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AR15.COM
12/16/2015 1:20:41 PM EDT
12/16/2015 1:22:13 PM EDT
[#1]
= 87

So easy
12/16/2015 1:22:18 PM EDT
[#2]
I know you move the 4 outside and then use substitution. I always get lost on square roots. Rewriting to the 1/2 doesn't really help me either.
12/16/2015 1:23:10 PM EDT
[#3]
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Quoted:
= 87
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fuck it was so simple why didn't I think of that. If I can't figure it out I am writing 87 down and calling it a day
12/16/2015 1:23:34 PM EDT
[#4]
Potato(e).
12/16/2015 1:24:11 PM EDT
[#5]
It's been a while, but I think "integration by parts" is what you seek
12/16/2015 1:25:08 PM EDT
[#6]
32/3
12/16/2015 1:26:36 PM EDT
[#7]
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Quoted:
It's been a while, but I think "integration by parts" is what you seek
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finding definite integral between 2 x values using anti derivatives is what I am doing. My book is traded and my professor doesn't give a fuck about actually helping anyone. That tenure thing....
12/16/2015 1:26:54 PM EDT
[#8]
Am I the only who gets pissed at bs like sqrt(4-x^2)? That trolling got fucking old in Algebra. Yet they still pull that shit in Calculus. You're not fooling anyone.
12/16/2015 1:27:36 PM EDT
[#9]
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Quoted:
32/3
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hmm I came up with this but - and you can't have a - area so I figured I fucked up.
12/16/2015 1:27:58 PM EDT
[#10]
12/16/2015 1:29:52 PM EDT
[#11]
My teacher asked me, "Who helps you with your math homework?" and I said "Arfcom." she said "Who on earth is Arfcom?"

So I swept the leg.....
12/16/2015 1:30:14 PM EDT
[#12]
Quote History
Quoted:


finding definite integral between 2 x values using anti derivatives is what I am doing. My book is traded and my professor doesn't give a fuck about actually helping anyone. That tenure thing....
View Quote View All Quotes
View All Quotes
Quote History
Quoted:
Quoted:
It's been a while, but I think "integration by parts" is what you seek


finding definite integral between 2 x values using anti derivatives is what I am doing. My book is traded and my professor doesn't give a fuck about actually helping anyone. That tenure thing....


That's the correct path.
12/16/2015 1:31:32 PM EDT
[#13]
Quote History
Quoted:
Am I the only who gets pissed at bs like sqrt(4-x^2)? That trolling got fucking old in Algebra. Yet they still pull that shit in Calculus. You're not fooling anyone.
View Quote


Business major?


Use Paul's Notes:  http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx .  That's about the limit of my sympathy for a student without a text book.

12/16/2015 1:32:26 PM EDT
[#14]
It's called integration by substitution.

Set u = 4 - x^2
So:  du = -2x

Multiply the original first term by -2 * -1/2 (i.e. "multiply by 1") to get S -2 * -4/2 * x * (4 - x^2)^1/2 dx

Sub in your u and du terms to get S -4/2 * u^1/2 du which I know you know how to solve.

Integrate, replace your u term with the x variable, then calculate the value of the formula.


Do your own homework with the information above.
12/16/2015 1:34:11 PM EDT
[#15]
You could just use mathematica...

If it didn't cost thousands of dollars to use
12/16/2015 1:34:34 PM EDT
[#16]
Quote History
Quoted:
It's called integration by substitution.

Set u = 4 - x^2
So:  du = -2x

Multiply the original first term by -2 * -1/2 (i.e. "multiply by 1") to get S -2 * -4/2 * x * (4 - x^2)^1/2 dx

Sub in your u and du terms to get S -4/2 * u^1/2 du which I know you know how to solve.

Integrate, replace your u term with the x variable, then calculate the value of the formula.


Do your own homework with the information above.
View Quote


got it. I see where I fucked up. I raised to a -1/2 power instead of a positive 1/2.


ARFCOM fixing the worlds problems one at a time
12/16/2015 1:36:04 PM EDT
[#17]
You have to integrate
12/16/2015 1:37:27 PM EDT
[#18]
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Quoted:
ARFCOM fixing the worlds problems one at a time
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12/16/2015 1:38:14 PM EDT
[#19]
288
12/16/2015 1:51:47 PM EDT
[#20]

Is this for Cal 1, 2, or 3?

Edit: Jesus, I just went through all my Cal 2 notes and now I want to stab myself...
12/16/2015 1:56:23 PM EDT
[#21]
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Quoted:
288
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I believe u are looking for the large UHFO thread with that number.
12/16/2015 1:59:50 PM EDT
[#22]




S20 4xv4-x2 = dx



 
12/16/2015 2:08:43 PM EDT
[#23]
pull out the "4", you already know that, then you have 4 integral x sqrt (4-x^2) dx

now do u substitution

u= 4-x^2
du= -2x dx
dx= -(1/2x) du

that should get you to inputing the boundary limits of 0 and 2.
12/16/2015 2:09:54 PM EDT
[#24]
I rather do a CQ CQ CQ DX...
12/16/2015 2:10:07 PM EDT
[#25]
I'm not trying to ruin your edumacation but just use Photomath.

12/16/2015 2:14:03 PM EDT
[#26]
I got nuffin.
If I helped you out, You wouldn't learn.