Relation between Derivatives and Integration

Question

What does it mean to take the INTEGRATION OF A DERIVATIVE FUNCTION? (i know it gives you the original function,but still don't know what it actually means)Does it mean that we are summing the slopes at a point, but integration means area under the curve i.e summation of all the rectangles as "n tends to infinity and the sub-intervals become smaller and smaller" whereas differentiation means slope at a point.

Also what does it mean to take the DERIVATIVE OF A INTEGRATION FUNCTION?



I know all of this falls under "The fundamental theorem of calculus" but still i haven't understood it and what it actually means? thank you.

Answer 1

This might not fully answer your question, but maybe we can clear things up a little bit.

First, you're exactly right. The derivative is the slope at a single point (slope of the tangent line) and the integral is area under a curve. On the surface those seem like completely unrelated processes.

That's what makes the Fundamental Theorem so special. It says that integration is actually just derivatives in reverse. (That may not be the most technically correct explanation, but it gets to the heart of the matter).

Without the Fundamental Theorem, we would have to calculate area under a curve using limits and Riemann Sums (as you allude to in your question). The Fundamental Theorem says we can accomplish the same thing using anti-derivatives.

What is an anti-derivative? It's an UN-derivative, the derivative in reverse. For example, the derivative of x² is 2x, so the anti-derivative of 2x is x². ^*(It's actually x² + C, but I'll explain that below.) The derivative of 4x⁵ is 20x⁴ so the anti-derivative of 20x⁴ is 4x^{5 *}(again, see footnote below). It's often worded this way: Given f(x), the anti-derivative is the function F(x) such that F'(x) = f(x).

Back to the Fundamental Theorem, it says that if you want to find the area under f(x) between points a and b, instead of using a Riemann Sum, you can find F(x) -- the anti-derivative of f(x) -- and then calculate F(b) - F(a).

For example, let's find the area under f(x)=x² between x=1 and x=2. The anti-derivative of x^2 is *F(x) = ¹⁄₃x³ (because the derivative of ¹⁄₃x³ is x² -- just do your power rule in reverse) F(2) = ⁸⁄₃ and F(1) = ¹⁄₃. The area under f(x)=x² between x=1 and x=2 is F(2) - F(1) = ⁷⁄₃.

So taking the integral of the derivative (or the derivative of the integral) is sort of like squaring a number and then taking the square root. They are opposite processes that essentially undo each other.

^*Now for the nitty-gritty stuff we ignored earlier. As you know, the derivative of a constant is zero (by its very definition, a constant has no rate of change). So the derivative of x² is 2x, but so is the derivative of x² +3 or x² -7. So if you're given 2x and asked to find its anti-derivative, it could be x² or x² +3 or x² -7 or x²±[any constant]. So to cover our bases, we say the anti-derivative is x² + C (where C represents any constant). We can essentially ignore that C when doing "definite integrals" (integrals with upper and lower limits where we're finding a specific area because the C's will cancel out when we subtract F(a) from F(b).

Answer 2

Here's an OUTLINE of the proof:

defintion of derivative

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limit of [f(x+h) - f(x)]/h as h-->0

is the limit of the diffence quotient...

defintion of integral

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These are the Reimann sums.

That is the integral is h*f(x) + h*f(X1) + h*f(x2) + ... h*f(Xn)

where X1,X2,...,Xn are any points inside the width of each rectangle,

as N grows infinitely LARGE.

The h factors out of the Reimann sums and cancels the h in the denominator

of the difference quotent in the derivative

The definition of the difference of these Reimann sums features the liberty

of choosing x1,X2, .... Xn inside each of the N sub-intervals.

Defining this sum as a telescoping sequence:

f(x) + f(x1) - f(x) + f(x2) - f(x1) + f(x3) - f(x2) + ... + f(Xn) - f(Xn-1) + f(X+h) - f(Xn)

the only surviving term is f(X+h) = f(x) as h tends to ZER0