Question

find the integral of (x^2)(1-x)^1/2, i get that you times it with 1/3 and you set 1-x to zero and get answer then plug that in, but i am lost with the formula help please

Answer 1

u=1-x

Answer 2

yes and du is \[x^{2}\]

Answer 3

no

Answer 4

du=-dx

Answer 5

u=1-x

so x=1-u
and \(x^2=(1-u)^2=1-2u-u^2\)

Answer 6

so then (1-2u-u^2)du?

Answer 7

no

Answer 8

\[x^2=1-2u-u^2\]
\[1-x=u\]
\[dx=-du\]

combine

Answer 9

1-x=u
\[\sqrt{1-x}=\sqrt{u}\]

Answer 10

ok, i see there is (1-2u-u^2)(u^1/2)(du), if am correct,

Answer 11

typo there: (1-2u+u^2) not -u^2

Answer 12

yes..now distribute the \(u^{1/2}\)

yes...typo from me above too

Answer 13

you are still missing the negative sign from the dx=-du

Answer 14

ah, yes missed that

Answer 15

\[\int x^2(1-x)^{1/2}dx\]
\[=\int (1-2u+u^2)(u^{1/2})(-du)\]

Answer 16

thank you for the help, i figured the question, and by the way data is my favorite from star trek next gen

Answer 17

good, because he is awesome ;)

Answer 18

lol