Hello agian.

how do u do the substitution t = x ^ 1-2

for dx / 3 + square root x

with upper limit as 1

and lower as 0

Thank you.

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- May 3rd 2006, 06:46 AMsimonsubstitution
Hello agian.

how do u do the substitution t = x ^ 1-2

for dx / 3 + square root x

with upper limit as 1

and lower as 0

Thank you. - May 3rd 2006, 07:51 AMtopsquarkQuote:

Originally Posted by**simon**

Now set:

$\displaystyle t = \sqrt x $

Thus

$\displaystyle dt = \frac{1}{2} \frac{1}{\sqrt x} dx$

or

$\displaystyle dx = 2 \sqrt x dt = 2 t \, dt$

For the limits on the integral:

Upper limit: $\displaystyle t = \sqrt 1 = 1$

Lower limit: $\displaystyle t = \sqrt 0 = 0$

Thus:

$\displaystyle \int_0^1 \frac{dx}{3+\sqrt x}$

$\displaystyle = \int_0^1 \frac{2 t \, dt}{3 + t} =2 \int_0^1 dt \frac{t}{t+3}=2 \int_0^1 dt \left ( 1 - \frac{3}{t+3} \right )$

$\displaystyle =2t|_0^1 - 6 \, ln|t+3| |_0^1 = 2 - 6 \, ln4 + 6 \, ln 3$

$\displaystyle = 2 + 6 \, ln \left ( \frac{3}{4} \right ) $

-Dan - May 3rd 2006, 08:06 AMCaptainBlackQuote:

Originally Posted by**topsquark**

exceptional performance in interpreting strangely/obscurely phrased

questions :D

RonL - May 3rd 2006, 08:18 AMNatasha1Quote:

Originally Posted by**CaptainBlack**

- May 3rd 2006, 12:14 PMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - May 3rd 2006, 01:20 PMThePerfectHacker
CaptainBlack, CaptainBlack! Did you hear' what topsquark-the fizizist said 'bout you! Do not tell him I said that...

- May 3rd 2006, 08:05 PMCaptainBlackQuote:

Originally Posted by**topsquark**

RonL

(Fishing for an award of the Math Help Form Medal for obscure humour :eek: )

RonL - May 4th 2006, 03:03 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan