How do you find the average value of the function #u(x) = 10xsin(x^2)# on the interval #[ 0, sqrtpi ]#?

Question

How do you find the average value of the function #u(x) = 10xsin(x^2)# on the interval  #[ 0, sqrtpi ]#?

Charles Arocha · Accepted Answer

The average value is #10/sqrtpi#. [Average rate of change is #0#]

The average value requested in the question is: #1/(sqrtpi - 0) int_0^(sqrtpi) 10xsin(x^2) dx# # = 5/sqrtpi int_0^(sqrtpi) sin(x^2) (2x)dx# # = 5/sqrtpi[-cos(x^2)]_0^sqrtpi# # = 5/sqrtpi [ -cos((sqrtpi)^2)- -cos(0^2)]# # = 5/sqrtpi [2] = 10/sqrtpi# An alternative query The topic "Average Rate of Change..." is where the question was posted, and it differs from the average value. Throughout this period, this function's average rate of change is #(f(sqrtpi)-f(0))/(sqrtpi-0) = (10sqrtpisin((sqrtpi)^2) - 10(0)sin(0^2))/(sqrtpi - 0)# # = 0/sqrtpi =0#

Ezekiel Alexander · Answer

undefined To find the average value of the function $ u(x) = 10x\sin(x^2) $ on the interval $[0, \sqrt{\pi}]$, you need to compute the definite integral of the function over the given interval and then divide the result by the length of the interval.

1. Compute the definite integral of $ u(x) $ over the interval $[0, \sqrt{\pi}]$ using the Fundamental Theorem of Calculus.
2. Find the length of the interval, which is $ \sqrt{\pi} - 0 = \sqrt{\pi} $.
3. Divide the result from step 1 by the length of the interval found in step 2.

$$ \text{Average value} = \frac{1}{\sqrt{\pi}} \int_{0}^{\sqrt{\pi}} 10x\sin(x^2) \,dx $$