Friday, June 10, 2011
Wednesday, June 8, 2011
integration by parts
1. introductory to integration by parts (wikipedia definition)
Integration by parts is a rule that transforms the integral of products of functions into other simpler integrals. The rule can be derived in one line by simply integrating the product rule of differentiation.
If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx, then integration by parts states that
When to use it:
- When the integrand is the product of two functions and substitution does not work
- If f(x) is continuous and differentiable and if g(x) is continuous
Integration by parts derived:
We know that the equation used in integration by parts is derived from the product rule as all integrands that can be solved using this are composed of two separate functions...
(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)
integrating both sides:
f(x)g(x)=f(x)g'(x) dx+f'(x)g(x) dx
suppose there is a given interval from a to b:
abf(x)g'(x) dx=f(x)g(x)ab-abf'(x)g(x) dx
with common notation:
f(x)g(x)ab=f(b)g(b)-f(a)g(a)
using the product rule and the fundamental theorem of calculus the rule is shown to be true:
f(b)g(b)-f(a)g(a)=abddx(f(x)g(x)) dx
=abf'(x)g(x) dx+abf(x)g'(x) dx
rule stated in indefinite integral form:
f(x)g'(x) dx=f(x)g(x)-f'(x)g(x) dx
2. how integration by parts helps/quickens integrating
Integration by parts is a rule that transforms the integral of products of functions into other simpler integrals. The rule can be derived in one line by simply integrating the product rule of differentiation.
If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx, then integration by parts states that
When to use it:
- When the integrand is the product of two functions and substitution does not work
- If f(x) is continuous and differentiable and if g(x) is continuous
Integration by parts derived:
We know that the equation used in integration by parts is derived from the product rule as all integrands that can be solved using this are composed of two separate functions...
(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)
integrating both sides:
f(x)g(x)=f(x)g'(x) dx+f'(x)g(x) dx
suppose there is a given interval from a to b:
abf(x)g'(x) dx=f(x)g(x)ab-abf'(x)g(x) dx
with common notation:
f(x)g(x)ab=f(b)g(b)-f(a)g(a)
using the product rule and the fundamental theorem of calculus the rule is shown to be true:
f(b)g(b)-f(a)g(a)=abddx(f(x)g(x)) dx
=abf'(x)g(x) dx+abf(x)g'(x) dx
rule stated in indefinite integral form:
f(x)g'(x) dx=f(x)g(x)-f'(x)g(x) dx
2. how integration by parts helps/quickens integrating
This strategy separates a very complicated integral into two individual integrals which will become easier to evaluate in relation to the initial function.
3. steps/guidelines to integrating by parts
a. begin with the part of the function which has the easiest and most simplified derivative (preferably 1) and label it “u”
b. label the derivative of u “du”
c. let the remainder of the original function be “dv”
d. integrate dv to find “v”
e. sub values into
f. integrate u and v
g. use algebra to solve
4. give easy example using guidelines
xe2x dx=?
step a: u=x
step b: du=1dx
step c: dv=e2x
step d: e2x dx=12e2x
step e: xe2x dx=x(12e2x)-12e2x dx
step f:
xe2x dx=x(12e2x)-12e2x dx
=x2e2x-12e2x dx
=x2e2x-1212e2x+C
=x2e2x-14e2x+C
5. give tedious example using guidelines
excos(x) dx
step a: u=cos(x)
step b: du=-sin(x) dx
step c: dv=ex dx
step d: ex dx =ex
step e: excos(x) dx=excos(x)+exsin(x) dx
before step f we must integrate by parts again...
step a: u=sin(x)
step b: du=cos(x) dx
step c: dv=ex dx
step d: exdx=ex
step e: exsin(x) dx=exsin(x)-excos(x) dx
step f: exsin(x) dx=exsin(x)-excos(x) dx
excos(x) dx=excos(x)+exsin(x)-excos(x) dx
2excos(x) dx=ex(sin(x)+cos(x))+C
excos(x) dx=ex(sin(x)+cos(x))2+C
to further help your understanding visit:
8. turn on v1210 and call it good!
Force and Work
We started out by defining force to the class. We did this by talking about Newton’s 3 laws.
I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
III. For every action there is an equal and opposite reaction.
The first law was what we used at first by using the equation of the sum of the forces all equaling zero. We used this to show how to set up the equations as well as solving for any variable such as a specific type of force.
We went on to explain the different types of forces.
Force of gravity – This is the product of the mass of an object multiplied by the gravitational constant of 9.81 m/s^2. This is true for any object close to the Earth’s surface. This force is always directed directly down.
Normal Force – This is the force put on an object from the surface that it is on. It is always perpendicular to the surface. It can be calculated by using Newton’s 2nd law, the equation of the sum of the forces equals’ mass times acceleration.
Force of friction – This is the force that is caused by the restriction of an object from moving across a surface. It can calculated by the equation of normal force times mew. Mew is the coefficient of friction between the object and surface. It ranges from 0 to 1, higher being more restrictive.
The force of tension – This is a force given by a rope pulling on an object or an object hanging up a string or something that would have tension. This can be calculated again my Newton’s laws.
When solving for these forces using Newton’s laws we need to set it up correctly than it is simple algebra processes; at least in the beginning of learning physics. What we need to know is that the forces in the x direction and y direction are completely independent of each other. Therefore, we need to use the equation twice, once for each direction if it is needed to solve a problem. Also, we need to know that when we set on the sum of the forces on the left side of the equation we need to separate the forces directed in the upward and downward direction; or facing the right or to the left direction. This is why making a force diagram help; which means to draw all of the forces about a dot to show what directions each of them are facing.
All of these forces are measured in Newton’s
Next we went on to work. We explained that most problems can be answered by doing either forces or energy. We first described the types of energies and their equations.
Kinetic Energy – 1/2mv^2 ß this is used for an object that is in motion; m is mass and v is velocity
Gravitational Potential Energy – mgh ß This is for an object that is hanging by something, or just has the potential to fall due to the force of gravity, m is mass, g is 9.81, and h is the height of the object.
Thermal energy – Fx ß This is used to calculate something such as the work done by the force of friction. F is the force and x is delta x, so how far it travels.
Elastic Energy – 1/2kx^2 ß This is used for an object working against a spring. K is the spring constant for an ideal spring, and x again is how far the object travels while against the spring.
In solving a problem with energy it is a simple process at first. You need to select two different places in the movement of an object. Then you decide energies are present at those 2 spots. You set them equal to each other and then fill in the variables that you know and use algebra to solve for whichever variable you’re looking for.
We next went over practice problems and took the quiz
Here are links that help
http://www.physicsclassroom.com/class/newtlaws/u2l2a.cfm
http://theory.uwinnipeg.ca/physics/force/index.html
http://phun.physics.virginia.edu/topics/energy.html
I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
II. The relationship between an object's mass m, its acceleration a, and the applied force F is F = ma. Acceleration and force are vectors (as indicated by their symbols being displayed in slant bold font); in this law the direction of the force vector is the same as the direction of the acceleration vector.
III. For every action there is an equal and opposite reaction.
The first law was what we used at first by using the equation of the sum of the forces all equaling zero. We used this to show how to set up the equations as well as solving for any variable such as a specific type of force.
We went on to explain the different types of forces.
Force of gravity – This is the product of the mass of an object multiplied by the gravitational constant of 9.81 m/s^2. This is true for any object close to the Earth’s surface. This force is always directed directly down.
Normal Force – This is the force put on an object from the surface that it is on. It is always perpendicular to the surface. It can be calculated by using Newton’s 2nd law, the equation of the sum of the forces equals’ mass times acceleration.
Force of friction – This is the force that is caused by the restriction of an object from moving across a surface. It can calculated by the equation of normal force times mew. Mew is the coefficient of friction between the object and surface. It ranges from 0 to 1, higher being more restrictive.
The force of tension – This is a force given by a rope pulling on an object or an object hanging up a string or something that would have tension. This can be calculated again my Newton’s laws.
When solving for these forces using Newton’s laws we need to set it up correctly than it is simple algebra processes; at least in the beginning of learning physics. What we need to know is that the forces in the x direction and y direction are completely independent of each other. Therefore, we need to use the equation twice, once for each direction if it is needed to solve a problem. Also, we need to know that when we set on the sum of the forces on the left side of the equation we need to separate the forces directed in the upward and downward direction; or facing the right or to the left direction. This is why making a force diagram help; which means to draw all of the forces about a dot to show what directions each of them are facing.
All of these forces are measured in Newton’s
Next we went on to work. We explained that most problems can be answered by doing either forces or energy. We first described the types of energies and their equations.
Kinetic Energy – 1/2mv^2 ß this is used for an object that is in motion; m is mass and v is velocity
Gravitational Potential Energy – mgh ß This is for an object that is hanging by something, or just has the potential to fall due to the force of gravity, m is mass, g is 9.81, and h is the height of the object.
Thermal energy – Fx ß This is used to calculate something such as the work done by the force of friction. F is the force and x is delta x, so how far it travels.
Elastic Energy – 1/2kx^2 ß This is used for an object working against a spring. K is the spring constant for an ideal spring, and x again is how far the object travels while against the spring.
In solving a problem with energy it is a simple process at first. You need to select two different places in the movement of an object. Then you decide energies are present at those 2 spots. You set them equal to each other and then fill in the variables that you know and use algebra to solve for whichever variable you’re looking for.
We next went over practice problems and took the quiz
Here are links that help
http://www.physicsclassroom.com/class/newtlaws/u2l2a.cfm
http://theory.uwinnipeg.ca/physics/force/index.html
http://phun.physics.virginia.edu/topics/energy.html
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