Antiderivative formula?

five40

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Oct 4, 2004
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Ok, for some reason calc came to my head and I remembered how to do derivatives but not antiderivatives. I mean simple ones like x squared=1/3xcubed I can just do in my head but I can't remember how to do hard ones. I know there is some relatively simple formula to get the harder ones but I can't remember it for the life of me. Oh yeah this isn't homework. I've been out of school for more than a couple years now.
 

Gibson486

Lifer
Aug 9, 2000
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hard ones like the ones with sin and cos? You need a good trig background to do that stuff in your head.
 

hypn0tik

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Jul 5, 2005
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Depends. Some of the harder ones can be done in different ways such as substutition (including trignometric substitution) or integration by parts.
 

Fenixgoon

Lifer
Jun 30, 2003
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i just have them memorized, so it's all based on what function they throw at you
 

five40

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Oct 4, 2004
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Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.
 

hypn0tik

Diamond Member
Jul 5, 2005
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Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.

Are you sh!tting me? Simplify and integrate that biatch.

Edit: Typo.
 

five40

Golden Member
Oct 4, 2004
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Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.

Are you sh!tting me? Simplify and integrate that biatch.

Edit: Typo.

Like I said...I haven't done this in a long while. How do I integrate it damn it.
 

sao123

Lifer
May 27, 2002
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Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x
 

hypn0tik

Diamond Member
Jul 5, 2005
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Originally posted by: five40
Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.

Are you sh!tting me? Simplify and integrate that biatch.

Edit: Typo.

Like I said...I haven't done this in a long while. How do I integrate it damn it.

Common factor out another x in the numerator to get (x^3(x^6+4))/x^3 = x^6 + 4.

The integral is trivial.
 

five40

Golden Member
Oct 4, 2004
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Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )
 

five40

Golden Member
Oct 4, 2004
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Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.

Are you sh!tting me? Simplify and integrate that biatch.

Edit: Typo.

Like I said...I haven't done this in a long while. How do I integrate it damn it.

Common factor out another x in the numerator to get (x^3(x^6+4))/x^3 = x^6 + 4.

The integral is trivial.

Yeah my bad. Once I actually did the math I saw it was yet another easy one.
 

Fenixgoon

Lifer
Jun 30, 2003
33,285
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Originally posted by: five40
Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )

same deal.. expand, then integrate
 

hypn0tik

Diamond Member
Jul 5, 2005
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Originally posted by: five40
Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )

That simplifies to x^4/(x^2 + 8)

You need to bust out a Trig. substitution.
 

five40

Golden Member
Oct 4, 2004
1,875
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Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )

That simplifies to x^4/(x^2 + 8)

You need to bust out a Trig. substitution.

I believe that's the part I don't remember. Integration by substitution.
 

hypn0tik

Diamond Member
Jul 5, 2005
5,866
2
0
Originally posted by: five40
Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )

That simplifies to x^4/(x^2 + 8)

You need to bust out a Trig. substitution.

I believe that's the part I don't remember. Integration by substitution.

Ah, in this case draw a right angled triangle. Label one of the sides as x, the other as sqrt(8). Then hypotenuse then becomes sqrt(x^2 + 8). If you set up the triangle the way I did, you'll get:

cos(t) = sqrt(8)/sqrt(x^2 + 8) --> 1/8 * cos^2 (t) = 1/(x^2 + 8)
tan(t) = x/sqrt(8) --> 64 tan^4 (t) = x^4

Also, tan(t) = x/sqrt(8) --> 1/cos^2 (t) dt = 1/sqrt(8) dx

Isolate for dx and substitute in. You end up getting an ugly Trig. integral.
 

five40

Golden Member
Oct 4, 2004
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Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: hypn0tik
Originally posted by: five40
Originally posted by: sao123
Originally posted by: five40
Originally posted by: Fenixgoon
i just have them memorized, so it's all based on what function they throw at you

Yeah but what do you have memorized. I can't remember it. Like (x^2(x^7+4x))/x^3. I wouldn't know where to start.


1)Simplify X^6+4
2)integrate the sum (X^7)/7 + 4x

Yeah that one was way to easy. I just typed one off of my head. I looked simplifying and it became very easy. Ok....x^5/(x^3+8x )

That simplifies to x^4/(x^2 + 8)

You need to bust out a Trig. substitution.

I believe that's the part I don't remember. Integration by substitution.

Ah, in this case draw a right angled triangle. Label one of the sides as x, the other as sqrt(8). Then hypotenuse then becomes sqrt(x^2 + 8). If you set up the triangle the way I did, you'll get:

cos(t) = sqrt(8)/sqrt(x^2 + 8) --> 1/8 * cos^2 (t) = 1/(x^2 + 8)
tan(t) = x/sqrt(8) --> 64 tan^4 (t) = x^4

Also, tan(t) = x/sqrt(8) --> 1/cos^2 (t) dt = 1/sqrt(8) dx

Isolate for dx and substitute in. You end up getting an ugly Trig. integral.

Thanks. That's all I needed and it triggered everything back. Now I'll be able to sleep without my mind racing.