Question

How to integrate v^2 (1-v) ^6

let u=1-v
then i get :
(1-u)^2(u^6)
what to do next?

Answer 1

\[\large \int\limits v^2(1-v)^6dv\]is that your question?

Answer 2

\[\large {u=1-v \ \\v= 1-u}\]

Answer 3

\[\large dv = -du\]

Answer 4

\[\large - \int\limits [(1-u)^2\cdot u^6]du\]

Answer 5

Expand (1-u)^2 then multiply u^6 inside the expanded terms . Then do as usual integration .

Answer 6

if in case of the (1-u) has very big index?

Answer 7

Dont forget -ve sign

Answer 8

What does expanding (1-u)^2 give you, first?

Answer 9

1-2u +u^2 then all of the term times by u^6

Answer 10

Yes.

Answer 11

but if i have v^8 * (v-1)^6 ?

Answer 12

\[\large - \int\limits\limits [(1-2u+u^2)\cdot u^6]du\]

Answer 13

Like ??? u have made substitution to reduce the index of (1-v)^6 to something u had already known

Answer 14

i mean 1-v --- in the baracket

Answer 15

oh

Answer 16

Then expand [(V-1)^3]^2

Answer 17

Now,what do you get when you multiply u^6 into all these terms?

Answer 18

i get it, thanks guys

Answer 19

Ok :)

Answer 20

It becomes lengthy but ultimately u"ll be able to solve it . Just use what u already knew :)

Answer 21

you could also integrate by parts using a tabular method

Answer 22

for efficiency:
http://people.whitman.edu/~hundledr/courses/M244S07/IntByParts.pdf