Proving F'(x)= f(x) using the definition of integral?

In summary: That is, F'(x)= f(x) for all x in [a,b]. Then, for any arbitrary partition of [a,b], the corresponding Riemann sums are equal to F(b)- F(a). Taking the limit as the partition becomes finer and finer, the Riemann sums converge to \int_a^b f(x)dx. Since F(b)- F(a) is equal to those Riemann sums, F(b)- F(a)= \int_a^b f(x)dx which is what you were trying to prove.In summary, the conversation discusses proving a statement using the definition of the integral. The approach involves considering an equidistant
  • #1
irresistible
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Hey guys,
Can you help me prove this?

Suppose that f:[a.b] -> R is integrable and that F:[a,b]->R is a differentiable function such thet F'(x)= f(x) for all xLaTeX Code: \\in [a,b].
Prove from the definition of the integral that;

F(b)-F(a) =LaTeX Code: \\int f(x) dx ( integral going from a to b)

I can prove this using the Fundamental theorem of calculus;however, this question specifically asks that we use the definition of integral to prove this:

I'm thinking that I have to use the "partition" prepositions to prove this.
Any ideas?
Thank you in advance guys!
 
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  • #2
You could start by considering an equidistant partition
[tex]\mathcal P = \{ x_0 = a, x_1 = a + \Delta, x_2 = a + 2 \Delta, \cdots, x_N = a + N \Delta \} [/tex] where [tex]\Delta = \frac{b - a}{N}[/tex]
and writing down the left and right Riemann sums. Then I think it's fairly straightforward manipulations of series and limit taking.
 
  • #3
I presume the "definition of the integral" you are talking about is "[itex]\int_a^b f(x)dx[/itex] is the area of the region bounded by y= f(x), y= 0, x= a, x= b" (I am assuming here that f(x) is greater than or equal to 0 and that f is continuous. You may want to generalize that.)

Let u be the minimum value of f(x) on [a,b] and U be the maximum value. Then that area lies between the areas of the two rectangles [itex]u(b-a)\le \int_a^b f(x)dx\le U(b-a)[/itex] since f is continuous there must exist a value [itex]\overline{x}[/itex], between a and b, such that [itex]\int_a^b f(x)dx= f(\overline{x})(b-a)[/itex]. (That's the "integral mean value theorem".)

Now define [tex]F(x)= \int_a^x f(t)dt[/tex]. Then [tex]F(x+h)= \int_a^{x+h} f(t)dt[/tex] and [tex]F(x+h)- F(x)= \int_a^{x+h}f(t)dt- \int_a^x f(t)dt= \int_x^{x+h}f(t)dt[/tex]

Applying the integral mean value theorem to that, for every non-zero h there must exist [itex]\overline{x}[/itex] between x and x+ h such that [tex]F(x+h)- F(x)= f(\overline{x})((x+h)- x)= f(\overline{x})h[/tex] and so
[tex]\frac{F(x+h)- F(x)}{h}= f(\overline{x})[/tex]
Since [itex]\overline{x}[/itex] always lies between x and x+h, as h goes to 0, [itex]\overline{x}[/itex] goes to x and, since f is continuous we have
[tex]\lim_{h\rightarrow 0}\frac{F(x+h)- F(x)}{h}= F'(x)= f(x)[/tex]
 

Related to Proving F'(x)= f(x) using the definition of integral?

1. What is the definition of an integral?

The definition of an integral is the limit of a sum of infinitely many small areas. It represents the total area under a curve on a graph.

2. How is the integral related to the derivative?

The integral and derivative are inverse operations. The derivative represents the rate of change of a function at a specific point, while the integral represents the accumulation of a function over a given interval.

3. Why is it important to prove F'(x)= f(x) using the definition of integral?

Proving this relationship using the definition of integral is important because it provides a rigorous and logical explanation for the connection between the derivative and integral. It also helps to solidify our understanding of these concepts and their applications in mathematics and science.

4. What is the process for proving F'(x)= f(x) using the definition of integral?

The process involves using the definition of integral to set up an equation, then using algebraic and calculus principles to manipulate the equation until it matches the form of F'(x)= f(x). This typically involves breaking the integral into smaller parts and applying the fundamental theorem of calculus.

5. Can this relationship hold for all functions?

Yes, this relationship holds for all continuous functions. However, some functions may require more complex techniques to prove this relationship, such as using Riemann sums or other integration methods.

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