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Q: How many positive integral factors of 60 are multiples of 6?
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Discuss line integral and its real life applications?

A line integral can evaluate scalar and vector field functions along a curve/path. When applied on vector field, line integral is considered as measure of the total effect of the vector field along a specific curve whereas in scalar field application, the line integral is interpreted as the area under the field carved out by a particular curve.Line integral has many applications in physics. In mechanics, line integral is used to determine work done by a force in moving an object along a curve. In circuit analysis, it is used for calculating voltage.


How many number between 400 and 500 which is exactly divisible by 12 15 and 20?

First, find the prime factors of 12,15,20:12=2*2*315=3*520=2*2*5Now, compute the LCM by multiplying the factors (count duplicates once):LCM=2*2*3*5=60Only the multiples of the LCM will be divisible by 12, 15, and 20:60,120,180,240,300,360,420,480,540...{420,480}


How is integral calculus used in daily life?

The different aspects of calculus are used in the real world every day. In business, specialists look at the derivatives of trends that can help them predict the future of stocks and markets. Architects commissioned for a job are given a budget and they use optimization to calculate the best amount of material they can get with that budget and space in a building they are designing. The Integral is used to show area under a curve. The indefinite integral is the antiderivative of a function. For these types of professions the integral is their Bible, metaphorically speaking. The watch the trends, convert the data into a quantitative function and then use the integral to predict the future of a company or simply use it with differentiation for an optimization problem. Their are many other uses as well that we use, sometimes subconsciously, in everyday life; these are just a couple of examples.


What is the anti-derivative of e-x2?

I believe the questioner means e^(-x^2), which is perhaps the most famous of many functions which do not have anti-derivatives which can be expressed by elementary functions. The definite integral from minus infinity to plus infinity, however, is known: It is sqrt(pi). The antiderivative to e^(-2x) is, (-*e^(-2x)/2) Though the anti-derivative (integral) of many functions cannot be expressed in elementary forms, a variety of functions exist only as solutions to certain "unsolvable" integrals. the equation erf(x), also known as the error function, equals (2/sqrt(pi))*integral e(-t^2) dt from 0 to x. As mentioned before, this cannot be expressed through basic mathematical functions, but it can be expressed as an infinite series. If the question is the antiderivative of e - x2, the answer is e*x - x3/3


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.