indefinate integral

[SoundTheSurrender]

So here is the problem...

/ (z)/ root (z+5) dz

it can be turned to

/ (z)(z+5)^(-1/2)



u = (z+5)

du = 1

dx = 1

Then I'm lost from here because I'm not sure what to do.

Thanks.

 

[ruffilb]

Sorry. My brain's out on vacation this summer.

I wish I could help

 

[state 08]

does " / " this mean INTEGRAL?

 

[SoundTheSurrender]

ya

 

[The Batt?sai]

Originally posted by: ruffilb
Sorry. My brain's out on vacation this summer.

I wish I could help

miine too

 

[Born2bwire]

\frac{2}{3}(z+5)^{1.5}-10\sqrt{z+5}

 

[SoundTheSurrender]

(2/3) * (z+5)^(3/2) - 10(z+5)^(-1/2)

thats nuts, i wasn't even close.

 

[thesurge]

Originally posted by: Born2bwire
\frac{2}{3}(z+5)^{1.5}-10\sqrt{z+5}

Latex much?

 

[Elderly Newt]

Originally posted by: djmihow
So here is the problem...

/ (z)/ root (z+5) dz

it can be turned to

/ (z)(z+5)^(-1/2)



u = (z+5)

du = 1

dx = 1

Then I'm lost from here because I'm not sure what to do.

Thanks.

ok... your dx=1 should be dz=1
then, you take your expression
/ (z)(z+5)^(-1/2) dz
and substitute your u and dz in and you get
/ (z)u^(-1/2)
now you have a z in there!
so go back to your u equation
u = z+5
and solve for z
z = u-5
substitute that for z and you get
/ (u-5)u^(-1/2)
and then... i think you distribute that.. and you get
/ u^(1/2) - 5u^(-1/2)
and now you can do basic intergration on that

hmm ok... i hope thats right.. somebody pleeeaase correct me if i did something wrong!

 

[aux]

Originally posted by: Elderly Newt

Originally posted by: djmihow
So here is the problem...

/ (z)/ root (z+5) dz

it can be turned to

/ (z)(z+5)^(-1/2)



u = (z+5)

du = 1

dx = 1

Then I'm lost from here because I'm not sure what to do.

Thanks.

ok... your dx=1 should be dz=1
then, you take your expression
/ (z)(z+5)^(-1/2) dz
and substitute your u and dz in and you get
/ (z)(u^-1/2)
now you have an a z in there!
so go back to your u equation
u = z+5
and solve for z
z = u-5
substitute that for z and you get
/ (u-5)(u^-1/2)
and then... i think you distribute that.. and you get
/ (u^1/2)-5u^-1/2
and now you can do basic intergration on that

hmm ok... i hope thats right.. somebody pleeeaase correct me if i did something wrong!

Looks about right, but the notation can be confusing for the OP.
u^-1/2 is better written as u^(-1/2) or, (the same) 1/sqrt(u), or (using OP's notation) 1/root(u)

 

[Eeezee]

indefinate = indefinite

 

[SoundTheSurrender]

Thank you so much, I was able to do it! I really appreciate this, this class is so hard and summer makes it even harder

 

[SoundTheSurrender]

I have this problem.

(1+x^2)/(x^2) Dx

so.. do I change it to

(1+x^2)(x^-2) ?

I don't know what to do with d and du... they don't form right.

 

[hypn0tik]

Originally posted by: djmihow
I have this problem.

(1+x^2)/(x^2) Dx

so.. do I change it to

(1+x^2)(x^-2) ?

I don't know what to do with d and du... they don't form right.

Split up the numerator.

(1+x^2)/x^2 = 1/x^2 + x^2/x^2 = 1/x^2 + 1

The rest should be straight forward.

Edit: I guess if it makes it easier to integrate, you can write it as x^-2 + 1. Either way, the result is the same.

 

[SoundTheSurrender]

Wow, I can't believe I missed that, thank you.

 

[SoundTheSurrender]

so it should be.

1x^(-2)+1dx

1 / x^-2 dx + 1 / dx / = represents the integrate sign

then

x^(-1)/1 +1x

1/x + 1x

but it should be 1/x - 1x.... I don't know how to get a negative?

according to my calculator, it should = 4.8 because it goes from 1 to 5...

 

[hypn0tik]

The integral of 1/x^2 + 1 dx = -1/x + x

You may be missing a negative sign in your original integral.

 

[MichaelD]

Originally posted by: aux

Originally posted by: Elderly Newt

Originally posted by: djmihow
So here is the problem...

/ (z)/ root (z+5) dz

it can be turned to

/ (z)(z+5)^(-1/2)



u = (z+5)

du = 1

dx = 1

Then I'm lost from here because I'm not sure what to do.

Thanks.

ok... your dx=1 should be dz=1
then, you take your expression
/ (z)(z+5)^(-1/2) dz
and substitute your u and dz in and you get
/ (z)(u^-1/2)
now you have an a z in there!
so go back to your u equation
u = z+5
and solve for z
z = u-5
substitute that for z and you get
/ (u-5)(u^-1/2)
and then... i think you distribute that.. and you get
/ (u^1/2)-5u^-1/2
and now you can do basic intergration on that

hmm ok... i hope thats right.. somebody pleeeaase correct me if i did something wrong!

Looks about right, but the notation can be confusing for the OP.
u^-1/2 is better written as u^(-1/2) or, (the same) 1/sqrt(u), or (using OP's notation) 1/root(u)

The two of you just killed what few remaining brain cells I had left. I know that people like you design the bridges I drive over every day...thank you for being so smart!

For the life of me; I'll never understand these....THINGS that you figure out with such ease.

You read the problem. You thought about it. You solved it.

To me, I see this:

Take the equation Y%w3/>^550.ee.
Add 1.56 kilograms of apples
Subtract one French Apple Pie of unknown weight
A dog of unknown size pees on it
A full moon shines ONLY 44% of it's available ultraviolet light on it
The square is 0)98(8*&6?@>>NULL
Enter your result here =

 

[Shimmishim]

woot woot calculus ftl!

 

[JohnCU]

we need mathematical symbols put into atot for homework help.

 

[6shiw1]

lol, calculus? I thought most of AT would have been familiar with integration as a sophomore in highschool.

 

[archcommus]

I just finished calc 3 last semester so after that my brain automatically purged all calc knowledge.

Sorry, can't help.

 

[xSauronx]

Originally posted by: 6shiw1
lol, calculus? I thought most of AT would have been familiar with integration as a sophomore in highschool.

heh, i didnt even get to finish my geometry classes. my math ability is pathetic.

 

[aux]

Originally posted by: JohnCU
we need mathematical symbols put into atot for homework help.

and "help with my homework" forum