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gksriharsha

What is integration? What is differenciation? What do we actually do by doing that?Explain please

  • one year ago
  • one year ago

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  1. ParthKohli
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    Integration is the inverse of differentiation. Differentiation is the inverse of integration.

    • one year ago
  2. terenzreignz
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    Differentiation, in a nutshell, is getting the rate of change of one variable, with respect to another. In effect, dy/dx is simply how quickly y changes given a change in x.

    • one year ago
  3. terenzreignz
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    Graphically, the derivative is the slope of the tangent line at a point. The steeper, the bigger, of course.|dw:1361026112848:dw| Make do with this, I suck at drawing!

    • one year ago
  4. terenzreignz
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    |dw:1361026152021:dw| So, this is the tangent line at this point, and its slope is the value of the derivative. As you can see, at this point, the function is increasing, and so the derivative, or the slope of this tangent line, is positive.

    • one year ago
  5. terenzreignz
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    |dw:1361026217803:dw| Over here, the slope of the tangent line is negative, and you can see that the function is decreasing at this point.

    • one year ago
  6. terenzreignz
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    There are times when the slope of the tangent line is zero, the could usually mean that you're at a relative extremum (either a maximum or a minimum). In any case, it means the function is 'stable' at that point.|dw:1361026305680:dw|

    • one year ago
  7. terenzreignz
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    Sticking to two variables, y = f(x), the derivative of y with respect to x, written dy/dx, is more often than not, also a function of x. dy/dx = f'(x) that's how it's usually written.

    • one year ago
  8. terenzreignz
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    Let y = f(x) The indefinite integral of f(x), written \[\large \int\limits f(x)dx\] is a function F(x) such that F'(x) = f(x), or as @ParthKohli mentioned, the inverse of differentiation.

    • one year ago
  9. terenzreignz
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    I say it is *a* function F(x) and not *the* function F(x) because the indefinite integral of a function is never unique. All indefinite integrals of functions only differ by constants, though.

    • one year ago
  10. terenzreignz
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    Now, if there's an indefinite integral, you can bet there's something called the "definite Integral". It has a different concept to it. Let's go back to the curve, and add in the cartesian plane.

    • one year ago
  11. terenzreignz
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    |dw:1361026761048:dw|

    • one year ago
  12. terenzreignz
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    Say you want to find this area, right here... |dw:1361026815625:dw|

    • one year ago
  13. terenzreignz
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    The curve is our function, y = f(x) a and b are where the function intersects the x-axis. The definite integral is the area of the region bounded by the curve of y=f(x) and the lines x=a and x=b Denoted by \[\huge \int\limits_{a}^{b}f(x)dx\]

    • one year ago
  14. terenzreignz
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    Now, see, graphically, the derivative is the slope of the tangent line, the indefinite integral is the function whose derivative is f(x). So the indefinite integral has something to do with derivatives While the definite integral has something to do with the area under a curve. Why are they both called integrals? :P

    • one year ago
  15. terenzreignz
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    The answer has to do with what's called the fundamental theorem of calculus. There are two of them, actually, but they say similar things, essentially. |dw:1361027070141:dw| The theorems make it so that the definite integral of the curve from point a to point b, this is equal to the difference of an antiderivative evaluated from points b to a. If that confuses you, then just consider this elegant way to put it: \[\huge \int\limits_{a}^{b}f(x)dx=F(b) - F(a)\] Where \[\huge \frac{d}{dx}F(x)=F'(x)=f(x)\]

    • one year ago
  16. terenzreignz
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    And there you go. Who'd have thought two seemingly unrelated things such as the slope of the tangent line, and the area under a curve, are actually very much related? :) I hope you enjoyed this rather drab online lecture from me :P

    • one year ago
  17. gksriharsha
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    yes!

    • one year ago
  18. terenzreignz
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    Great :)

    • one year ago
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