Question

integral of 1/(squaroot(3-x))

Answer 1

the answer is -2(sqrt(3-x)) in my book but they skipped all the steps in this "example"

Answer 2

Looks like you can make a nice substitution here, have any guesses? Also remember that you can rewrite 1/sqrt(a) as simply a^(-1/2) so it looks like other derivative/integration rules you already know.

Answer 3

if i do that then i get\[\frac{ (3-x)^{-\frac{ 3 }{ 2 }} }{ -\frac{ 3}{ 2 } }\]

Answer 4

Wait, how did you get there? Show more of your steps so I can help you out.

Answer 5

i can't even explain how i got there but thats what my "logic" drove me to

Answer 6

\[\frac{ 1 }{ \sqrt{3-x} }=(3-x)^{-\frac{ 1 }{ 2 }} \] So the 1/2 comes from being a square root, and the negative is what you get from it being in the bottom of a fraction and moved to the top.

If you want me to explain that more, I can. But the point of doing that is because it looks a lot like other stuff you can take the derivative and integral of, like:

x^2 or 4x^3 or whatever, just examples that (3-x)^(-1/2) kind of looks like.

But now you need to change something with a substitution to make it easier to integrate, any guesses or questions?

Answer 7

wait i can explain it quickly... i let u=3-x then -du=1
\[-\int\limits_{}^{}u ^{ \frac{-1}{2}} = \frac{ u ^{\frac{ -3 }{2 }} }{ \frac{ -3 }{ 2 } }\]

Answer 8

thats what i used to get there

Answer 9

Ahh, I see. What you're doing is going the wrong direction though. You need to add 1 to the exponent, not subtract 1. You subtract 1 when you take the derivative, such as from x^3 you get 3x^2.

Answer 10

oh i see where i went wrong, then i'll get