Transcript A Real Use For Taking an Integral

Finding the Area and the Volume of portions under curves.
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Normally, to find the area under a single
curve, you could use Riemann's sums.
However our way is faster and less
aggravating.
There are 3 steps to finding the area under a
single curve.
1) Find your intervals
2) Take an integral
3) Solve using your intervals
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Y=x2
[0,1]
x-axis
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X=y2
[0,1]
y-axis
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To find the area under 2 curves, the steps are
similar as the ones used to find the area
under just a single curve.
In the method, you have to establish which
variable you are using.
If the functions you are using are in terms of X
then you subtract the function on the bottom
from the function on top.
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Your intervals are the numbers at the top and
bottom of the integral sign. They are the
numbers that you are limiting your function
to.
This is helpful for finding the area because
you need to have a finite portion of the graph
that you are trying to find.
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To find your intervals, you must simply set
the 2 of your functions equal to each other.
Then you simplify them so you have your
intervals.
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y=x2
y=
[0,1]
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Y=3x
y=x3+2x2
Give the intervals and the solution
first quadrant.
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In these functions, you will be in terms of Y.
Since you are in terms of Y, your intervals
change from being on the x-axis to on the Yaxis.
You figure out which function is further on
the left of the plain and subtract it from the
one further on the right
Right – Left.
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X= y2-6y
X=-Y
[0,5]
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x=y2-5y
y-axis
Give your intervals, and solution
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Put graph
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if you try and find the area of sine between 02 then you will end up getting zero.
That’s because the area that you get that is
above the x-axis, mathematically, will cancel
out the other.
This is why you take half the interval and then
multiply the final product by 2.
So go from 0- and multiply your solution by
2.
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y=x3
y=x
x=1 x=-1
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y=x2
x=1
y=-2x4
x=-1
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y=x2-4
y=-x2-2x
Give the interval, and the solution.
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x=y2
x=y+2
Give the intervals, and the solution.
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Y=sinx
[0,2]
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To find the volume of a solid of revolution,
you will need to know four things: the
function, what axis you are rotating the
function around, the interval which you are
finding the volume and what equation to use.
You use the disk method when the whole interval of
the function is touching the axis you are rotating it
around.
 The equation for the disk method is πb∫a[f(x)]2dx
 The function you are finding the volume for must
have its variables equaling the axis you are rotating
it around.
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Also, if the axis you are rotating the function around
is the x-axis, your function should be equal to y, and
if you are rotating the function around the y-axis,
your function must equal x.
 For example, if you were rotating x2 around the xaxis, it would appear as y=x2 and if it was around the
y-axis, you would have to solve for x.
 Once you plug the function into the equation, you
take its integral and solve at the interval.
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Here is an example using the disk method
around the x-axis:
π0∫4[f(x2)]2dx
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Here is an example using the disk method
around the y-axis:
π0∫4[f(√y)]2dx
y=1/(√x) x=1 x=4 about x-axis
The washer method is used when there are two
functions and there will be a hole in the revolution
of the solid.
 You always subtract the inside function from the
outside function.
 The equation for the washer method is πb∫a[(f(x))2(g(x))2]dx
 Depending on the axis of rotation, the function will
be equal to the opposite variable of the axis.
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For example, if you are rotating the volume about
the y-axis, the functions would both be equal to x,
and vice versa.
 If you were finding the volume y=x2 minus y=x3
about the x-axis, you wouldn’t have to change the
variables. But if you were rotating them about the yaxis, you would have to solve for x.
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Here is an example using the washer method
about the x-axis:
π0∫4[(ex)2-(x)2]dx
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Here is an example using the washer method
about the y-axis:
π0∫1[(y)2-(√y)2]dx
π0∫4[(x3)2-(x)2] dx about x-axis
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The shell method is used when you are solving for the
volume of one function is going to touch the axis of
revolution at one point or when the function would be a
disk method around the opposite axis.
The equation for the shell method is
2πb∫a[radius*height]dx
The radius of the function is how wide it is and its height
is the function. Usually the radius is x or y, depending on
what you set the function equal to.
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If the function was y=√x, the radius of the
function is x and the height is √x.
When using the shell method, the function must
be equal to the opposite axis you are revolving
the function around.
For example, if your function was y=√x, you
would use the shell method if you were revolving
it around the x-axis, otherwise you would use
the disk method.
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Here is an example of using the shell method
about the y-axis:
2π0∫1[x(2x-x2)]dx
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Here is an example using the shell method
about the x-axis:
2π0∫3[y(√y)]dx
2π0∫4[x(x2-4x)]dx about the y-axis
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