To calculate the average value of f(x) = √x on the interval [0, C], we need to evaluate the definite integral of √x with respect to x from 0 to C and divide it by the length of the interval (C - 0):

Average value = (1/(C - 0)) * ∫[0, C] √x dx

Let's calculate the average value for the given values of C:

For C = 1:
Average value = (1/(1 - 0)) * ∫[0, 1] √x dx
= (1/1) * [(2/3)x^(3/2)] from 0 to 1
= (2/3) * (1^(3/2) - 0^(3/2))
= (2/3) * (1 - 0)
= 2/3

For C = 9:
Average value = (1/(9 - 0)) * ∫[0, 9] √x dx
= (1/9) * [(2/3)x^(3/2)] from 0 to 9
= (2/3) * (9^(3/2) - 0^(3/2))
= (2/3) * 9
= 6

For C = 81:
Average value = (1/(81 - 0)) * ∫[0, 81] √x dx
= (1/81) * [(2/3)x^(3/2)] from 0 to 81
= (2/3) * (81^(3/2) - 0^(3/2))
= (2/3) * (81^(3/2))
= 2/3 * 81^(3/2)

For C = 100:
Average value = (1/(100 - 0)) * ∫[0, 100] √x dx
= (1/100) * [(2/3)x^(3/2)] from 0 to 100
= (2/3) * (100^(3/2) - 0^(3/2))
= (2/3) * (100^(3/2))
= 2/3 * 100^(3/2)

The pattern we observe is that as C increases, the average value also increases. Specifically, we see that the average value is:

2/3 for C = 1
6 for C = 9
2/3 * 81^(3/2) for C = 81
2/3 * 100^(3/2) for C = 100

Got a question with my answer?

Message Me

Cheapmath.com is completely free!

Asking question is free. Getting answers is free.

+99

Continue