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Core2Student

Guest

Definite Integrals and Area

I'm stuck on this question, I've found out a few parts of it, but not sure how to do the last little bit.

I have the curve

and the normal

which intersects with the curve at the points (3,2) and (5,0).

What I need to find is the area between the curve and the normal.

The curve crosses the x-axis at the points (4,0) and (5,0), but this means there are parts below and above the x-axis that I need to find.

I have found that the area between the curve and the x-axis is 1/6.

And the area between x=3, x=5, the normal and the x-axis is 2.

However, how do you find the remaining area between the normal and the curve?

Thank you in advance.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Definite Integrals and Area

I am working on your problem,
and realized that if you "lift" or
"push" or "translate" both equations
upward by 2 y-units, then you
won't have the parts below the
x-axis anymore.  Everything
moves together, so the picture
looks identical.  This is how to
do it.  Subtract 2 (always think
backwards for translating, so
subtract where you want to go)
to y to move the graph up 2 units.
y - 2 = x^2 - 9x + 20
y = x^2 - 9x + 22
And the line becomes
y - 2 = 5 - x
y = 7 - x
And then you integrate if you've
learnt some calc and do both
definite integrals and subtract them,
after you make a graph and show
one is above the other on the range
of 3 to 5.

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igloo myrtilles fourmis

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Definite Integrals and Area

John + Core2Student, when computing the area between two curves, it matters not whether the curves themselves cross the x-axis, the only thing you have to worry about is that the area you are computing is such that one curve is always above the other one, if the intersect eachother within the area you are computing, then you have to split it into two integrals as you would if you were computing the area under a single curve that crossed the x-axis.

The trick, is rather than computing the areas of the two curves and the x-axis seperately, and then subtracting, you instead compute the area between the difference of the two curves themselves, and the x-axis, aka.

the fact that the value is negative simply means that the graph of y = 5-x is above the graph of y = x^2-9x+20 between x = 3 and x = 5

the area is then the absolute value of this:

Last edited by luca-deltodesco (2009-04-03 08:22:17)

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The End Of All Things To Come.

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Definite Integrals and Area

Yes, I get the same answer, 1  1/3.
Here's the 2-units upward on y way.
Lower line:
(1/3)x^3 - (9/2)x^2 + 22x
at 5, 125/3-

25*9/2+22*5=39.1666667
at 3, 9 - 40.5 + 66=75-

40.5=34.5 at x=5
def.int=39.167-

34.5=4.66667

Upper straight line:
7x - (1/2)x^2
at 5, 35 - 12.5, or 22.5
at 3, 21 - 4.5, or 16.5
def.int.=22.5-16.5=6

Area = 6 - 4.666667 = 1.33333

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igloo myrtilles fourmis

Core2Student

Guest

Re: Definite Integrals and Area

luca-deltodesco: Thank you for your answer, I think that that is a lot simpler way. However, we weren't taught to do it that way, so I will definitely try it with the questions in the exercise. ^_^ Thank you very much.

John E. Franklin: I've gone through your solution and it makes sense. Thank you very much as well. =P

Both of your assistance has been greatly appreciated.