What is the average value of function (x-3)^2 in [2,5]?

Question

i am thinking the answer of the book is wrong, but could someone check the answer, thanks so much !!!

Answer 1

integrate (x-3)^2 between 2 and 5
then divide this answer by 5-2=3

∫ (x-3)^2 dx
= (1/3)*(x-3)^3 between 2 and 5
= (1/3)(2^3 - (-1)^3)
=9/3
=3
then
y(av) = [∫ (x-3)^2 dx] / [5-2] = 3/3 = 1

the end
.

Answer 2

int (x-3)^2 for x = 2 to 5

u = x-3
du = dx

int u^2 du = u^3/3
= (x-3)^3/3, from 2 to 5

= (5-3)^3/3 - (2-3)^3/3
= 8/3 + 1/3
= 3. this is the total area.

SQUASH this into a rectangle by dividing by (5-2) or 3.

So average value = 1.

Answer 3

∫ (x-3) ² dx = ⅓ x ((x-9)x +27) + C

The above integral from 2 to 5 = 11,66667 - 8,666667 = 3

The average value of function = 3/(5-2) = 1
-

Answer 4

(x-3)^2 in [2,5]

substituting 2 we get the value=1
" 3 we get the value=0
" 4 " " " " =1
" 5 " " " " =4

therefore the average value is=(1+0+1+4)/4=1.5 :-)