Area between y = x^2 - 4x, y = x - 4, from x = 1 to x = 4

Blitze105

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Aug 28, 2008
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Hello here are the equations:

y = x^2 - 4x
y = x - 4

The bounds are [1,4]

I set this up like this:

(integral) [1,4] (x - 4) - (x^2 - 4x)

(integral) [1,4] (-x^2 + 5x - 4)

Am i correct so far?

------------

Also i have another problem:

f(x)= sin(x) g(x) = cos(x) (pi)/4 < x < 5(pi)/4

I need to solve for each intersecting point... but i forgot how I'd go about manipulating that.
 
Re: Area between 2 curves

The first one looks good.

The second one, just graph them both and look at the intersection points in the given range.
 
Re: Area between 2 curves

galactus said:
The first one looks good.

The second one, just graph them both and look at the intersection points in the given range.

Thank you very much.

I have one more problem if you could take a glance at it as well...

y = x^3 y = 0 x = 1 about the line y = 2

My problem is the "about y=2" we haven't covered this yet. I am guessing i would set it up as about the same but subtract 2 somehow before i integrate. i have read a little bit about it and the problem comes in because all of the problem examples i find are all in Y = something and this has an x=1.. not sure how i should do that.

Could some one show me this as well i would appreciate it greatly.
 

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Re: Area between 2 curves

You do not state as such, but I would assume that 'about the line y=2' refers to a solid of revolution.

Are you studying integration yet?. Because that is what it entails. That is, assuming I am interpreting the problem correctly.

We revolve the given functions about the line y=2 and get a solid we find the volume of. It is one of the coolest things about calculus.

Using washers, your volume would be given by \(\displaystyle {\pi}\int_{0}^{1}(x^{6}-4x^{3})dx\)

Using shells: \(\displaystyle 2{\pi}\int_{0}^{1}(y-2)(y^{\frac{1}{3}}-1)dy\)
 
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