hey guys i've a funny one here

[kingjewel1]

i'm trying to work out the area between

1/x^4-x^4 and x^2 and the x axis and the y axis

1/x^4-x^4 and x^2

 

can you give me a hand please?

i've integrated between 0 and 0.9 but it gives me a negative area. :S

what am i doing wrong?

 

eventually i wanted to work out the area between -x^2 and the quartic.

-x^2

 

aren't these curves beautiful

[JaKiri]

Which bit? The small, visably finite bits?

[JaKiri]

If it is, then integrate x^2 between 0 and the x value of the intercept with respect to x.

 

This will give you this area;

http://img140.exs.cx/my.php?loc=img140ℑ=grapha6sz.jpg

 

Then, integrate 1/x^4 - x^4 between 0 and the y value of the intercept with respect to y.

 

This will give you this area;

http://img133.exs.cx/img133/6342/graphba5su.jpg

 

Then, you need to subtract this rectangle;

http://img133.exs.cx/img133/5333/graphc2se.jpg

 

to give you this area;

http://img133.exs.cx/img133/5889/graphd7ds.jpg

 

You now have the 2 pieces of the right hand side, add them together and multiply by 2 (to get the LHS), and you're done.

[The Rebel]

i'm trying to work out the area between

1/x^4-x^4 and x^2 and the x axis and the y axis

1/x^4-x^4 and x^2

 

can you give me a hand please?

i've integrated between 0 and 0.9 but it gives me a negative area. :S

what am i doing wrong?

 

eventually i wanted to work out the area between -x^2 and the quartic.

-x^2

 

aren't these curves beautiful

 

I would split the area into two shapes resting against the x-axis, and breaking by a line parallel to the y-axis at the intersecting point, and add the totals.

[kingjewel1]

you see, i've been setting the functions equal to each other thus

1/x^4-x^4=x^2

and integrating this with respect to x between 0 and 0.9 which is it's positive intersecting point.

 

why can't i do this?

[JaKiri]

you see' date=' i've been setting the functions equal to each other thus

[b']1/x^4-x^4=x^2[/b]

and integrating this with respect to x between 0 and 0.9 which is it's positive intersecting point.

 

why can't i do this?

 

Becuase you can't do that in integration. Just look at the 1/x^4 - x^4 curve in isolation; is integrating that between 0 and 0.9 really what you want to do?

[kingjewel1]

i have examples where i've equalled two intersecting functions and then integrated the result... i'll see if i can dig some out. I was wanting to find the total area bounded by the quartic and the quadratic. I integrated between 0 and 0.9 because x=0.9 is where the two functions intersect in the positive quadrant. somebody mentioned integrating with respect to y but then i'm not sure how to do that. making y the subject from this guy or would i not just need to int(1/(1/x^4-x^4))

 

. y=1/x^4-x^4-x^2 how do i make y subject here?

excuse my ignorance

[The Rebel]

If it is' date=' then integrate x^2 between 0 and the x value of the intercept with respect to x.

 

This will give you this area;

http://img140.exs.cx/my.php?loc=img140ℑ=grapha6sz.jpg

 

[/quote']

 

Couldn't you then simply find the area bounded by the quartic and the x-axis between x=0.9 and x=1. To give the the are shown in:

 

http://img133.exs.cx/img133/5889/graphd7ds.jpg

 

Then add these two together.

[kingjewel1]

yeah sure you could do that. it's just that i wanted to do it all in one go by joining the functions. just like eg area between x^3 and x^2 in the positive quadrant X= 0 and 1 intx^3-x^2dx betweent 0 and 1 =x^4/4-x^3/3 sub in 1 = 1/12

Just the same as doing intx^3dx - intx^2dx between 0 and 1

 

why doesn't this method work for the other functions? this is buggin me

[The Rebel]

yeah sure you could do that. it's just that i wanted to do it all in one go by joining the functions. just like eg area between x^3 and x^2 in the positive quadrant X= 0 and 1 intx^3-x^2dx betweent 0 and 1 =x^4/4-x^3/3 sub in 1 = 1/12

Just the same as doing intx^3dx - intx^2dx between 0 and 1

 

why doesn't this method work for the other functions? this is buggin me

 

I very much doubt this is possible to put into a general formula. As functions become more complex' date=' there could produce any number of intersects between each function and the axis, in any four of the quandrants. Therefore generating any number of shapes to find the area of. The example of [math']x^{3}-x^{2}[/math] is easy enough because of its simple shape. Then take an example of circle [math]x^{2}+y^{2}=1[/math] placed on top of a high amplitude, high frequency sine curve. There could be many different areas formed, some "in the air" away from the axis within regions of areas that sit on the axis.

 

I think with any integral, the functions have to be visualised before the calculations can take place.

[kingjewel1]

I take your point "The Rebel". I actually have the functions drawn out infront of me to make sure they do intersect where i thought. Like this: functions <click>. I calculated the area in two stages in the end.

Thanks guys for all your help

[The Rebel]

I take your point "The Rebel". I actually have the functions drawn out infront of me to make sure they do intersect where i thought. Like this: functions <click>. I calculated the area in two stages in the end.

Thanks guys for all your help

 

There's a very good reason (I thought) for the name The Rebel, maybe one day I'll properly introduce myself on the "official introduce yourself" thread.