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indefinate integral

S

SoundTheSurrender

Diamond Member
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.
 
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.
Click to expand...

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!
 
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.
Click to expand...

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!
Click to expand...

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)



 
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.
Click to expand...

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.
 
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...


 
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.
Click to expand...

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!
Click to expand...

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)
Click to expand...


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 =


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

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