/18

= 16/9 or 1.78 (approx)

Therefore, the average value of the function f(x) = √(x+2) on the interval [1,7] is approximately 1.78.