Integral as Antiderivative: A Mathematical Proof

[anantchowdhary]

Is there a mathematical proof that can prove that the integral is the antiderivative?

 

[Doc Al]

fundamental theorem of calculus

Look up the fundamental theorem of calculus.

 

[anantchowdhary]

umm...ive tried to study it...but my question is a little different...i meant to say..

if we take out the area of the graph of a function f(x) bound at certain limits, then..is there any way to prove that the indefinite integral gives us the antiderivative of the function

or rather..the area bound under two limits of the function is simply the limits applied to the antiderivative of the function

 

[Kummer]

Given an integrable function $$f:[a,b]\mapsto \mathbb{R}$$ we define,
$$g(x) = \int_a^x f(t) dt$$ for all $$t\in [a,b]$$.
Then, $$g$$ is a continuous function on $$[a,b]$$. And furthermore if $$f$$ is continuous at $$t_0 \in (a,b)$$ then $$g$$ is differentiable at $$t_0$$ with $$g'(x_0)=f(x_0)$$

Are you asking this?

 

[mathwonk]

suppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,

i.e.f(x)h < A(x+h)-A(x) < f(x+h)h.

satisfy yourself of this by drawing a picture.

now the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.

by the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).now in creasing is not n eeded ubt makes it easier.

 

[anantchowdhary]

thanks..ill give it a thought

 

[HallsofIvy]

A little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely "below" the graph. Let A_n be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that $A_n \le A$ where A is the area of that region (the "area below the graph").

Now do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking Aⁿ to be the total area of all those rectangles, we must have $A \le A^n$.

That is, for all n, $A_n \le A \le A^n$

If f HAS an integral (if f is integrable) then, by definition, the limits of A_n and Aⁿ must be the same- and equal to the integral of f from a to b. Since A is always "trapped" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.

That's pretty much the proof given in any Calculus book.

 

[Gib Z]

"is there any way to prove that the indefinite integral gives us the antiderivative of the function"

I seemed to understand this as - ' is there any way to prove F(x), where F'(x)=f(x), is the same function as the function given by $$\int f(x) dx$$.' The fundamental theorem of Calculus shows when we put upper and lower bounds, b and a, on the integral, it results in F(b) - F(a), however to the specific question as I understood, the dropping of bounds is just a nice notation for the integral/antiderivatve function.

 

[anantchowdhary]

A little more generally- assume f(x)> 0 for a< x< b. For any number n, divide the interval from a to b into n equal sub-intervals (each of length (b-a)/n)). Construct on each a rectangle having height equal to the minimum value of f on the interval- that is, each rectangle is completely "below" the graph. Let A_n be the total area of those rectangles. Since each rectangle is contained in the region below the graph of f, it is obvious that $A_n \le A$ where A is the area of that region (the "area below the graph").

Now do exactly the same thing except taking the height to be the maximum value of f in each interval. Now, the top of each interval is above the graph of f so the region under f is completely contained in the union of all the rectangles. Taking Aⁿ to be the total area of all those rectangles, we must have $A \le A^n$.

That is, for all n, $A_n \le A \le A^n$

If f HAS an integral (if f is integrable) then, by definition, the limits of A_n and Aⁿ must be the same- and equal to the integral of f from a to b. Since A is always "trapped" between those two values, the two limits must be equal to A: The area is equal to the integral of f from a to b.

That's pretty much the proof given in any Calculus book.

umm...i tried this out..i Followed pretty much of it

but how do we prove that dA/dx=f(x)?

thanks

 

[mathwonk]

notice the proof i gave for the increasing case, which is due to newtton, does not need the deep theorem that a continuous function always has a max and a min on a closed bounded interval.

it also covers (when used piecewise) all polynomials, con tinuous rational functions, and continuous trig functions, that occur in practice. hence there is

no reason for books to omit this proof or banish it to an appendix.

 

[HallsofIvy]

Define F(x) to be
$$\int_a^x f(t)dt$$
For f(x)> 0 we can interpret that as the area between the graph y= f(x) and the x-axis, from a to the fixed value x. Then F(x+h) is
$$\int a^{x+h}f(t)dt$$
the area between the graph y= f(x) and x-axis, from a to the fixed value x+h.

Now F(x+h)- F(x) is the area between the graph y= f(x), between x and x+h. It's not difficult to see that is equal to the area of the rectangle with base x to x+h and height f(x^*) where x^* is some value between x and x+h: That is F(x+h)- F(x)= hf(x^*). Then
$$\frac{F(x+h)- F(x)}{h}= f(x^*)$$
Taking the limit as h goes to 0 forces x^*, which is always between x and x+h, to go to x.
Therefore
$$\frac{dF}{dx}= \lim_{h\rightarrow 0} \frac{F(x+h)- F(x)}{h}= f(x)$$

 

[mathwonk]

let me combine halls discussion with mine in post 5.

he has shown, modulo the theorem that max and min exist,

that in my notation, if m(h) is the min of f on the interval [x,x+h], and if M(h) is the max,

that m(h)h < A(x+h)-A(x) < M(h)h.

where < means less than or equal.

Then again if we compute the derivative of A as the limit of
[A(x+h)-A(x)]/h

as h goes to 0, we see by the inequalities that [A(x+h)-A(x)]/h
is squeezed between

M(h) and m(h) for all h. since f is continuous, these numbers both apprioach f(x), so lim [A(x+h)-A(x)]/h
= dA/dx = f(x).

 

[mathwonk]

to prove the point halls says is "not difficult to see", uses the intermediate value theorem, in case you do not see it.

 

[anantchowdhary]

thanks a lot for the proof!

 

[HallsofIvy]

"It is easy to see ..." means "I think there is a proof but I can't remember it just now".

"Obvious to the most casual observer" means "I hope no one asks me to prove it"!

 

[HallsofIvy]

"It is easy to see ..." means "I think there is a proof but I can't remember it just now".

"Obvious to the most casual observer" means "I hope no one asks me to prove it"!

 

[ssd]

suppose f is aN INCREASING CONTINUOUS FUNCTION. then the area under the graph between x and x+h is between f(x)h and f(x+h)h,

i.e.f(x)h < A(x+h)-A(x) < f(x+h)h.

satisfy yourself of this by drawing a picture.

now the derivative of that area function is the limit of [A(x+h)-A(x)]/h, as h goes to zero.

by the inequality above, this limit iscaught b etween f(x) and f(x+h), for all h, which emans, since f is continuious, it equals f(x). i.e. dA/dx= f(x).


now in creasing is not n eeded ubt makes it easier.

This is a beauty... incomplete though. I believe,for a student new to calculus, it is enough to remove all mental barriers about the fact whether an integral is anti-derivative.

 

[jambaugh]

Technically the indefinite integral is defined to be the anti-derivative (or rather a variable anti-derivative depending on an arbitrary constant since "the" anti-derivative is not unique.)

The definite integral is a value rather than a function so you can't call it an anti-derivative.

That technical nit picking aside see your undergraduate calculus text and the two forms of the FTC.