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

1 Answer
Jim H
Aug 31, 2015

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

Explanation:

The question asks for the average value which 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#

A different question

The question was posted under the topic "Average Rate of Change . . . " which is different from the average value.

The average rate of change of this function on this interval is

#(f(sqrtpi)-f(0))/(sqrtpi-0) = (10sqrtpisin((sqrtpi)^2) - 10(0)sin(0^2))/(sqrtpi - 0)#

# = 0/sqrtpi =0#

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