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 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 antiderivatives. This part of the theorem has invaluable practical applications, because it markedly simplifies the computation of definite integrals.

The Fundamental Theorem of Calculus, Part 1

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

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by

$F (x) = \int_{a}^{x} f (t) d t$

Then, F is continuous on [a, b], differentiable on the open interval (a, b), and

$F' (x) = f (x)$

for all x in (a, b).

The Fundamental Theorem of Calculus, Part 2

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

Let $f$ be a continuous real-valued function defined on a closed interval [a, b]. Let F be an antiderivative of $f$ , that is one of the infinitely many functions such that, for all x in [a, b],

$f (x) = F' (x)$

Then

$\int_{a}^{b} f (x) d x = F (b) -F (a)$

Examples

As an example, suppose you need to calculate

$\int_{2}^{5} x^{2} d x .$

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

$\int_{2}^{5} x^{2} d x = F (5) - F (2) = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39$

Or, suppose you need to calculate

$\frac{d}{d x} \int_{0}^{x} t^{3} d t$

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

$\frac{d}{d x} \int_{0}^{x} t^{3} d t = \frac{d}{d x} F (x) - \frac{d}{d x} F (O) = \frac{d}{d x} \frac{x^{4}}{4} = x^{3} .$

But this result could have been found much more easily as

$\frac{d}{d x} \int_{0}^{x} t^{3} d t = f (x) \frac{d x}{d x} - f (O) \frac{d O}{d x} = x^{3} .$