Avg value of fun. over given interval..(please explain each step in detail bc I am lost:( ) f(x)=2x+1 on [0,4].. THANK YOU!!!!

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A function f(x) that is closed in an interval (a,b) the its average value defined by `1/(b-a)int_a^bf(x)dx`

For our question;

`f(x) = 2x+1`

`a = 0`

`b = 4`

Average value

`= 1/(4-0)int_0^4(2x+1)dx`

`= 1/4[x^2+x]_0^4`

`= 1/4(16+4-0-0)`

`= 5`

**So the average value of the function is 5.**

**Sources:**

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