The fundamental theorem of calculus when expressed most simply this states the integration? and differentiation are inverses of one another. i.e. if the function f(x) is continuous on some closed interval [a,b] containing the point c, then the function F(x) defined by:
is differentiable and its derivative is f(x). Proof: On the one hand, it is clear from the definition of the Riemann integral (see integration) that |
The fundamental theorem of calculus is the statement that the two central operations of calculus, differentiation and integration, are inverses of each other. This means that if a function is first integrated and then differentiated, the original function is retrieved. An important consequence of this, sometimes called the second fundamental theorem of calculus, allows one to compute integrals by using an antiderivative of the function to be integrated. |
(*) |I|infIf(x) ≤ ∫I f(x)dx ≤ |I|supIf(x) |
IntuitionIntuitively, the theorem simply says that if you know all the little instantanous changes in some quantity, then you may compute the overall change in the quantity by adding up all the little changes. |
where infIf(x) and |I|supIf(x) are, respectively, the greatest lower bound and the least upper bound of f(x) over the (closed) interval I, and |I| is the length of the interval I (ie, |[a,b]|=b-a.) Define |
To get a feeling for the statement, we will start with an example. Suppose you travel in a straight line, starting at time t = 0, and with varying speeds. If d(t) denotes the distance from the origin at time t and v(t) denotes the speed at time t, then v(t) is the instantaneous rate of change of d and is therefore the derivative of d. Suppose you know only v(t) from your speedometer, and you want to recover d(t). The fundamental theorem of calculus says that you should integrate v in order to get d. And this is exactly what you would have done, even without knowing that theorem: record the speed at regular intervals, maybe after 1 minute, 2 minutes, 3 minutes and so on, and then multiply the first speed with 1 minute to get an estimate for the distance covered in the first minute, then multiply the second speed with 1 minute to get the distance covered in the second minute etc., and then add all the distances up. In order to get an even better estimate of your current distance, you need to record the speeds at shorter time intervals. The limit as the length of the intervals approaches zero is exactly the definition of the integral of v. |
F(x):=∫[0,x]f(t)dt The mean value theorem states that |
Formal statementsStated formally, the theorem says that if the function f(x) is continuous on some closed interval [a, b], then the function F(x) defined by: x F(x) = ∫ f(t) dt a is differentiable on the whole interval and its derivative is f(x). |
(F(x)-F(0))/x=F'(t) for some t in [0,x]. |
Part II of the theorem gives an important method for computing the integral of the continuous function f: if F(x) is any antiderivative of the function f(x) (i.e. if F'(x) = f(x)), then b ∫ f(x) dx = F(b) - F(a) a As an example, suppose you need to calculate 5 ∫ x2 dx 2 Here, f(x) = x2 and we can use F(x) = 1/3 x3 as antiderivate. Therefore: 5 ∫ x2 dx = F(5) - F(2) = 125/3 - 8/3 = 117/3 = 39. 2 |
Using (*), we see that |
Generalizations |
inf[0,x]f(t) ≤ F'(t) ≤ |I|sup[0,x]f(t) |
We don't need to assume continuity of f on the whole interval. Part I of the theorem then says: if f is any Lebesgue integrable function on [a, b] and x0 is a number in [a, b] such that f is continuous at x0, then x F(x) = ∫ f(t) dt a is differentiable for x = x0 with F'(x0) = f(x0). |
Lastly, we use the continuity hypothesis (again) to get that 1) F'(t) is right continuous at 0 and 2) F'(0)=f(0). |
Part II of the theorem is true for any Lebesgue integrable function f which has an antiderivative F (not all integrable functions do, though). |
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The version of Taylor's theorem which expresses the error term as an integral can be seen as a generalization of the Fundamental Theorem. |
Part II of the Fundamental Theorem of Calculus states: <pre> |
There is a version of the theorem for complex functions: suppose U is an open set in C and f: U -> C is a function which has a holomorphic antiderivate F on U. Then for every curve γ : [a, b] -> U, the [curve integral]? can be computed as |
b ∫ f(x) dx = F(b) - F(a) a |
∫ f(z) dz = F(γ(b)) - F(γ(a)) γ |
</pre> |
The fundamental theorem can be generalized to curve and surface integrals in higher dimensions and on manifolds. The most powerful statement in this direction is Stoke's theorem. |
To get a feeling for the statement, we will start with an example. Suppose you travel in a straight line, starting at time t = 0, and with varying speeds. If d(t) denotes the distance from the origin at time t and v(t) denotes the speed at time t, then v(t) is the instantaneous rate of change of d and is therefore the derivative of d. Suppose you know only v(t) from your speedometer, and you want to recover d(t). The fundamental theorem of calculus says that you should integrate v in order to get d. And this is exactly what you would have done, even without knowing that theorem: record the speed at regular intervals, maybe after 1 minute, 2 minutes, 3 minutes and so on, and then multiply the first speed with 1 minute to get an estimate for the distance covered in the first minute, then multiply the second speed with 1 minute to get the distance covered in the second minute etc., and then add all the distances up. In order to get an even better estimate of your current distance, you need to record the speeds at shorter time intervals. The limit as the length of the intervals approaches zero is exactly the definition of the integral of v.
x F(x) = ∫ f(t) dt ais differentiable on the whole interval and its derivative is f(x).
Part II of the theorem gives an important method for computing the integral of the continuous function f: if F(x) is any antiderivative of the function f(x) (i.e. if F'(x) = f(x)), then
b ∫ f(x) dx = F(b) - F(a) aAs an example, suppose you need to calculate
5 ∫ x2 dx 2Here, f(x) = x2 and we can use F(x) = 1/3 x3 as antiderivate. Therefore:
5 ∫ x2 dx = F(5) - F(2) = 125/3 - 8/3 = 117/3 = 39. 2
We don't need to assume continuity of f on the whole interval. Part I of the theorem then says: if f is any Lebesgue integrable function on [a, b] and x0 is a number in [a, b] such that f is continuous at x0, then
x F(x) = ∫ f(t) dt ais differentiable for x = x0 with F'(x0) = f(x0).
Part II of the theorem is true for any Lebesgue integrable function f which has an antiderivative F (not all integrable functions do, though).
The version of Taylor's theorem which expresses the error term as an integral can be seen as a generalization of the Fundamental Theorem.
There is a version of the theorem for complex functions: suppose U is an open set in C and f: U -> C is a function which has a holomorphic antiderivate F on U. Then for every curve γ : [a, b] -> U, the [curve integral]? can be computed as
∫ f(z) dz = F(γ(b)) - F(γ(a)) γ
The fundamental theorem can be generalized to curve and surface integrals in higher dimensions and on manifolds. The most powerful statement in this direction is Stoke's theorem.