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The fundamental theorem of calculus specifies the relationship between the two central operations of Calculus , Differentiation and Integration .

The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an Indefinite Integration
More exactly, the theorem deals with Definite Integration with variable upper limit and arbitrarily selected lower limit. This particular kind of definite integration allows us to compute one of the infinitely many Antiderivatives of a function (except for those which do not have a zero). Hence, it is almost equivalent to Indefinite Integration , defined by most authors as an operation which yields any one of the possible antiderivatives of a function, including those without a zero. can be reversed by a differentiation.

The second part, sometimes called the second fundamental theorem of calculus, allows one to compute the Definite Integral of a function by using any one of its infinitely many Antiderivative s. This part of the theorem has invaluable practical applications, because it markedly simplifies the computation of definite integrals.

The first published statement and Proof of a restricted version of the fundamental theorem was by James Gregory (1638-1675)
See, e.g., Marlow Anderson, Victor J. Katz, Robin J. Wilson, ''Sherlock Holmes in Babylon and Other Tales of Mathematical History'', Mathematical Association of America, 2004, p. 114 .
. Isaac Newton (1643–1727) and Gottfried Leibniz (1646–1716) independently developed the theorem in its final form.


INTUITION

Intuitively, the theorem simply states that the sum of Infinitesimal changes in a quantity over time (or some other quantity) add up to the net change in the quantity.

To comprehend this statement, we will start with an example. Suppose a particle travels in a straight line with its position given by ''x''(''t'') where ''t'' is time. The derivative of this function is equal to the infinitesimal change in quantity per infinitesimal change in time (of course, the derivative itself is dependent on time). Let us define this change in distance per change in time as the speed ''v'' of the particle. In Leibniz's Notation :

: rac{\mathrm dx}{\mathrm dt} = v(t).

Rearranging This Equation , it follows that:

:\mathrm dx = v(t)\,\mathrm dt.

By the logic above, a change in ''x'', call it \Delta x, is the sum of the infinitesimal changes d''x''. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. As one can reasonably infer, this operation works in reverse as we can differentiate the result of our integral to recover the original derivative.


FORMAL STATEMENTS

There are two parts to the Fundamental Theorem of Calculus. Loosely put, the first part deals with the derivative of an Antiderivative , while the second part deals with the relationship between antiderivative and Definite Integral .


First part


This part is sometimes referred to as First Fundamental Theorem of Calculus.

Let ''f'' be a continuous real-valued function defined on a Closed Interval ''b'' . Let ''F'' be the function defined, for all ''x'' in ''b'' , by
:F(x) = \int_a^x f(t)\, \mathrm dt
Then, for every ''x'' in ''b'' ,
:F'(x) = f(x)\,.

The operation \int_a^x f(t) is a Definite Integral with variable upper limit, and its result ''F''(''x'') is one of the infinitely many Antiderivatives of ''f''.


Second part


This part is sometimes referred to as Second Fundamental Theorem of Calculus.

Let ''f'' be a continuous real-valued function defined on a Closed Interval ''b'' . Let ''F'' be an Antiderivative of ''f'', that is one of the infinitely many functions such that, for all ''x'' in ''b'' ,
:f(x) = F'(x)\,.
Then
:\int_a^b f(x)\,\mathrm dx = F(b) - F(a).


COROLLARY

Let ''f'' be a real-valued function defined on a Closed Interval ''b'' . Let ''F'' be a function such that, for all ''x'' in ''b'' ,
:f(x) = F'(x)\,
Then, for all ''x'' in ''b'' ,
:F(x) = \int_a^x f(t)\,\mathrm dt + F(a)
and
:f(x) = rac{\mathrm d}{\mathrm dx} \int_a^x f(t)\,\mathrm dt.


EXAMPLES

As an example, suppose you need to calculate

:\int_2^5 x^2\, \mathrm dx.

Here, f(x) = x^2 and we can use F(x) = {x^3\over 3} as the antiderivative. Therefore:

:\int_2^5 x^2\, \mathrm dx = F(5) - F(2) = {125 \over 3} - {8 \over 3} = {117 \over 3} = 39.

Or, more generally, suppose you need to calculate

:{\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt.

Here, f(t) = t^3 and we can use F(t) = {t^4 \over 4} as the antiderivative. Therefore:

:{\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt = {\mathrm d \over \mathrm dx} F(x) - {\mathrm d \over \mathrm dx} F(0) = {\mathrm d \over \mathrm dx} {x^4 \over 4} = x^3.

But this result could have been found much more easily as

:{\mathrm d \over \mathrm dx} \int_0^x t^3\, \mathrm dt = f(x) {\mathrm dx \over \mathrm dx} - f(0) {\mathrm d0 \over \mathrm dx} = x^3.


PROOF


Suppose that
:F(x) = \int_{a}^{x} f(t) \mathrm dt.

Let there be two numbers ''x''1 and ''x''1 + Δ''x'' in ''b'' . So we have
:F(x_1) = \int_{a}^{x_1} f(t) \mathrm dt
and
:F(x_1 + \Delta x) = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt.

Subtracting the two equations gives
:F(x_1 + \Delta x) - F(x_1) = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt - \int_{a}^{x_1} f(t) \mathrm dt. \qquad (1)

It can be shown that
:\int_{a}^{x_1} f(t) \mathrm dt + \int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt = \int_{a}^{x_1 + \Delta x} f(t) \mathrm dt.
:(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)
Manipulating this equation gives
:\int_{a}^{x_1 + \Delta x} f(t) \mathrm dt - \int_{a}^{x_1} f(t) \mathrm dt = \int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt.

Substituting the above into (1) results in
:F(x_1 + \Delta x) - F(x_1) = \int_{x_1}^{x_1 + \Delta x} f(t)\mathrm dt. \qquad (2)

According to the Mean Value Theorem for integration, there exists a ''c'' in ''x''1 + Δ''x'' such that
:\int_{x_1}^{x_1 + \Delta x} f(t) \mathrm dt = f(c) \Delta x .

Substituting the above into (2) we get
:F(x_1 + \Delta x) - F(x_1) = f(c) \Delta x \,.

Dividing both sides by Δ''x'' gives
: rac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = f(c).
:Notice that the expression on the left side of the equation is Newton's Difference Quotient for ''F'' at ''x''1.

Take the limit as Δ''x'' → 0 on both sides of the equation.
:\lim_{\Delta x o 0} rac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = \lim_{\Delta x o 0} f(c).

The expression on the left side of the equation is the definition of the derivative of ''F'' at ''x''1.
:F'(x_1) = \lim_{\Delta x o 0} f(c). \qquad (3)

To find the other limit, we will use the Squeeze Theorem . The number ''c'' is in the interval ''x''1 + Δ''x'' , so ''x''1 ≤ ''c'' ≤ ''x''1 + Δ''x''.

Also, \lim_{\Delta x o 0} x_1 = x_1 and \lim_{\Delta x o 0} x_1 + \Delta x = x_1.

Therefore, according to the squeeze theorem,
:\lim_{\Delta x o 0} c = x_1.

Substituting into (3), we get
:F'(x_1) = \lim_{c o x_1} f(c).

The function ''f'' is continuous at ''c'', so the limit can be taken inside the function. Therefore, we get
:F'(x_1) = f(x_1) \,.
which completes the proof.

(Leithold et al, 1996)


Alternative proof

This is a limit proof by Riemann Sums .

Let ''f'' be continuous on the interval ''b'' , and let ''F'' be an antiderivative of ''f''. Begin with the quantity
:F(b) - F(a)\,.

Let there be numbers

x


such that

:a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b.

It follows that
:F(b) - F(a) = F(x_n) - F(x_0) \,.

Now, we add each ''F''(''x''''i'') along with its additive inverse, so that the resulting quantity is equal:
:\begin{matrix} F(b) - F(a) & = & F(x_n)\,+\, + F(x_1) \,-\,F(x_0) \, \
& = & {Link without Title} \,+\, {Link without Title} \,+\, {Link without Title} \, \end{matrix}

The above quantity can be written as the following sum:
:F(b) - F(a) = \sum_{i=1}^n - F(x_{i-1}) \qquad (1)

Next we will employ the Mean Value Theorem . Stated briefly,

Let ''F'' be continuous on the closed interval ''b'' and differentiable on the open interval (''a'', ''b''). Then there exists some ''c'' in (''a'', ''b'') such that
:F'(c) = rac{F(b) - F(a)}{b - a}.

It follows that
:F'(c)(b - a) = F(b) - F(a). \,

The function ''F'' is differentiable on the interval ''b'' ; therefore, it is also differentiable and continuous on each interval ''x''''i''-1. Therefore, according to the mean value theorem (above),
:F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}). \,

Substituting the above into (1), we get
:F(b) - F(a) = \sum_{i=1}^n - x_{i-1}) .

The assumption implies F'(c_i) = f(c_i). Also, x_i - x_{i-1} can be expressed as \Delta x of partition i.

:F(b) - F(a) = \sum_{i=1}^n x_i) \qquad (2)

Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem , describes an approximation of the curve section it is drawn over. Also notice that \Delta x_i does not need to be the same for any value of i, or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with n rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve.

By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann Integral . That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity.

So, we take the limit on both sides of (2). This gives us
  :<math>F(b) - F(a) \lim_{\ \Delta \ o 0} \sum_{i=1}^n x_i) </math>