Question

ok i got my answer to be 75/3....
integral 1 to 2 (1+2y)^2dy

Answer 1

It should be 49/3

Answer 2

not sure where i went wrong

Answer 3

integral 4y^2 + 4y + 1
4y^3/3 +4y^2/2 + y

plug in 1 and 2

Answer 4

heres what i did/.....\[\int\limits_{1}^{2} (1+2y)^{2} dy\]
\[(1+2y)(1+2y)= 1+4y=4y ^{2}\]

Answer 5

1y+(4y^2/2)+(4y^3/3) with integral 1 to 2

Answer 6

i got the first part to 13/3+62/3...

Answer 7

i meant minus!!!!ugghhh now i see

Answer 8

[32/3+8+2]-[4/3+2+1]

Answer 9

ok now i know thanks i see where i got mess up