# U - Substitution for integrals

• June 13th 2008, 09:58 PM
JonathanEyoon
U - Substitution for integrals
lower boundary = 0

upper boundary = 4

x / (sqrt 1 + 2x)

I've let u = 1 + 2x

du = xdx

new boundaries = 1 to 9

u^-1/2 du. If I integrate it, it will 2u^1/2

2(9)^1/2 - 2(1)^1/2 = 4

This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem (Wondering). The answer should be 10/3
• June 13th 2008, 10:03 PM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
lower boundary = 0

upper boundary = 4

x / (sqrt 1 + 2x)

I've let u = 1 + 2x

du = xdx

new boundaries = 1 to 9

u^-1/2 du. If I integrate it, it will 2u^1/2

2(9)^1/2 - 2(1)^1/2 = 4

This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem (Wondering). The answer should be 10/3

$u=1+2x \to du=2dx$ it does not equal xdx
• June 13th 2008, 10:05 PM
Mathstud28
Quote:

Originally Posted by JonathanEyoon
lower boundary = 0

upper boundary = 4

x / (sqrt 1 + 2x)

I've let u = 1 + 2x

du = xdx

new boundaries = 1 to 9

u^-1/2 du. If I integrate it, it will 2u^1/2

2(9)^1/2 - 2(1)^1/2 = 4

This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem (Wondering). The answer should be 10/3

$\int_0^{4}\frac{\varphi}{1+2\varphi}d\varphi$

Let $\lambda=\sqrt{1+2\varphi}\Rightarrow\varphi=\frac{ \lambda^2-1}{2}$

So $d\varphi=\lambda{d\lambda}$

And seeing that at $\lambda(0)=\sqrt{1+2(0)}=1$

and $\lambda(4)=\sqrt{1+2(4)}=3$

We have

$\frac{1}{2}\int_1^{3}\frac{(\lambda^2-1)\lambda}{\lambda}d\lambda=\frac{1}{2}\bigg[\frac{\lambda^3}{3}-\lambda\bigg]\bigg|_{1}^{3}=\frac{10}{3}$
• June 13th 2008, 10:08 PM
JonathanEyoon
Quote:

Originally Posted by TheEmptySet
$u=1+2x \to du=2dx$ it does not equal xdx

(Sleepy) forgive me, i've been studying all day and it's late now. I should have definitely caught that. I'll rework the problem and try again.
• June 13th 2008, 10:12 PM
TheEmptySet
This is a fun different sub try

$t^2=1+2x \implies 2tdt=2dx \iff tdt=dx$

Solving the first for x we get $t^2=1+2x \iff x=\frac{t^2-1}{2}$

$\int_{0}^{4}\frac{x}{\sqrt{1+2x}}dx=\int_{1}^{3}\f rac{\frac{t^2-1}{2}}{\sqrt{t^2}}tdt=\frac{1}{2}\int_{1}^{3}t^2-1dt=\frac{1}{2}\left( \frac{1}{3}t^3-t\right)\bigg|_{1}^{3}=\frac{10}{3}$
• June 13th 2008, 10:13 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
$\int_0^{4}\frac{\varphi}{1+2\varphi}d\varphi$

Let $\lambda=\sqrt{1+2\varphi}\Rightarrow\varphi=\frac{ \lambda^2-1}{2}$

So $d\varphi=\lambda{d\lambda}$

And seeing that at $\lambda(0)=\sqrt{1+2(0)}=1$

and $\lambda(4)=\sqrt{1+2(4)}=3$

We have

$\frac{1}{2}\int_1^{3}\frac{(\lambda^2-1)\lambda}{\lambda}d\lambda=\frac{1}{2}\bigg[\frac{\lambda^3}{3}-\lambda\bigg]\bigg|_{1}^{3}=\frac{10}{3}$

Quote:

Originally Posted by TheEmptySet
This is a fun different sub try

$t^2=1+2x \implies 2tdt=2dx \iff tdt=dx$

Solving the first for x we get $t^2=1+2x \iff x=\frac{t^2-1}{2}$

$\int_{0}^{4}\frac{x}{\sqrt{1+2x}}dx=\int_{1}^{3}\f rac{\frac{t^2-1}{2}}{\sqrt{t^2}}tdt=\frac{1}{2}\int_{1}^{3}t^2-1dt=\frac{1}{2}\left( \frac{1}{3}t^3-t\right)\bigg|_{1}^{3}=\frac{10}{3}$

I think I have seen this method done before somewhere...hmm(Rofl)
• June 13th 2008, 10:14 PM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
(Sleepy) forgive me, i've been studying all day and it's late now. I should have definitely caught that. I'll rework the problem and try again.

I know you were on this morning before I went to work and you are still on now that I'm home. (Crying)
• June 13th 2008, 10:18 PM
JonathanEyoon
question :

Could I just substitute how I did originally?

u = 1 + 2x

du / 2 = dx

As for the last substitution for the x in the numerator, can I just let x = (u - 1) / 2?
• June 13th 2008, 10:19 PM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
question :

Could I just substitute how I did originally?

u = 1 + 2x

du / 2 = dx

As for the last substitution for the x in the numerator, can I just let x = (u - 1) / 2?

Yes that is what you need to do!

Nice(Clapping)
• June 13th 2008, 10:20 PM
Chris L T521
Quote:

Originally Posted by JonathanEyoon
lower boundary = 0

upper boundary = 4

x / (sqrt 1 + 2x)

I've let u = 1 + 2x

du = xdx

new boundaries = 1 to 9

u^-1/2 du. If I integrate it, it will 2u^1/2

2(9)^1/2 - 2(1)^1/2 = 4

This is what I keep getting but apparently it's wrong.....I'm not understanding how else to do this problem (Wondering). The answer should be 10/3

Let $\zeta=1+2x\implies x=\frac{1}{2}(\zeta-1)$

Thus, $\,d\zeta=2\,dx\implies\,dx=\frac{1}{2}\,d\zeta$

Limits change as well.. $\zeta(4)=9, \ \zeta{0}=1$

Thus, our integral is

$\frac{1}{4}\int_1^9 \frac{\zeta-1}{\sqrt{\zeta}}\,d\zeta=\frac{1}{4}\int_1^9 \bigg[\sqrt{\zeta}-(\zeta)^{-\frac{1}{2}}\bigg]$

$=\frac{1}{4}\left.\bigg[\frac{2}{3}\zeta^{\frac{3}{2}}-2\sqrt{\zeta}\bigg]\right|_1^9=\frac{1}{4}\bigg(12+\frac{4}{3}\bigg)= \color{red}\boxed{\frac{10}{3}}$
• June 13th 2008, 10:21 PM
JonathanEyoon
Yippee~ So the new boundaries will still remain the same as what I originally had above right?

New lower = 1 and New upper = 9?
• June 13th 2008, 10:22 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
Let $\zeta=1+2x\implies x=\frac{1}{2}(\zeta-1)$

Thus, $\,d\zeta=2\,dx\implies\,dx=\frac{1}{2}\,d\zeta$

Limits change as well.. $\zeta(4)=9, \ \zeta{0}=1$

Thus, our integral is

$\frac{1}{4}\int_1^9 \frac{\zeta-1}{\sqrt{\zeta}}\,d\zeta=\frac{1}{4}\int_1^9 \bigg[\sqrt{\zeta}-(\zeta)^{-\frac{1}{2}}\bigg]$

$=\frac{1}{4}\left.\bigg[\frac{2}{3}\zeta^{\frac{3}{2}}-2\sqrt{\zeta}\bigg]\right|_1^9=\frac{1}{4}\bigg(12+\frac{4}{3}\bigg)= \color{red}\boxed{\frac{10}{3}}$

I say we reform! No more using x or y as a variable. Viva la $\varphi$!
• June 13th 2008, 10:23 PM
TheEmptySet
Quote:

Originally Posted by JonathanEyoon
Yippee~ So the new boundaries will still remain the same as what I originally had above right?

New lower = 1 and New upper = 9?

Yep...
• June 13th 2008, 10:25 PM
Chris L T521
Quote:

Originally Posted by Mathstud28
I say we reform! No more using x or y as a variable. Viva la $\varphi$!

Haha. I'm gonna use this one from now on. Viva la $\varrho$! :D