Area enclosed between y= -6x^2+10x and the line y= -8x+12

[apple2357]

Ok,

If i want to find the area enclosed between y = -6x^2+18x-12 and the x axis. I can integrate between x=1 and x=2 and answer equals 1.

If i now think about the area enclosed between y= -6x^2+10x and the line y= -8x+12, from the reasoning above algebraically this should be the same answer as above between x=1 and x=2. But looking at it graphically i have an area above and an area below the x axis , that doesn't seem to make sense to me?

Does this make sense to anyone??!!

 

[HallsofIvy]

For the second problem "above and below the x-axis" is irrelevant. All that matters is the relative positions of the parabola and the straight line.

 

[apple2357]

So if i am trying to work out the area below, i dont need to consider it in stages and thinking carefully about when it goes beneath the x axis?

 

[pka]

So if i am trying to work out the area below, i dont need to consider it in stages and thinking carefully about when it goes beneath the x axis?

The integral \(\displaystyle \int_1^2 { \left[ {( - 6{x^2} + 18x - 12) - ( - 8x + 12)} \right]dx} \) gives the shaded area.
Why is that enough?

 

[apple2357]

Does it? Because some of it is beneath the x axis and you have to consider signs and do it in two sections or so i thought?

 

[Steven G]

Does it? Because some of it is beneath the x axis and you have to consider signs and do it in two sections or so i thought?

The area between the two curves is what it is, that is, it does not depend on where the area is located.

Consider this: What if to both functions we add 100. This will make the area in question above the x-axis (if I am wrong then just add a bigger number than 100). Now when you compute the integral the +100s cancel out! This should make it clear that it does not matter if part of the area is above and below the x-axis. Is that clear??

 

[apple2357]

Ah, that makes sense! I hadn't thought about translating both functions.
Presumably you can do this with any enclosed area between two functions?

It is just because i saw this online and the suggested solution as below. Are you telling me this is un-necessary?

 

[Steven G]

Ah, that makes sense! I hadn't thought about translating both functions.
Presumably you can do this with any enclosed area between two functions?

It is just because i saw this online and the suggested solution as below. Are you telling me this is un-necessary?

You can translate, rotate etc any area as long as you do not change the size of the area. As I said before, the size of the area is what it is and the location does not matter. The fact that if you raise or lower the area does not have an effect on the area, since the added constants cancel out, should be enough for you to see that a vertical shift does not change the area. Now what about a horizontal shift? Does a horizontal shift change the size of the area? How about a rotation or a stretching?

 

[apple2357]

You can translate, rotate etc any area as long as you do not change the size of the area. As I said before, the size of the area is what it is and the location does not matter. The fact that if you raise or lower the area does not have an effect on the area, since the added constants cancel out, should be enough for you to see that a vertical shift does not change the area. Now what about a horizontal shift? Does a horizontal shift change the size of the area? How about a rotation or a stretching?

A horizontal shift isn't going to change but you will need to integrate between different limits ( dependent on the shift?) , though it is not particularly helpful as the area still falls beneath the x axis?

Stretching certainly will, for example the 'x' term could get replaced by 'kx' and that looks messy?

Not sure about rotation?? My hunch says area is preserved, but not sure why?

 

[pka]

Does it? Because some of it is beneath the x axis and you have to consider signs and do it in two sections or so i thought?

Yes it does! You said the answer is \(\displaystyle 1\). Well look HERE.

Surely in your calculus notes you have the upper curve minus lower curve method of finding the area between two curves?

 

[sinx]

So if i am trying to work out the area below, i dont need to consider it in stages and thinking carefully about when it goes beneath the x axis?

you only have to be careful to set up the area integration correctly;
if you don't set it up right, a negative area below the x-axis can subtract from the area above;
or/ you will get an answer that is 1/2 what it should be.
however, if you set up the integration correctly, with correct limits and signs, this will not be a problem.

example; say you mirror this particular parabola across the x axis,and you want the area between the two, from x=0 to 1.
y=-6x²+10x is the top.
y=6x²-10x is on the bottom.

*The area is not int [-6x²+10x] dx, with limits of x=0,1
it is A=int [(-6x²+10x)-(6x²-10x)]dx with limits of x=0,1
this integral gives the total area between the parabolas.

*In this integral, you are saying the lower limit is the x axis, i.e. y=0.
and you will (only) get the area that is between y=0 and the (upper) parabola

the best way to think of it is; An area is a double integration.
dA=dydx
and y (int dy) is the limits applied to int dy, which should be 2 equations in x. One eqn is the upper limit, the other is the lower limit.
so if you are only integrating one equation, you are saying the second equation is y=0, the x axis.