What is the average value of a function #y=50x+5# on the interval #[3,9]#?

Question

2149 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-30T22:14:06

#1/(9-3)int_3^9(50x+5)dx=305#

The average value of a function #f(x)# on the interval #[a,b]# is given by:

#"average value"=1/(b-a)int_a^bf(x)dx#

So here, the average value is:

#1/(9-3)int_3^9(50x+5)dx#

#=5/6int_3^9(10x+1)dx#

Integrating term by term:

#=5/6[5x^2+x]_3^9#

#=5/6[(5(9)^2+9)-(5(3)^2+3)]#

#=5/6(414-48)#

#=305#

Answer 2 · 2017-04-30T22:20:49

#avg = 305#

Use the average function #avg = 1/(b-a) int_a^b f(x) dx#

#avg = 1/(9-3) int_3^9 (50x + 5)dx = 1/6 int_3^9 (50x + 5)dx#

#avg = 1/6 [(50x^2)/2 + 5x]_3^9 = 1/6 [25x^2 + 5x]_3^9#

#avg = 1/6 [(25 *9^2 + 5*9) - (25 * 3^2 + 5*3)]#

#avg = 1/6 (2070 - 240)#

#avg = 305#