6.4 Exponential Growth and Decay
6.5 Population Growth (slope fields)
6.6 Numerical Methods
Chapter 7
7.1 Integral as Net Change
7.2 Areas in the Plane
7.3 Volumes
7.5 Applicatoins from Science and Statistics
Chapter 8
8.1 L' Hopital's Rule
8.3 Improper Integrals
8.4 Parital Fractions and Integral Tables
6.3 Integration by Parts
What is it?
With derivatives when there are two functions multiplied together, a special proceedure must be followed (if you don't remember go back to the product rule for derivatives d/dx (uv) = u dv/dx + v du/dx). The same is true for integrals. Using the formula, even an idiot can integrate two functions multiplied together, although there are some tricks in some instances.
The formula
With two functions u and v:
°u dv = uv - °v du
How it works:
When you have two functions multiplied together (ie. 2x2 ex), you can let u equal one variable and v equal the other variable. Then use the formula which yields an answer which is more easy to solve, since you now are integrating v and du.
For example
°2x ex
Let:
u = 2x
du = 2
dv =ex
v = ex
°u dv = uv - °v du
°2x ex = 2x ex - °2ex
now, the integral is easy to solve.
°2x ex = 2x ex - 2ex + C
So what about definite Integrals (finding the area within a given range)
To find the area of a similar equation with two functions (2x2 ex), then you would solve it the same way, then plug in the limits to find the exact area.
°from 2 to 0 2x ex
See the above integration by parts.
°2x ex = 2x ex - 2ex = [2ex (x - 1) ]2 to 0
°2x ex = [2e2 (2 -1)] - [2e0 (0-1)]
°2x ex = = 2e2 + 2
Tablular Integration
What is it?
Often times, with equations like x4ex, you must repeat the use of the formula many times. A good way to handle this is to use tablular integration.
Example
°x2exdx
There are two functions x2 and ex, we can call them f(x) and g(x). The formula for integration by parts is °u dv = uv - °v du, so the point is to take the integral of one, and the derivative of the other. We can make a table to keep track.
Skip the first row for f(x) and the sign and put g(x) in the first row. Then put f(x) in the second row, and take the derivative of the row above it starting with the third row until you get to 0. Then put a plus sign in the second row of the sign column and alternate plus and minus for as many rows as possible. Then take the integral of g(x) in the row above for each row in the g(x) column.
| sign | f (x) derivatives | g(x) integrals | ||
| skip this space | skip this space | ex | ||
| + | x2 | °ex = ex -----> | ex | |
| - | d/dx x2 = 2x --> | 2x | °ex = ex -----> | ex |
| + | d/dx 2x = 2 ---> | 2 | °ex = ex -----> | ex |
| - | d/dx 2 = 0 ---> | 0 | °ex = ex -----> | ex |
Now, to use the table, we just combine each row.
°x2exdx = +x2 ex - 2xex + 2 ex- 0 ex = x2 ex - 2xex + 2 ex + C
6.4 Exponential Growth and Decay
What is it?
Exponential Growth or decay concerns two properties, an initial value, and a rate at which a quantity (for instance reproducing bacteria or savings with interest). Obviously, exponential growth or decay involves an increase or decrease in quantity in an exponential fashion, as opposed to a linear change. In general, this means that the increase is relatively small at first, then over time, the change is greater and greater in magnitude.
The formula: y = yo ekt
The Law of Exponential Change says:
If y changes at a rate proportional to the amount present (dy/dt = ky) and y= yo when t = 0, then
y = yo ekt
where k > 0 represents the growth and k < 0 represents decay.
The number k is the rate constant of the equation.
What the Law of Exponential Change really means
The quantity of things (bacteria, money, etc.) at any given time is called y. dy/dt is the change in the quantity over time. yo is the quantity of things in the initial conditions. For example, if you start out with 1000 dollars, yo = 1000. The second property is the rate of change, which of course is exponential. If k is negative, then the quanity of things (y) will decrease over time (when t is bigger). If k is positive, then the quanity of things (y) will increase over time. While y (the quantity of items) and t (time) are variables, k (the rate of change) and yo (the initial quanitity of things) are constants which do not change.
Real life uses of Exponential change and the formulas
What's the deal with population growth?
Population growth is different because the number of individuals in a populations is a discontinuous function of time. The reason is because there is no such thing as half a person, people are always whole, and therefore the number of people is always a positive whole number. Thus, slope fields are applied to this type of growth.
What is a slope field?
Slope fields are a graphical representation of the change in some quantity at any time. Each whole number on the x coordinate and y coordinate has a line which is slanted to represent the growth. A positive slant (/) indicates an increase, and a negative slant (\) indicates a decrease at that time. The degree of slant indicates the degree of change. The more horizontal the line, the less the rate of change.
6.6 Numerical Methods
Euler's Method
What is it?
Euler's method uses linearization to approximate. Remember that the formula for linearization is:
L(x) = y(xo) + y' (xo)(x-xo)
or
L(x) = y0 + f (xo,yo) (x-xo)
Of course, an approximatin does not yield the exact value, but L(x) is pretty close to y(x), which is the actual value, if the interval is small.
Linearization is used in Euler's Method by stringing together a set of linearizations to approximate an entire curve over a longer area, since linearization is only useful over short areas.
Thus, we know the initial coordinates of xo and yo, we use linearization to find a new y1. Since we know dx, we can also find x1. Then we can repeat the process as a string of linearizations, now using the new x1 and y1 variables.
Example:
given the following information, we will do three approximations using Euler's method:
y' = 1 + y
y(0) = 1
starting at xo = 0 with dx = 0.1
From the given information we know that
xo = 0
y(0) = yo = 1
We can find all the x values:
x1 = xo + dx
dx = .1
x1 = 0 + .1 = .1
x2 = x1 + .1 = .2
x3 = x0 + 3dx = .3
To find the y values:
y1 = y0 + f (xo,yo) (x-xo)
= yo + (1 +yo) dx
= 1 + (1 + 1)(.1) = 1.2
y2 = y1 + f (x1,y1) (x-x1)
= y1 + (1 +y1) dx
= 1.2 + (1 + 1.2)(.1) = 1.42
y3 = y2 + f (x2,y2) (x-x2)
= y2 + (1 +y2) dx
= 1.42 + (1 + 1.42)(.1) = 1.662
This is a numerical way of finding an approximate value of y at a given value of x knowing only the slope formula.
Improved Euler's Method
How do we improve Euler's Method?
The way to improve Euler's Method is to take the average of two slopes. We first estimate yn as in the original Euler method, but denote it zn. Then take the average of f (xn-1, yn-1) and f (xn, zn) in the next step.
zn = yn-1 + f (xn-1, yn-1) dx,
yn = yn-1 + [f(xn-1, yn-1) + f(xn, zn) / 2] dx
Chapter 7
In this chapter, we will learn how to apply integrals to solve certain problems.
7.1 Integral as Net Change
Some applications in Physics are:
1) Finding the displacement knowing the velocity (change in distance over time): Take the integral of the velocity formula.
2) Finding the total distance traveled : take the integral of the absolute value of the velocity.
3) Consumption over time - growth decay and food consumption over time
4) Work - force over time
7.2 Areas in the Plane
In this section, we will learn how to find the area between two functions. It is best learned by thinking of the funcitons graphically. We are trying to find the area between their curves. However, since the curves are rounded and not straight, we need to use integration to find the area between curves, instead of some simple geometric formula. This technique can be important when calculating some value involving two functions of rate of change.
How do we do it?
If f and g are continuous with f (x) ¦ g(x) throughout [a,b], then the area between the curves y = f(x) and y = g(x) from a to b is the integral of [f-g] from a to b,
A = ° from a to b [f(x) - g(x)] dx
What it really means?
All this means that if you have two continuous functions, to find the area between the two curves you just subtract the curve underneath from the curve above. Of course, this takes some knowledge of the curves to find out which one is on top and which is on the bottom and where they might intersect and change position.
Partitioning
Sometimes, the curves may require different sections, since in different areas, different curves may be on the top, or the bottom.
For example, to find he area in the first quadrant that is bounded above by y = x1/2 and the line y = x - 2, you would need two regions, from 0 to 2 and 2 to 4 on the x axis.
The reason is because from 0 to 2, the bottom curve is the line y = 0 and from 2 to 4, the bottom curve is y = x - 2.
When finding the area between two curves, the curves on the top and bottom must remain the same between any region.
Integrating with Respect to y
Why is it?
Sometimes, it might be easier or necessary to find the area between two curves by integrating. Take for instance two horizontal lines, or curves which fail the vertical line test.
How do you do it?
It is the same proceedure as integrating with respect to x, however, the curve functions must be put in terms of y.
For instance, y = x + 4 in terms of y is x = y - 4. You would then integrate y - 4 with boundaries coresponding to the y axis.
7.3 Volumes
Everything we learned in 7.2 is all well and good for curves on flat two dimentional graphs, but in real life, physical objects are in three dimentions. We need to find the area in a volume. Fortunately, we can do this with integrals, but unfortunately, sometimes it can be very difficult.
How do we do it?
The volume of a solid of known integrable corss sectino area A(x) from x = a to x = b is the integral of A from a to b,
V = ° from a to b of A(x) dx.
What it really means: Since already know how to integrate in two dimentions, we can think of a three dimentional volume as two dimentions times another dimention. Thus we slice the volume into a set of two dimentional sheets, which together form the entire volume. Think of cutting a carrot into very small thing slices. If you could cut the slices so it has no thickness, only height and width, then each slice would be one cross section. If you added up the area of each cross section, you would get the entire volume. You find the area of a cross section, by integrating it as a two dimentional plate. In this case, each slice of carrot makes up the whole carrot stick.
General method
1) Sketch the solid and a typical cross section: this may not seem necessary, but visualization is the most important step. You can not begin without knowing what the shape is and what formulas to use to derive the cross sections.
2) Find the formula for one cross section. The volume of a cube is x3, but a cross section is x2, a two dimentional square.
3) Find the limits of integration. This means we have to make sure that the cross section has the same formula for any region. A three dimentional volume may be made up of several different two dimentional cross secions and we can seperate them into regions and add the regions up to find the total volume.
4) Finally, integrate A(x) to find the volume.
The Other Method: Cylindrical Shells
When the first method, which can be refered to as the "discs" method, will not work, we can use the cylindrical shells method.
What is it?
Another way to find the volume of a 3 dimentional shape is with the cylindrical shells method. Basically, it is like it sounds. We try to see how many cylinders exactly can fit inside the shape. Then we find out the volume of each cylinder and add them up. It is most useful when the axis of revolution is perpendicular to the axis containing the natural interval of integration. In other words, the shape should seem round as if we could fit circular cans inside of it.
How do we do it?
1) Sketch the shape and draw a line segment in the region that is parallel to the axis of revolution. This is the height at any given point, of a cylinder which fits inside the shape. Now, the distance from this line to the axis of revolution is the radius. Imagine if you filled the entire region with more parallel lines, it would be all the cylinders that fit in the volume.
2) Identify the imits of integration.
3) Integrate to find the volume
Example:
The region bounded by y = x1/2, the x-axis, and the line x = 4 is revolved around the x-axis to generate a solid. Find the volume.
So first we would sketch the diagram. We realize that the integration must be in respect to y, since the thickness of a shell is dy. Thus, the height is derived from 4 - x1/2 . x1/2 = y so x = y2. Thus, in terms of y, the height is 4 - y2. The radius is the distance from any point y to the radius, which is the line y = 0. So the radius is y - 0 = 0. Thus the general formula for one cylinder is 2 pi (height)(radius). In this case, it is 2 pi (4 -y2) (y). The integral is from 0 to 2, since x = 4 intersects y = x 1/2 at y = 2. Thus:
V = ° from 0 to 2 of 2 pi (4 -y2) (y) dy = 8y.
7.5 Applications from Science and Statistics
From the beginning, we have attempted to use these integration techniques to solve real life problems. Here we can see how they can be applied to work, fluid force and fluid pressure and normal probablities (statistics).
Work
As you can remember, work is the force times displacement. We can solve some more complicated work problems with our new skills, such as cases when the amount of force changes over time.
Example:
If a leaky bucket weighs 22 N when it is empty. It is lifted from the ground at a constant rate to a point 20 m above the ground by a rope weighing .4 N (m). The bucket starts with 70 N of water but is leaks at a constnt rate until all the water is gone when it reaches the top. Find the work done.
As you can see, the leaky bucket causes a lot of problems for us, since there is different amount of water and thus it takes less force when the water is gone.
First we do the force to lift the bucket, since the bucket always weighs the same.
Work = force times distance = 22 N x 20 m = 440 J
Next we do the force to lift the water, which is much harder since the water leaks. As the bucket is raised, there is less water to carry, less force is necessary and thus the work done is affected.
Basically, we need a formula to find the amount of water in the bucket at any given distance from the ground. We can set up the following proportion, with the variable x as the distance from the ground.
At any given time, the proportion of water will be (20 - x) / 20. So, when the bucket is zero meters from the ground, (20 - 0) / 20 = 1 or 100 percent. When it is 10 meters from the ground, it is (20 - 10) / 20 which is 1/2 or 50 percent. So now, we need to find the formula for the force, which is just the original weight, times the proportion, since as water leaks out, the force needed to pull the water up is whatever percentage of the original weight that is left. The original weight is 70 N
Thus we have: F (x) = 70 (20-x) / 20, which in simplified terms:
= 70 (1 - x/20)
= 70 - x 70/20
= 70 - 3.5 x
We are trying to find work, which is the integral of F (x) which we take from 0 to 20, since x represents the distance.
W = ° from 0 to 20 of (70 - 3.5x) dx = [70x - 1.75 x2 ] evaluated from 0 to 20
W = [70 (20) - 1.75 (202) ] - [0 -0] = 1400 - 700 = 700 J
Finally, we need to find the amount of work necessary to raise the rope. At first, you may think that the rope is like the bucket, but rather it is like the water. As you raise the rope, there is less rope to raise. For example, after you have raised the rope 10 meters, there is only 10 meters of rope left to raise. The rope is 20 meters long so the amount of rope left to lift is always 20 - x, where x is the distance of the bucket from the ground (or the amount of rope already raised).
The force in terms of the distance from the ground (x) is equal to weight of rope times proportion of the rope left. The original weight of the rope is .4 N. Thus:
F (x) = 0.4 (20 -x ) dx
Work is the integral of the force in the givin interval of distance (x).
W = ° from 0 to 20 of 0.4 (20 - x) dx
W = [ 8x - .2 x2 ] evaluated from 0 to 20 = (8(20) - .2(202)) - (0-0) = 160 - 80 = 80 J
Now just add all the work together from the bucket, the water and the rope to find the total work:
440 + 700 + 80 = 1220 J
Fluid Force and Fluid Pressure
Another example of how changes in force affect the overall calculation in Physics is the principle of fluid force and fluid pressure. Think of a dam which holds water back. It must be stronger at the bottom than at the top, because there is more force exerted on the bottom. The reason is because the water above exerts an additional amount of pressure.
fluid pressure is force per unit area and is equal to the wieght times the height of the liquid:
p = wh
Thus, to solve fluid force and fluid pressure problems, we use the same technique as with work. We use this formula to find the pressure depending on the conditions given. Then we integrate.
Normal Probabilities
Many principles in statistics involve the bell curve, or the normal curve. It gives the probability under average conditions. Using principles of integration, finding the area under a probability curve will give you then percentage which coorelates to the inteval selected. So for example, for one standard deviation on a normal curve, the area under the curve yields .65, or 60 percent. The total area under the curve is always equal to 1. Thus, the integral of any interval will yield a fraction between 0 and 1.
Chapter 8
This chapter deals with infinite limits and more advanced concepts based on what we have already learned, but now adding in the concept of infinity.
8.1 L'Hopital's Rule
What is it?
L'Hopital (pronounced LO PITAL)'s rule is an example of SOME OTHER METHOD. If you remember back when we were trying to solve limits, there were three options. Under certain circumstances, we can not just substitute in the limit which x approaches in to the equation. L'Hopitals is used when both f(x) and g(x) are zero at x = a, where a is the limit which x approaches.
This is a problem because
lim f(x)
x -> a g(x) will give you 0/0 when you substitue in for f(x) and g(x). Thus we have L'Hopitals to help us in this situation.
How do you do it?
The first theorem
Suppose that f(a) = g(a) = 0, that f'(a) and g'(a) exist and that g'(a) g 0. Then the formula says that
lim f(x) = f ' (a)
x->a g(x) g' (a)
So for example:
If we want to solve
lim (1 + x)1/2 - 1
x -> 0 x
Then by substitution we would get 0/0. Note that this method is only possible when by substituting we get 0/0.
So we use L'Hopitals Rule and take the derivative of the top and divide by the derivative of the bottom.
d/dx (1 + x)1/2 - 1 = 1 / 2 (1 + x)1/2
d/dx (x) = 1
So the derivative of the top, over the derivative of the bottom is
1 / 2 (1 + x)1/2
1
Now we can substitute in and get 1 / 2 (1 + 0)1/2 = 1 / 2 (1) = 1/2
1 1
The Second Variation of L'Hopital's Rule
Suppose that f (a) = g(a) = 0, the same conditions as before, and that f and g are differentialble on an open interval I containing a, and that g ' (x) g 0 on I if x g a. Then:
lim f (x) = lim f ' (x)
x -> a g(x) x -> a g' (x)
What it really means:
It is the same idea, but instead, the condition is that in the open interval called I in which a is a part of, g ' (x) is not zero anywhere on I except at a, which is the limit.
Indeterminate Forms
What are they?
Indeterminate forms are quotents such as º/ º, º times zero and º times º. Like 0/0, L'Hoptials can also be used in these situations.
8.3 Improper Integrals
Ok, so you know how to find the area under the curve in a definate area. That is easy, so we want to give ourselves a bit of a challenge and find the area under a curve with an infinate area.
Take for example the curve y = e -x/2 in the first quadrant. It may seem like the area under the curve is infinate but it is not. To solve this problem, we would take the integral of the function from 0 to x = b
A (b) = ° from 0 to b of e -x/2 dx = [-2e-x/2] evaluated from 0 to b. Which equals
-2 e -b/2 - (-2 e -0/2) = - 2 e -b/2 + 2 because e0 = 1
The next step in the proceedure is to take the limit of the integral, A (b).
lim A (b) = lim (-2e -b/2 + 2) = 2
b -> º b-> º
Because the limit as e approaches negative infinity is zero, and 0 + 2 = 2.
Thus the area under the curve is 2.
So what is an improper integral?
An improper integral is an integral with infinite limit of integration. So,
Here is how to solve them under three conditions:
1) if f(x) is continuous on [a,º), then
° from º to a of f (x) dx = lim ° from a to b f(x) dx
b -> º
2) if f(x) is continuous on (-º ,b], then
° from b to -º of f (x) dx = lim ° from a to b f(x) dx
a -> - º
3) if f(x) is continuous on (-º , º), then
° from -º to º) of f (x) dx = ° from c to - º of f(x) dx + ° from º to c of f(x) dx, where c is any real number.
Converging and diverging
In short, the limit is finite if the improper integral converges. If there is no limit, then the improper integral diverges, in other words, the area is not finite.
Improper Integrals with Infinate Discontinuties
What is it?
It sounds very complicated, but really just a classification of one type of impropert integral.
Integrals of functions that become infinite at a point within the inteerval of intergration are also improper integrals.
Here is how to solve them under three conditions:
1) if f(x) is continuous on (a,b], then
° from a to b of f (x) dx = lim ° from c to b f(x) dx
c -> a+
2) if f(x) is continuous on [a ,b), then
° from a to b f f (x) dx = lim ° from c to a f(x) dx
c -> b-
3) if f(x) is continuous on [a, c) U (c , b) then
° from a to b of f (x) dx = ° from c to a of f(x) dx + ° from c to b of f(x) dx.
Here are the ways to test for convergence and divergence:
Direct comparison test
Let f and g be continuous on [a,º) with 0 — f(x) — g(x) for all x ¦ a. Then
1) ° from a to º of f (x) dx converges if ° from a to º of g (x) dx converges
2) ° from a to º of g (x) dx diverges if ° from a to º of f (x) dx diverges
Limit Comparison Test
If the positive functions of f and g are continuous on [a, º) and if
lim f(x) = L , 0 < L < º,
x-> º g(x)
then
° from a to º of f (x) dx and ° from a to º of g (x) dx both converge or diverge
8.4 Partial Fractions and Integral Tables
Partial Fractions- what are they?
A partial fraction is when you have two fractions added
together. Any rational functoin can be writen as the sum of basic fractions,
called partial fractions. The reason it is important is because then we can take
a function, seperate it into two fractions, and integrate each of the fractions,
which may be easier.
For example, to take the integral of a complicated fraction like
° (5x -3) / (x2 -2x -3) dx
To create partial fractions, we can factor the denominator.
x2 -2x -3 = (x+1)(x-3)
So we set it up so that
(5x -3) = A + B
(x2 -2x -3) (x+1) (x-3)
Obiously, we have to find out what A and B are.
So, we take
5x -3 = A(x-3) + B(x+1)
= (A + B)x - 3A + B
Then we can form two linear equations from these:
A + B = 5
-3A + B = -3
Then we do a simple substitution and find that A = 2 and B = 3
Now we can substitute back in and solve
°(5x -3) = ° 2 +° 3
(x2 -2x -3) (x+1) (x-3)
Which is = 2 ln | x + 1 | + 3 ln | x - 3 | + C
Integral Tables
What are they?
Integral tables are used to help solve partial fractions. Partial fractions that are decompositions of a rational function can be integrated using analytic methods. The integral tables (found in Appendix A7 of Finney, Demana, Waits and Kennedy's Calculus - Graphical, Numerical, Algebraic book) lists the solutions to solving most integrals.
For example:
° [1 / (x2 + 1)] + [1 / (x2 - 2x + 5)] dx
We can seperate it into two integrals.
° 1 / (x2 + 1) dx + ° 1 / (x2 - 2x + 5) dx
If you look in the tables, it tells you that
°dx / (a2+ x2) = 1/a tan -1 x/a + c
So if you set a = 1 then a2 = 1
and you get that °1 / 1 + x2 = 1/1 tan -1 x/1 + c = tan -1 x + c
For the xsecond part we can complete the square
x2 - 2x + 5 = x2 - 2x + 1 + 4 = (x - 1)2 + 4
So therefore using the same formula as above, we get
°1 / (x2 - 2x + 5) = °1/ (x - 1)2 + 4
we use u substitution where u = x -1, du = dx
= °du/ (u2 + 4)
remember that we are using
= °dx / (a2+ x2) = 1/a tan -1 x/a + c
So a = 2 because 22 = 4,
1/2 tan -1 u/2 + c
Substituting back for u we get:
1/2 tan -1(x-1)/2 + c
So combined we get
tan -1 x + 1/2 tan -1(x-1)/2 + c