Question

another question @jim_thompson5910 @billj5

Answer 1

I have to find the average value for this:

Answer 2

\[\int\limits_{0}^{-3}\frac{ 4 }{ (2x+6)^{2} }\]

Answer 3

you'll use u-substitution here

Answer 4

since b is less then a then i have to switch them and do \[-\int\limits_{-3}^{0}\]

Answer 5

yes, correct

Answer 6

but then for my 1/b-a what do i use for b and what do i use for a?

Answer 7

I think it doesn't matter really because the "b-a" portion should be positive anyway

|b-a| = |a-b|

Answer 8

I really can't figure this out, can you help me? @jim_thompson5910

Answer 9

\[\int\limits_{0}^{-3}\frac{ 4 }{ (2x+6)^2 }\] and I gave to find the avg value and mean value

Answer 10

did you try u-substitution?

Answer 11

my teacher never told us to do that for this problem

Answer 12

ok so let's replace 2x+6 with some simpler variable, often that's the letter u

Answer 13

let u = 2x+6

if u = 2x+6, then du/dx = 2, agreed?

Answer 14

yes

Answer 15

that means

du/dx = 2

du = 2*dx

I multiplied both sides by dx

Answer 16

then multiply both sides by 2

du = 2*dx

2du = 4dx