how is it useful in our lives? any real-world applications?
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Imagine you have a pipe. Water flows through the pipe. For the first hour, water flows through at 3 gallons/hour. For the second, it flows through at 4 gallons/hour. For the third, it flows through at 2 gallons/hour.
How much water flowed through the pipe?
Well, this is an easy problem. 3 gallons/hour times 1 hour = 3 gallons in the first hour, 4 gallons/hour times 1 hour = 4 gallons in the second hour, and 2 gallons/hour times 1 hour = 2 gallons in the third hour. That is, 3 + 4 + 2 = 9 gallons total.
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But what if the pipe works a little differently? What if instead of working at the same speed for a full hour, the flow were always changing? What if, say, the flow starts at 0 and smoothly increases to 30 gallons/hour after three hours? So after half an hour it’s flowing at 5 gallons/hr, after an hour it’s flowing at 10 gallons/hour, and so on, but it’s smoothly changing between those points. How much water does the pipe pump in those three hours?
It’s not necessarily obvious how you’d solve this problem with basic math tools. Because the rate is always changing, you can’t just use the basic (amount) = (rate) x (time) formula, because there’s never a window of time to multiply by. It only pumps at 10 gallons/hour for a single instant at exactly 1 hour, but it was pumping slightly less just before, and slightly more just afterward.
In a sense, what you’d “like” to do is be able to solve this the same way you solved the first problem, by cutting things up into intervals where the flow is constant and then just adding them up. If you could do that, the problem would be easy. And it turns out that integration is effectively that: it lets you cut the time up into *infinitely small* intervals, treat the flow as constant during those intervals, and then *add up infinitely many of those infinitely small intervals* to get a sensible answer.
The answer, as it turns out, is integral from t=0 to t=3 of the flow rate (which, since it goes from 0 to 30 in 3 hours, is 10t gallons/sec). We compute: integral(t from 0 to 3) 10t dt = 5t^(2)|(t=0 to t=3) = 5(3)^2 – 5(0)^2 = 45 gallons. Easy.
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Integration has more applications than I could list. But to name just a few:
* It allows you to compute the area or volume of objects that are more complicated than geometry would help you with. You can find something like the area under a parabola, prove the area of a circle formula from geometry, or otherwise handle shapes geometry can’t.
* It allows you to handle continuous changes in things. Position over time is the integral of velocity. The inflation since 1980 is the integral from 1980 to today of all the rates of inflation over that time. The total water pumped by our pipe is the integral of the rate at which the water is flowing.
* To find things like center of gravity in physics or center of lift in aviation.
* To compute the amount of energy required to move an object against some opposing force, like a wind or an electric field.
* To find the length of complicated curves, or even to define what we mean by “length of a curve”.
Integrals are a foundational part of almost every computational field. In fact, many formulas you’re probably familiar with (things like x(t) = -(1/2)at^2 + v0*t + x_0 for movement under constant acceleration, or even KE = 1/2 mv^2 for kinetic energy) are actually just special cases of integrals.
I’d argue integrals probably have more “bang for your buck” in terms of applications than anything else in all of math, actually. They’re often pretty easy to compute and can solve a huge number of seemingly unrelated problems.
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