# substitution

• May 3rd 2006, 06:46 AM
simon
substitution
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 AM
topsquark
Quote:

Originally Posted by simon
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.

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

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 AM
CaptainBlack
Quote:

Originally Posted by topsquark
$\displaystyle \int_0^1 \frac{dx}{3+\sqrt x}$

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

Please award yourself the Math Help Forum Medal for clairvoyance, for
exceptional performance in interpreting strangely/obscurely phrased
questions :D

RonL
• May 3rd 2006, 08:18 AM
Natasha1
Quote:

Originally Posted by CaptainBlack
Please award yourself the Math Help Forum Medal for clairvoyance, for
exceptional performance in interpreting strangely/obscurely phrased
questions :D

RonL

:D
• May 3rd 2006, 12:14 PM
topsquark
Quote:

Originally Posted by CaptainBlack
Please award yourself the Math Help Forum Medal for clairvoyance, for
exceptional performance in interpreting strangely/obscurely phrased
questions :D

RonL

(Everybody: Don't tell CaptainBlack, but I knew he was going to say that! ;) )

-Dan
• May 3rd 2006, 01:20 PM
ThePerfectHacker
CaptainBlack, CaptainBlack! Did you hear' what topsquark-the fizizist said 'bout you! Do not tell him I said that...
• May 3rd 2006, 08:05 PM
CaptainBlack
Quote:

Originally Posted by topsquark
(Everybody: Don't tell CaptainBlack, but I knew he was going to say that! ;) )

-Dan

I suppose you are now expecting to be awarded Oak Leaves to your medal. :D

RonL

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

RonL
• May 4th 2006, 03:03 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I suppose you are now expecting to be awarded Oak Leaves to your medal. :D

RonL

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

RonL

(Snort!) I'll be more likely to get the Poison Oak Leaves Award...

-Dan