Question

help please number 3 this is what i did

Answer 1

http://assets.openstudy.com/updates/attachments/5760b450e4b0c9d3c486c58c-marcelie-1465955422098-hwsction.jpg

Answer 2

@satellite73

Answer 3

did you integrate from 1 to 5?

Answer 4

yes but i got lost

Answer 5

\[\pi \int\limits_{1}^{5} ( \sqrt{x-1})^2 dx\]

Answer 6

\[\pi \int _1^5(\sqrt{x-1})^2dx\] is it
they have made it very easy for you since when you square you just get \[\pi\int_1^5(x-1)dx\]

Answer 7

yeah, what you wrote
how lost can you be?

Answer 8

oh so the sq root got cancled with the 2 right ?

Answer 9

my alegbra is rusty D:

Answer 10

i guess you could call it "cancel" lol

Answer 11

yaas lool

Answer 12

it is just that \(\sqrt{\text{whatever}}^2=\text{whatever}\)

Answer 13

well actually the absolute value of whatever, but here everything is positive anyways

Answer 14

ah okay thanks you

Answer 15

yw

Answer 16

can you help me with 5 please :D

Answer 17

Drawing sketch would helps: