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Calc II Help

M

manlymatt83

Lifer
Hi All,

Ok, here's my latest problem. And I'm sick as a dog... but still gotta do calc!

f(x) = x^2 - 4x + 3
g(x) = -x^2 + 2x + 3

It says "setup the definite integral that gives the area of the region"

In the graph, g(x) is higher than f(x), so I suppose I want to do g(x) - f(x)

therefore:

int from 0 to 3 of (-x^2 + 2x +3 - x^2 -4x + 3)

After I combine terms, I get:

int from 0 to 3 of ( (-2x^2 - 2x + 6) dx )

The book gets:

int from 0 to 3 of ( (-2x^2+6x) dx )

So the only discrepency is my -2x + 6 they claim is only +6x...

any ideas?
 
... my head hurts just looking at all that. 🙁

Just remember you will never use that unless you become a rocket scientist and even then who wants to become one? 😛
 
They already found the antiderivative, now all you need to do is apply the second Fundamental Law of calculus and find the difference between F(1) and F(0) where F(x) = [x^3/3 + x^2/2 + x ] BAR 1 - 0
 
Originally posted by: mjuszczak
I'm working on a problem where the example goes from:

The integral from 0 - 1 of: [ ( x^2 + 2 ) - ( - x ) ] dx

to

= [ x^3/3 + x^2/2 + x ] BAR 1 - 0


What are they doing here? The integral sign is dissapearing and all of a sudden a bar appears at the end from 1 to 0... are they getting the anti derivatives? Combining? What are they doing? Its the only step I don't understand.
Click to expand...

The BAR means that you plug in 1 to what is in the parentheses and then you plug in 0 to what is in the parantheses and then you subtract them.
 
the bar means evaluate from 1 to 0. that means first plug in 1 to the entire expression, then plug in 0 and subtract it from what you got when you plugged in 1.
 
I just don't understand what THEY are doing.

How do they go from the first step to the second as I showed in the OP? Did they fill 1 and 0 in?
 
Originally posted by: mjuszczak
I just don't understand what THEY are doing.

How do they go from the first step to the second as I showed in the OP? Did they fill 1 and 0 in?
Click to expand...

all they did was the integral of the expression x^2+x+2 which comes out to be x^3/3+x^2/2+2x

your answer is wrong...it should be 2x, not just plain x. the integral of 2dx is 2x.
 
Originally posted by: chuckywang
This is basic calculus.
Click to expand...
That was helpful.

And why did you completely ignore the proof I sent to you about the vector function (x,y,z) being equal to vector function (?,0,0) in spherical coordinates?
 
Originally posted by: mjuszczak
I'm working on a problem where the example goes from:

The integral from 0 - 1 of: [ ( x^2 + 2 ) - ( - x ) ] dx

to

= [ x^3/3 + x^2/2 + x ] BAR 1 - 0


What are they doing here? The integral sign is dissapearing and all of a sudden a bar appears at the end from 1 to 0... are they getting the anti derivatives? Combining? What are they doing? Its the only step I don't understand.
Click to expand...

x^3/3 + x^2/2 + 2x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + 2x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 2*1) - (0^3/3 + 0^2/2 + 2*0) and you will get your answer.

-Tom

EDIT: Didn't bother checking the integration, but Bigsm00th is right, should be 2x instead of x.
 
Originally posted by: Random Variable
Originally posted by: chuckywang
This is basic calculus.
Click to expand...
That was hekpful.

And why did you completely ignore the proof I sent to you about the vector function (x,y,z) being equal to vector function (?,0,0) in spherical coordinates?
Click to expand...

send that to me
 
The answer is wrong. it's x^3/3 + x^2/2 + 2x


Instead of integrating, look at the answer and find the derivative, may be that will be easier for you to learn.
 
Originally posted by: Bigsm00th
Originally posted by: mjuszczak
I just don't understand what THEY are doing.

How do they go from the first step to the second as I showed in the OP? Did they fill 1 and 0 in?
Click to expand...

all they did was the integral of the expression x^2+x+2 which comes out to be x^3/3+x^2/2+2x

your answer is wrong...it should be 2x, not just plain x. the integral of 2dx is 2x.
Click to expand...

Does the integral = the antiderivate?
 
Originally posted by: Soccer55
Originally posted by: mjuszczak
I'm working on a problem where the example goes from:

The integral from 0 - 1 of: [ ( x^2 + 2 ) - ( - x ) ] dx

to

= [ x^3/3 + x^2/2 + x ] BAR 1 - 0


What are they doing here? The integral sign is dissapearing and all of a sudden a bar appears at the end from 1 to 0... are they getting the anti derivatives? Combining? What are they doing? Its the only step I don't understand.
Click to expand...

x^3/3 + x^2/2 + x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 1) - (0^3/3 + 0^2/2 + 0) and you will get your answer.

-Tom
Click to expand...

actually no it isnt.

d/dx(x^3/3) = x^2

d/dx(x^2/2) = x

d/dx(x) = 1

x^2+x+1 is not the original expression. the answer he has listed is wrong.

edit: i see you caught the error soccer55.
 
Originally posted by: Random Variable
Originally posted by: Bigsm00th
Originally posted by: Random Variable
Originally posted by: chuckywang
This is basic calculus.
Click to expand...
That was hekpful.

And why did you completely ignore the proof I sent to you about the vector function (x,y,z) being equal to vector function (?,0,0) in spherical coordinates?
Click to expand...

send that to me
Click to expand...

Why?
Click to expand...

because i am curious. who pissed in your cornflakes?
 
Originally posted by: Bigsm00th
Originally posted by: Soccer55
Originally posted by: mjuszczak
I'm working on a problem where the example goes from:

The integral from 0 - 1 of: [ ( x^2 + 2 ) - ( - x ) ] dx

to

= [ x^3/3 + x^2/2 + x ] BAR 1 - 0


What are they doing here? The integral sign is dissapearing and all of a sudden a bar appears at the end from 1 to 0... are they getting the anti derivatives? Combining? What are they doing? Its the only step I don't understand.
Click to expand...

x^3/3 + x^2/2 + x is the anti-derivative of ( x^2 + 2 ) - ( - x ) with respect to x. Since you're doing a definite integral (the 0 to 1 thing), you have to evaluate your integral at both 0 and 1 and subtract them. That is what "[ x^3/3 + x^2/2 + x ] BAR 1 - 0" means. So you need to do (1^3/3 + 1^2/2 + 1) - (0^3/3 + 0^2/2 + 0) and you will get your answer.

-Tom
Click to expand...

actually no it isnt.

d/dx(x^3/3) = x^2

d/dx(x^2/2) = x

d/dx(x) = 1

x^2+x+1 is not the original expression. the answer he has listed is wrong.

edit: i see you caught the error soccer55.
Click to expand...

Heh, yeah.....probably should have checked the work in the OP before responding, but I was being lazy. 😛

-Tom
 
Originally posted by: mjuszczak
Originally posted by: Bigsm00th
Originally posted by: mjuszczak
I just don't understand what THEY are doing.

How do they go from the first step to the second as I showed in the OP? Did they fill 1 and 0 in?
Click to expand...

all they did was the integral of the expression x^2+x+2 which comes out to be x^3/3+x^2/2+2x

your answer is wrong...it should be 2x, not just plain x. the integral of 2dx is 2x.
Click to expand...

Does the integral = the antiderivate?
Click to expand...

not quite. an antiderivative is actually an indefinite integral, where an integral has defined limits of integration. they are close enough that people will understand what you mean unless they are being a tight-ass on purpose. dont use them interchangeably though if you are trying to be exactly correct.
 
Originally posted by: Bigsm00th
Originally posted by: Random Variable
Originally posted by: Bigsm00th
Originally posted by: Random Variable
Originally posted by: chuckywang
This is basic calculus.
Click to expand...
That was hekpful.

And why did you completely ignore the proof I sent to you about the vector function (x,y,z) being equal to vector function (?,0,0) in spherical coordinates?
Click to expand...

send that to me
Click to expand...

Why?
Click to expand...

because i am curious. who pissed in your cornflakes?
Click to expand...

A monkey.

The magnitude of the vector in the ?-direction = x*sin?cosf + y*sin?sinf + z*cos?
= ?sin?cosfsin?cosf + ?sin?sinfsin?sinf + ?cos?cos?
= ?sin²?(cos²f + sin²f) + ?cos²?
= ?sin²? + pcos²? = ?

The magnitude of the vector in the ?-direction = x*cos?cosf + y*cos?sinf -z*sin?
=?sin?cosfcos?cosf + ?sin?sinfcos?sinf - ?cos?sin?
=psin?cos?(cos²f + sin²f) - pcos?sin? = 0

The magnitude of the vector in the f-direction = -x*sinf + y*cosf + z*0
= -?sin?cosfsinf + ?sin?sinfcosf = 0

Therefore vector function (x,y,z) in cartesian coordinates is the same as vector function (?,0,0) in spherical coordinates.
 
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