Question

If the substitution u = sqrt(x-1) is made, then the integral from 2 to 5 [(sqrt(x-1)/x]dx = ?

Answer 1

\[u = \sqrt{x-1}\]
\[\int\limits_{2}^{5} \frac{ \sqrt{x-1} }{ x }dx\]

Answer 2

\[du = \frac{ 1 }{ 2\sqrt{x-1} }dx\]
=>
\[dx = 2\sqrt{x-1} du \] is that right?

Answer 3

but how would i put that x in the denominator in terms of u?

Answer 4

would it be x = u^2 +1 ? then plug into the denominator? oh btw, don't evaluate. just set up.
-_- wow, i just kinda answered my own question. how sad

Answer 5

You could figure it out faster than I could even begin to understand how to do it :D

Answer 6

You have to change the limits of integration, though.

Answer 7

sorry.
yeah i keep getting
\[\int\limits_{2}^{5} \frac{ 2u }{ u^2 +1 }du \]
can you explain how or why the limits change?

Answer 8

Well, seeing as you're no longer integrating with respect to x,
and it WAS x that initially ran from (in this case) 2 to 5, this time it's u that runs....
from u(2) to u(5).
That is to say, u, evaluated at x = 2, to u, evaluated at x = 5.

Answer 9

when you set x =...., x =1,---> u =... you must change the limit from x to u. because the original one respect to x , after changing, the function respect to u

Answer 10

but why would you evaluate it at u = 2 and u = 5? i'm not really understanding that part :(

Answer 11

Here, let's illustrate the chain rule, and here, F'(x) = f(x)
\[\Large \int\limits_{a}^{b}F\circ u(x)u'(x)dx\]

Answer 12

yeah i understand that.

Answer 13

Now, we let u=u(x)
du=u'(x)dx
\[\Large \int\limits_{a}^{b}F(u)du\]

Answer 14

ohhh....
that makes sense
thanks :)

Answer 15

Aww, I made a mistake in representing... no matter. Suppose F is the derivative of some function G. Then this would be
\[\Large G(u)]_{x=a}^{x=b}=G(u(a))-G(u(b))=\int\limits_{u(a)}^{u(b)}F(u)du\]

Answer 16

That's done :)

Answer 17

ok, thanks again :)

Answer 18

did you figure out what your new integral will look like

Answer 19

\[\int\limits_{1}^{2} \frac{ 2u^2 }{ u^2+1 }du \]
i got the answer, but it took me a little while to figure out why.
:/

Answer 20

good

Answer 21

Well-played :)

Answer 22

GG :)

sorry again. i understand what you did but why is it G(u(a)) - G(u(b)) ? I thought it's supposed to be G(u(b)) - G(u(a)) ? by the Fundamental Theroem of Calculus ?

Answer 23

Yeah. Typo. Good thing you caught that :)\[\Large G(u)]_{x=a}^{x=b}=G(u(b))-G(u(a))=\int\limits_{u(a)}^{u(b)}F(u)du\]
I'd give you another medal, but I can't :D

Brilliant error spotting ^.^

Answer 24

okay, whew. :o
aww:3
haha, thanks.