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Derivative and integral (inverse operations?)

  1. Feb 3, 2010 #1
    I'm learning calculus, but I'm still a beginner on that.

    So the concepts I actually have are the following:

    The derivative is the slope of the line tangent to the curve of a function at a given point (so we need 1 argument: the x you want the slope).

    The integral is the area under the curve of a function between two given points (so we need 2 arguments: the start and the end).

    So, considering my concepts are right, continue reading.

    I know too that integral and derivative are inverse operations of calculus (like multiplication and division).

    But I can't really understand how can the slope of a tangent line to the curve be the inverse of the area under the curve. And I can't even understand how these two operations are comparable this way because the first one needs one argument and the second one needs two arguments.

    Someone please help me understanding this...

    Thanks in advance.
     
  2. jcsd
  3. Feb 3, 2010 #2
    Well, first of all don't confuse the definite integral (which is the area) and the indefinite integral (which is the anti-differentiation). Second of all, I don't see what that has to do with the argument thing.

    Let's be more precise. If we choose a fixed point a, and integrate f(x) from a to some arbitrary point x, then we got a function of x:

    [tex]F(x)=\int^{x}_{a}f(s)ds[/tex]

    Notice that s is a dummy variable, like an indexer. It dies off once the integral is computed, and we're left with an expression of x. Now the fundamental theorem of calculus tells us that whenever F(x) is continuous then F'(x)=f(x).

    Let's understand it intuitively, with slopes and area, assuming our functions are nice and smooth. The theorem tells us that the slope of F(x), that is, its rate of change, is f(x). F(x) measures the area under the graph of f(x), so it's differential change [tex]F(x+\delta x)-F(x)[/tex] will be the area of the tiny rectangle under f(x), or [tex]f(x)\delta x[/tex]
    And since [tex]\frac{F(x+\delta x)-F(x)}{\delta x}[/tex] is an approximation of the derivative, as you let dx go to 0, you have that F'(x)=f(x). Nice.
     
  4. Feb 3, 2010 #3
    Thank you for the reply elibj123. It helped me alot but I still have some doubts about what you said.

    What's the use of the anti-differentiation? It's just used in the calculation of the definite integral?

    So, the definite integral is not the inverse of derivative? (Only the anti-differentiation is?)

    I didn't understand the dummy variable thing. What you mean by "compute" the integral? Is it the proccess of anti-differentiation?

    I'm a bit confused ;S
     
  5. Feb 4, 2010 #4
    After you've established that indefinite integration is anti-differentiation, it is used to compute definite integrals:

    [tex]\int^{b}_{a}f(x)dx=F(b)-F(a)[/tex]

    Where F(x) is the anti-derivitive.

    The derivative is an object accepting a function and returning a new function. (it's called an operator on the functions, or a functional)
    The definite integral is an object accepting a function and returning a number. So just conceptually you can understand that they are not inverse operations.


    Look at the integral

    [tex]\int^{b}_{a}f(x)dx[/tex]

    I could just swap x with another letter (not a&b to avoid confusion) and it will remain the same:

    [tex]\int^{b}_{a}f(\theta)d\theta[/tex]

    Since the definition of f remains the same. After you compute the integral (no matter what technique you have to do that, or even if it cannot be done), the variable of integration disappears, leaving with an expression of a & b.

    Now I let the upper bound be x, and I'm left with an expression of a & x. Since a is some fixed point (let it be 0), the result is a new function of x, which happens to be the anti-derivative. Actually you see that for different values of a I get different functions, which shows the fact that f(x) has infinitely many anti-derivatives.
     
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