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Integration: Sines, Cosines and U-Substitution

Q:  ∫cos5(x)sin4(x)dx

A: Some books have tables and charts to memorize on how to integrate these types of problems: what to do if the power is odd on one but even on the other, etc (called reduction formulas)… Boring… Who has time, space or desire to memorize formulas? Let’s solve the damn problem.

First, I see that there are sines and cosines, and we know that one is the derivative of the other (more or less).  This tells me that u-substitution is likely going to come up.

Remember: whenever you see a function and its derivative present in a problem, you want to be thinking u-substitution!

I also know that by using the Pythagorean Identity, sin2(x)+cos2(x)=1, I can convert an “even number” of cosines to sines and vice versa.

So, now that I know u-substitution is most likely, I want to leave behind one function to be “du” and the rest should be converted to “u’s”.

Cosine is literally the “odd man out”.  There is an odd number of cosines, so I will leave one cosine behind to eventually serve as du and convert the rest like so:

∫cos5(x)sin4(x)dx

∫cos(x)*cos4(x)*sin4(x)dx

∫cos(x)*(cos2(x))2*sin4(x)dx

∫cos(x)*(1-sin2(x))2*sin4(x)dx

Now, I can do a fairly clean u-substitution:

Let u = sin(x)

Then, du = cos(x)dx

Continue reading Integration: Sines, Cosines and U-Substitution

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Integration: Disk (Washer Method) vs Shell Method

Here is an example to help you understand and visualize the difference between the disk method and the shell method.  I will do the same problem twice: once using disk method and once using shell method.  Keep in mind: not all problems are equally as easily solved with both methods — that’s why we have multiple methods!  Some problems may be easy to do with the shell method but nearly impossible with the disk method, or vice versa.

If you haven’t read the blog posts that discuss the basic differences between the disk method and the shell method, read those first:

To get the most out of this problem, grab a pen and paper and do the problem along with me: draw each picture and write each equation.

Q:  Find the volume obtained by rotating the area contained by y = √(x), y = 0, and x = 4 around the y-axis.

A:  First, draw a picture of the area that is to be rotated.

integration1

integration2

 

Visualize the rotation.  Actually give the shape a name to help solidify it in your mind.  I think of this shape as a cinder cone:

cindercone

Obviously this isn’t a perfect visual match, but giving a real image to an abstract shape really helps with mentally processing the steps.

Method 1:  Disk (washer) Method.  Remember, the disk and washer method are the same thing.  In the disk method, we visualize stacked circles (or pancakes, as I like to say).  The washer method is the same stacked pancakes with holes (like washers or donuts).

I like to complete every problem with the same thought process.  Follow these steps for all volume/integration problems and you will get the hang of it!

Step 1.  Determine if this is a dx or a dy problem.

Which way are we stacking our pancakes?

integration3

Pancakes are being stacked vertically, so this is a “dy problem”.  This means the limits of integration and the equations used will all be in terms of y.

?? ____?____  dy

Step 2.  Find the limits of integration.

Where does the pancake stacking start?  Where does it end?  Remember — in terms of y since this is a dy problem.

Pancakes start at y=0 and end at the place where y = √(x) and x = 4 intersect (the top of the cinder cone).  These bounds intersect at the point (4, 2).

So, pancake stacking starts at y=0 and end at y=2.

02 ____?____  dy

Step 3.  Find the equation of the areas that are being stacked.

We’ve pre-decided that we are going to use the disk (washer) method.  There are definitely washers happening, because there is a big outer circle minus a hole (to create the cinder cone).  So, we need to stack outer pancakes minus inner pancakes (big circles minus small circles):

Area = πR2 – πr2

Let’s start with the area of the big circles, which I’ve called πR2.  What is the radius, R, of the big circles?  Is it changing throughout the problem or is it constant?

integration3

The radius of the larger circles is constant.  The larger circles have a radius of 4 throughout the entire cinder cone:  R = 4

Now look at the smaller circles.  Is the radius, r, constant or changing?

integration4

Notice the radius of the smaller circles is changing.  This radius is a horizontal distance, starting from the y-axis and moving on out.  The radius is x: r = x.

BUT WAIT…. remember, this is a dy problem.  All equations need to be in terms of y!  So, using r = x will not help.  We need to find a way to represent x in terms of y.  Fortunately, we have the equation to help: y = √(x), so, x = y2

Good.  So, R = 4 and r = y2

Our equation is now complete and ready to solve:

02 π(4)2 – π(y2)2 dy

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425)

Now… Are we ready to solve this same problem using the shell method?

Method 2:  Shell method.  Instead of visualizing stacked pancakes to create our cinder cone, we will visualize stacked “Russian Dolls” (cylinders).  We are going to stack these cylinders so tightly together that the lateral area of the cylinders will stack to create volume.

integration5

Back to our steps:

Step 1.  Determine if this is a dx or a dy problem.

The dolls are being stacked inside of each other and on outward, expanding along the x-axis.  This is a “dx problem”.  This means the limits of integration and the equations used will all be in terms of x.

?? ____?____  dx

Step 2.  Find the limits of integration.

Where does the cylinder stacking start and where does it end?  In other words: what is the radius of our smallest cylinder and what is the radius of our largest cylinder?

The smallest cylinder has a radius of x=0 and the largest cylinder has a radius of x=4

04 ____?____  dx

Step 3.  Find the equation of the areas that are being stacked.

We are stacking lateral areas of a cylinder, which has equation: 2πrh (r is the radius of the cylinder, h is the height).

Let’s figure out the radius, r, of a random cylinder/doll in our shape.  The radius does change, so it is a variable.

integration6

But, it does not necessarily depend on the functions.  The radius is simply an x-value that continues to grow until we hit the wall of x = 4.

So, r = x.

Now, let’s figure out the height, h, of a random cylinder/doll in our shape.  The height is changing and it is definitely affected by the functions.  The height is the y-value of the bounding function: h = y…. BUT WAIT… remember, this is a dx problem — no y’s allowed. So, use the equation: y = √(x).  Therefore, h = √(x)

So, we have:

04  2π(x)(√(x)) dx

I will leave the solving to you… but as a final answer I get 128π/5 (approx = 80.425) — no coincidence that this is the same answer obtained by method 1!

Same shape, two methods, same answer.  Phew.

 

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When to use Disk Method versus Shell Method, Part 2

To start, read: When to use Disk Method versus Shell Method, Part 1 to get a general visual.  This is a little more detailed:

Volume should be thought of as infinitely stacked area. In the disk method, you are infinitely stacking circles (think pancakes). In the shell method, you are infinitely stacking lateral surface areas of cylinders (think “Russian Dolls” that stack inside of each other)

The Disk Method: Since you are stacking pancakes, the general formula that you will be integrating is π*r2. If the radius of each disk is changing throughout the shape, the radius, r, will be a function, dependent on either y or x, depending on how you are rotating.

disk_method

The Shell Method: Since you are stacking lateral areas of cylinders, the general formula that you will be integrating is 2*π*r*h (lateral area of cylinder formula). The radius, r, will be a simple function involving an x or a y. The height, h, will depend on the functions that are being rotated.

shell

Now that you have an understanding of the concept, view this example… I solve a volume problem using the disk method.  Then, I solve the exact same problem using the shell method:   Integration Example: Disk Method vs. Shell Method

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Integration and Marginal Cost

Q:  Suppose the marginal cost is:  MC = 12 + 8x with C(10) = 1020.  Find the fixed and total cost of producing 80 units.

A:

We gotta know that the marginal cost is derivative of the cost.  So, to find the cost function, we need to integrate the marginal cost function:

∫12 + 8x dx = 12x + 4x2 + C

So, the cost function, C(x) = 12x + 4x2 + C

And we know that C(10) = 1020.  So, we can plug in these values to solve for the constant C like so:

Continue reading Integration and Marginal Cost

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Integration Examples

Q:

 ∫14*(12 – 4x)(12x – 2x2)6 dx

Answer:

Since I see that (12 – 4x) is the derivative of (12x – 2x2), my instinct tells me to use u-substitution.

So, let

u = 12x – 2x2

then, compute du:

du = 12 – 4x dx

Now, we can directly substitute in u and du into our problem:

Continue reading Integration Examples

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When to use Disk Method versus Shell Method, Part 1

Q:  When should I use a disk / washer method versus a shell / cylinder method for integration?

Answer:

First, the visual difference: 

The disk / washer method is used when you can think of your shape as “stacked pancakes” (the washer method is just removing any center “holes” from these pancakes).

The shell / cylinder method is used when you can think of your shape as “stacked Russian Dolls” (you know, those dolls that stack inside of each other, and you keep opening them up to find a small doll inside, etc..)

A mathematical difference:

Example 1:  Rotate around the x-axis to create a volume.

If you rotate around the x-axis and use the disk method:  You will be stacking your pancakes horizontally (with respect to x).  Therefore, your limits and functions of integration will be in terms of “x” (it will be a “dx” problem).  The radius of your disk will need to be in terms of x.

disk_method

If you rotate around the x-axis and use the shell method:  You will be stacking your cylinders vertically.  Your limits and functions of integration will all be in terms of “y” (it will be a “dy” problem).  The height of your cylinder will need to be in terms of y.  The radius of your cylinder will most likely just be “y” itself – though not guaranteed.

shell_method

Example 2:  Say you are rotating an area around the y-axis to create a volume.

If you rotate around the y-axis and use the disk method:  You will be stacking your pancakes vertically (with respect to y).  Therefore, your limits and functions of integration will be in terms of “y” (it will be a “dy” problem).  The radius of your disk will need to be in terms of y.

If you rotate around the y-axis and use the shell method:  You will be stacking your cylinders horizontally.  Your limits and functions of integration will all be in terms of “x” (it will be a “dx” problem).  The height of your cylinder will need to be in terms of x.  The radius of your cylinder will most likely just be “x” itself – though not guaranteed.

Which method to pick?

Take into account the visual of your shape.  Does it look like pancakes stacked on top of each other or cylinders nested inside of each other?

Take into account the math.  Is it easier to put the functions in terms of y or in terms of x?  That will guide you on which to pick (depending on your axis of rotation).

This was a start to understanding.  Next read: When to use Disk Method versus Shell Method, Part 2.

Also check out this example…  I do one problem using the disk method.  Then, I do the same problem using the shell method:  Integration Example: Disk Method vs. Shell Method