Throughout this handout, assume that f is a continuous function on [a, b].

FTC1 FTC2
(Proof)

[rev. 11/8/2002]

Assume that the FTC1 method of finding definite integrals is valid, i.e., , where G is any antiderivative of f. Let d [a, b] be a constant, and let G be the designated antiderivative. Then

(Q.E.D.)

FTC2 FTC1
(Proof)

[rev. 11/8/2002]

Assume that the FTC2 conclusion is valid, i.e., the derivative of an “accumulator function” of x (where the accumulator is defined as a definite integral with variable upper endpoint x) is merely the integrand evaluated at x. Let d [a, b] be a constant, and let H be the accumulator function where . By FTC2, therefore, H is an antiderivative of f. Since any two antiderivatives differ by a constant, any antiderivative G can be written as G(x) = H(x) + C, where C is a constant. Equivalently, H(x) = G(x) − C. Then

(Q.E.D.)