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A definite integral can be defined as the limit as the norm of the partition approaches zero of a function value times a difference in its independent variable. Since I do not have access to a proper symbol library on here, I will try to write it out as best as I can:
lim as p->0 of sum from i=1 to n of f(xi)*(delta-x).
You may know about Riemann sums, which, in functions of one variable, approximate an integral using rectangular partitions of the area under a function's curve that go from the axis in question to some defined value on the curve in height and are some defined width of independent variable.
For instance, to approximate the area between the function y=4-x2 and the x-axis, the following could be done:
- The function has zeroes at x=-2 and x=2, so the total length of x to be partitioned is 4 units.
- You can choose any way to divide up this 4-unit length, but the most common way is to pick even intervals based on a desired number of rectangles. If you were asked to approximate this area with four rectangles, each partition would be of width 1 unit, since 4/4=1. The use of more rectangles causes the approximation to get more accurate. In truth you do not need to use equal partitions, but for most math classes equal partitions are asked for.
- Using the above 4-rectangle approach, you now have four equal intervals, [-2,-1], [-1,0], [0,1], [1,2]. You can choose any x-value on these intervals to plug into the function to get its corresponding y-value that will serve as the "height" of each rectangle. Generally if you choose the left-most value of one interval, you choose the left-most value of all intervals, but once again this is not necessarily the only way to do it, simply the most easily approachable.
- By constructing four rectangles, all of width 1 and of height y=4-x2 where x is some x on each interval, you get four rectangles whose areas can be added together to approximate the definite integral of y=4-x2 from -2 to 2 with respect to x.
Now, that was kind of a long-winded review, but I want to make sure you know that before I continue, and I can refer back to that to simplify my explanation of how limits relate to integrals. As I said above, as the number of partitions you use increases, the approximation of the integral gets more and more accurate. It follows a behavior similar to a limit in that it approaches closer and closer to the actual value of the integral. But, as you increase the number of partitions, the width of the partitions gets smaller and smaller since you are fitting more rectangles into the same space. To get the true value of the integral, you would need an infinite number of zero-width rectangles. This is impossible to definitively calculate, but by taking the limit as the width of the partitions approaches zero, we can find the exact value of the integral.
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When you find particular integral then why you not add constantof integration?
Where you refer to a particular integral I will assume you mean a definite integral. To illustrate why there is no constant of integration in the result of a definite integral let me take a simple example. Consider the definite integral of 1 from 0 to 1. The antiderivative of this function is x + C, where C is the so-called constant of integration. Now to evaluate the definite integral we calculate the difference between the value of the antiderivative at the upper limit of integration and the value of it at the lower limit of integration: (1 + C) - (0 + C) = 1 The C's cancel out. Furthermore, they will cancel out no matter what the either antiderivatives happen to be or what the limits of integration happen to be.
What is the objective of concept of limit?
This makes it possible to give a precise formal definition of both the derivative, and the definite integral - which are both extremely useful concepts in math, physics, and engineering.
How you can defferentiate an integral?
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
How do you bisect the area of definate integral 1 to 6 of 1 divided by x sqared?
First, find the area under the curve y = 1/x2, with boundary lines x = 1 and x = 6, by calculating the integral of 1/x2 with lower limit 1, and upper limit 6. Then divide it by 2. (6)integral(1) of (1/x2) dx = (6)integral(1) of (x--2) dx = -x-1|(6),(1) = -1/x|(6)(1) = -1/6 +1 = 5/6. Thus, the half of the area under the curve is 5/12.
How do you calculate integrals that go to infinite?
You do what we call an "improper integral". I will denote the integral of f from a to b as intl a-b (f) here. so we define intl a-infinity (f) as lim b->infinity a-b(f) So it is a limit, and just like all other integrals, it may or may not exist (+/- infinity or infinite uncountable oscilations etc.) You have have to prove yourself though about its properties (it's easy since I reduced it to the regular integral) and you will see it's a perfectly fine definition. If you want examples, I have lots, message me.
How I solve the limit of the definite integral?
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/defintdirectory/DefInt.html
When you find particular integral then why you not add constantof integration?
Where you refer to a particular integral I will assume you mean a definite integral. To illustrate why there is no constant of integration in the result of a definite integral let me take a simple example. Consider the definite integral of 1 from 0 to 1. The antiderivative of this function is x + C, where C is the so-called constant of integration. Now to evaluate the definite integral we calculate the difference between the value of the antiderivative at the upper limit of integration and the value of it at the lower limit of integration: (1 + C) - (0 + C) = 1 The C's cancel out. Furthermore, they will cancel out no matter what the either antiderivatives happen to be or what the limits of integration happen to be.
What is the relationship between the SASRIA loss limit and the Companies Act, 1973?
explain the relationship between the sasria loss limit and the companies act 1973
What is the objective of concept of limit?
This makes it possible to give a precise formal definition of both the derivative, and the definite integral - which are both extremely useful concepts in math, physics, and engineering.
What is the relationship between Sasria Loss limits and the Companies aCT 1973?
explain the relationship between the sasria loss limit and the companies act 1973
Which layer of the atmosphere no definite outer limit?
Troposhphere
Use the concept of a limit to explain how you could find the exact value for the definite integral value for a section of your graph?
The definite integral value for a section of a graph is the area under the graph. To compute the area, one method is to add up the areas of the rectangles that can fit under the graph. By making the rectangles arbitrarily narrow, creating many of them, you can better and better approximate the area under the graph. The limit of this process is the summation of the areas (height times width, which is delta x) as delta x approaches zero. The deriviative of a function is the slope of the function. If you were to know the slope of a function at any point, you could calculate the value of the function at any arbitrary point by adding up the delta y's between two x's, again, as the limit of delta x approaches zero, and by knowing a starting value for x and y. Conversely, if you know the antideriviative of a function, the you know a function for which its deriviative is the first function, the function in question. This is exactly how integration works. You calculate the integral, or antideriviative, of a function. That, in itself, is called an indefinite integral, because you don't know the starting value, which is why there is always a +C term. To make it into a definite integral, you evaluate it at both x endpoints of the region, and subtract the first from the second. In this process, the +C's cancel out. The integral already contains an implicit dx, or delta x as delta x approaches zero, so this becomes the area under the graph.
How do you solve 5.4e0.06t using the fundamental theorem of calculus?
We need more information. Is there a limit or integral? The theorem states that the deivitive of an integral of a function is the function
How you can defferentiate an integral?
If the upper limit is a function of x and the lower limit is a constant, you can differentiate an integral using the Fudamental Theorem of Calculus. For example you can integrate Integral of [1,x^2] sin(t) dt as: sin(x^2) d/dx (x^2) = sin(x^2) (2x) = 2x sin(x^2) The lower limit of integration is 1 ( a constant). The upper limit of integration is a function of x, here x^2. The function being integrated is sin(t)
A traffic officer is compiling information about the relationship between the hour of the day and the speed over the limit at which the motorist is ticketed He computes a correlation coefficient of 0?
This would mean that the officer found no relationship between speeding and the hour of day.
What is the time limit for an adulterous relationship?
There is no time limit. If it's adultery, it's adultery.
Does the thermosphere have a definite outer limit?
The thermosphere does not have a definite outer limit as it gradually transitions into space. Its density decreases with altitude, and eventually merges with the exosphere, the outermost layer of Earth's atmosphere. The boundary between the thermosphere and the exosphere is not clearly defined.
