ConceptComplete

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) establishes the profound connection between differentiation and integration: they are inverse operations. Part I says integration followed by differentiation recovers the original function. Part II provides a practical method for evaluating definite integrals using antiderivatives. This is the cornerstone of calculus.


Statements

Theorem6.1FTC Part I

If f:[a,b]Rf : [a, b] \to \mathbb{R} is continuous, then the function

F(x)=axf(t)dtF(x) = \int_a^x f(t) \, dt

is differentiable on (a,b)(a, b) with F(x)=f(x)F'(x) = f(x) for all x(a,b)x \in (a, b).

Theorem6.2FTC Part II

If f:[a,b]Rf : [a, b] \to \mathbb{R} is continuous and FF is any antiderivative of ff (i.e., F=fF' = f), then

abf(x)dx=F(b)F(a).\int_a^b f(x) \, dx = F(b) - F(a).

RemarkRelationship

Part I constructs an antiderivative F(x)=axf(t)dtF(x) = \int_a^x f(t) \, dt for continuous ff. Part II uses any antiderivative to evaluate integrals. Together, they say: integration and differentiation are inverse operations.


Applications

ExampleEvaluating definite integrals

To compute 02(3x2+2x)dx\int_0^2 (3x^2 + 2x) \, dx, find an antiderivative: F(x)=x3+x2F(x) = x^3 + x^2. By FTC Part II,

02(3x2+2x)dx=F(2)F(0)=(8+4)0=12.\int_0^2 (3x^2 + 2x) \, dx = F(2) - F(0) = (8 + 4) - 0 = 12.

ExampleFTC Part I: derivative of an integral

Let G(x)=0xsin(t2)dtG(x) = \int_0^x \sin(t^2) \, dt. By FTC Part I, G(x)=sin(x2)G'(x) = \sin(x^2).

ExampleChain rule with FTC

Let H(x)=0x2et2dtH(x) = \int_0^{x^2} e^{-t^2} \, dt. By FTC Part I and the chain rule,

H(x)=e(x2)22x=2xex4.H'(x) = e^{-(x^2)^2} \cdot 2x = 2x e^{-x^4}.


Summary

The Fundamental Theorem of Calculus bridges differentiation and integration:

  • Part I: Differentiation undoes integration.
  • Part II: Antider ivatives compute definite integrals.
  • Applications: evaluating integrals, solving differential equations, physics.

See FTC Proof for the proof.