Question

Integration by substitution.
File attached. Spot mistakes.

Answer 1

I believe i'm correct until 3rd last step..

Answer 2

Is it: \[\int\limits n \sqrt{1-n}dn \] or \[\int\limits \sqrt[n]{1-n}dn\]

Answer 3

First one.

Answer 4

Um, idk if i'm wrong, but the problem seems shorter to me thatn what you did. What I did was used u-sub to substitute 1-n and then expanded the integrand to get two integrals...

Answer 5

@abb0t : im new at this that's why i prefer doing step by step.
Wolfram is giving me this:
http://www.wolframalpha.com/input/?i=integrate+xsqrt(1-x)+dx%2C+1-xt%5E2

Answer 6

That's fine, I will help you out step by step :)

Answer 7

Ty. :)

Answer 8

First you used
u = 1-n
du = -dn
Now, notice that you have now
\[\int\limits n \sqrt{u}du\] which is the same as: \[\int\limits n (u)^\frac{ 1 }{ 2 }du\]
You need to now find something to use and substitute for n. Notice that you have, above:
u = 1-n
rearrange using algebra to solve for n to get
n = 1-u

Answer 9

Now, you can substitute that to get:
\[\int\limits (1-u)u^\frac{ 1 }{ 2 }du\]
Distribute to make it easier to integrate. And I'm sure you know how to integrate that.

Answer 10

http://i3.kym-cdn.com/entries/icons/original/000/009/993/tumblr_m0wb2xz9Yh1r08e3p.jpg

Answer 11

After integration we should get:
\[\Large \int\limits u^{0.5} - u^{1.5}\to \frac{u^{1.5}}{1.5}- \frac{u^{2.5}}{2.5}\]

Answer 12

Ummm...yeah, HAHA. That's correct. Weird seeing it in decimal form tho. Lol

Answer 13

Lol. I like playing with decimals.
What's next?

Answer 14

Well, you're pretty much done. Just plug in the original u-sub (u = 1-n) and add a constant since this is an indefinite integral. Always put a constant at the end or else on an exam, prepare to lose points.

Answer 15

Please elaborate this: Just plug in the original u-sub (u = 1-n)

I'm sorry. This method is new for me but it's cooler. ;D

Answer 16

Well, the point of using u-substitution is to find something to substitute to make it easier to integrate. But "u" was your substitute for 1-n. which was the original function. So your final answer should be a representation of your original function, since you know that an integral is just the antiderivative of a function. Therefore, you must use "n" because that was your original function, and u was not in the original function.
I hope that makes sense :)

Answer 17

How are we going to write it??
Then i'll understand.

Answer 18

wait a minute, did you integrate yet? Lol. You don't plug it in until you integrate your function.

Answer 19

I integrated above in the decimals?

Answer 20

First, integrate:
\[\int\limits u^\frac{ 3 }{ 2 } du - \int\limits \sqrt{u}du\]

Answer 21

Oh wait, yeah you did. haha.

Answer 22

Sorry, decimals throw me off. Lol. Give me a min to write it out for you

Answer 23

Sure sure.

Answer 24

\[\frac{ 2 }{ 5 }(1-n)^\frac{ 5 }{ 2 }-\frac{ 2 }{ 3 }(1-n)^\frac{ 3 }{ 2 }+C\]

Answer 25

Notice, that all I did was simply plug in (1-n) for every 'u'
Since your original function was composed of 'n' then your final answer should be in terms of n.

Answer 26

Got till here.
But isn't this different from http://www.wolframalpha.com/input/?i=integrate+xsqrt(1-x)+dx%2C+1-xt%5E2 ?

Answer 27

Um, i dont understand what 1-xt^2 means after the integral? I am assuming t is a constant?

Answer 28

in other words it's like:
1 - x = u^2

Answer 29

That would mean you have to square both sides to get u, or in ur case 't'. But taking the derivative of "t". You don't get anything that you can use to substitute.
\[u = \sqrt{1-x}\]
\[du = -\frac{ 1 }{ 2\sqrt{1-x} }dx\]
and that would make no sense to use as a substitute. Unless they want you to multiply by it's conjugates, but that method seems less practical than simple substitution..

Answer 30

The question is this:
\[\int\limits x \sqrt{1-x}. dx , 1-x=t^2\]

Answer 31

I think they gave you the substitution to use, and you have to manipulate it to get it to look like something in the function. Hence, you would take the square root.

Answer 32

Yes! That's it

Answer 33

Here is an example problem:

Answer 34

Well, looking at your work, it looks correct to me. It looks like you used conjugates. Which is fine also. I just used my mehtod often and it works easier for me. But your work looks right.

Answer 35

Then why is wolfram giving a different answer?

Answer 36

Wolfram always simplifies the answers so that may be one reason, or misreads question specific questions.

Answer 37

Umm.. Yeah i guess.
Thank you.