A Real Use For Taking an Integral

Finding the Area and the Volume of portions under curves.

Area under a SINGLE curve. 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 intervals2) Take an integral3) Solve using your intervals

Examples!! Y=x2 [0,1] x-axis

The Answer!

Working with the Ys

X=y2 [0,1]y-axis

The Answer

Area Under 2 CurvesTop – Bottom 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.

What are intervals?

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.

How to Find your intervals 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.

Area under 2 Curves

y=x2 y= [0,1]

Answer

Examples

Y=3x y=x3+2x2

Give the intervals and the solution

first quadrant.

Answer

Right – Left

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 Y-axis.

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.

Examples

X= y2-6y X=-Y [0,5]

Answer

Examples x=y2-5y y-axis Give your intervals, and solution

Put graph

Symmetric Functions

if you try and find the area of sine between 0-2 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.

Examples

y=x3 y=x x=1 x=-1

Examples

y=x2 y=-2x4

x=1 x=-1

Answer

Try me! Work

y=x2-4 y=-x2-2x Give the interval, and the solution.

Answer Work

Try me 2! Work

x=y2 x=y+2 Give the intervals, and the solution.

Answer Work

Try me 3! Work

Y=sinx [0,2]

Answer Work

Volume of Solids of Revolution 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.

Disk Method 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.

Disk Method 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 x-axis, 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.

Disk Method

Here is an example using the disk method around the x-axis:

π0∫4[f(x2)]2dx

Disk Method

Disk Method

Here is an example using the disk method around the y-axis:

π0∫4[f(√y)]2dx

Disk Method

Try Me! Work

y=1/(√x) x=1 x=4 about x-axis

Answer Work

Washer Method 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.

Washer Method

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 y-axis, you would have to solve for x.

Washer Method

Here is an example using the washer method about the x-axis:

π0∫4[(ex)2-(x)2]dx

Washer Method

Washer Method

Here is an example using the washer method about the y-axis:

π0∫1[(y)2-(√y)2]dx

Washer Method

Try Me! Work

π0∫4[(x3)2-(x)2] dx about x-axis

Answer Work

Shell Method 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.

Shell Method 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.

Shell Method

Here is an example of using the shell method about the y-axis:

2π0∫1[x(2x-x2)]dx

Shell Method

Shell Method

Here is an example using the shell method about the x-axis:

2π0∫3[y(√y)]dx

Shell Method

Try Me! Work

2π0∫4[x(x2-4x)]dx about the y-axis

Answer work

Biblyography

http://mathdemos.gcsu.edu/mathdemos/shellmethod/gallery/gallery.html

http://www.mathgv.com/ http://www.musopen.org/music.php