What are integrals and how are they used in physics and calculus?

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Integrals are fundamental concepts in calculus and physics, used to calculate areas under curves and determine quantities like work done by variable forces. There are two main types of integrals: indefinite, which reverses differentiation to recover original functions, and definite, which calculates the area between a curve and the x-axis over specified limits. Integration involves breaking a problem into smaller parts and summing the results, improving accuracy with smaller subdivisions. This method is particularly useful in physics for analyzing varying forces, such as gravity, over distances. Understanding integrals enhances comprehension of both calculus and physical applications.
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Hi, my physics teacher has recently started putting integrals into all of his examples and notes, but we haven't gotten to integrals in calculus yet.

Could anyone basically tell me what integrals are,. and what they are for
 
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hi,

The same thing also happened to me when I took physics and calculus concurrently. This is an oversimplification, but hopefully it will enough for now.

There are two types of integration; definite and indefinite.

Indefinite integration (also known as anti-differentiation) does exactly what its name implies.

Say you have a function and take its derivative. If you integrate the derivative, you will get the original function back. For example:

y = x^2

\frac {dy} {dx} = 2x

\int 2x \,dx = x^2 + c

where c is an arbitrary constant.

Definite integration is often used to find area. If you have a geometric shape, you can use formulas to find area.

But say you have the above function, y = x^2, and you would like to know the area of the region bound between the x-axis and the curve, between the points x = 0, and x = 1.

You can exactly determine this area by evaluating a definite integral.

One of the many uses of this in physics is to determine the work done by a variable force. Anyway, that’s where I first saw integration.

Reading ahead in your calculus book or asking your math teacher for a sneak preview during his/her office hours might also help you better understand what you’re doing.
 
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That's excellent, Melinda! :) Just a small addendum - think of integration as breaking something up into a very large number of very small bits and adding the results together. For instance, in the case Melinda cited, a variable force - let's say you're trying to figure out how much work gravity does on an object falling from a certain height. Well, that's easy - work is force times distance, so you just multiply the weight of the object by the distance it falls, and boom. No pun intended.

However, suppose the height is so great that gravity can't be treated as a constant - what do you do then? What you might do is break it up - say, figure out what happens for the first quarter of the fall assuming gravity is constant, then refigure gravity and do it for the next quarter, and so on until you get it to the ground. Then you add up the four pieces.

That'll certainly be better than if you took it all at once, but it's still only an approximation. How do you know if the approximation is good enough? Clearly (and this can and should be proved), the smaller the pieces you take, the better your approximation will be. It seems reasonable to suppose that, if you could take an infinite number of pieces and add them all together, there would be no error. This is what integration does, in concept anyway - break the distance into infinitesimal pieces and add the results all together. Any time you have a situation where you're trying to figure the effects of something which varies over a range with which you're concerned, be it distance or time or anything else, integration is the primary tool for analysis.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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