AP Calculus AB and BC glossary

Fundamental Theorem of Calculus Part 2

Also called: FTC Part 2, Evaluation theorem

Part 2 of the Fundamental Theorem of Calculus says a definite integral equals any antiderivative evaluated at the upper bound minus the same antiderivative at the lower bound. It is what makes definite integrals computable without Riemann sums.

∫abf(x) dx=F(b)−F(a),where F′=f\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F' = f

Any antiderivative works, because the constant of integration cancels in the subtraction. That is why no plus C appears in a definite integral.

The theorem requires ff to be continuous on the closed interval. Applying it blindly across a vertical asymptote produces nonsense, sometimes even a negative value for an obviously positive area.

Change the bounds with the variable

After a substitution you must either convert the bounds to the new variable or convert back before evaluating. Mixing old bounds with a new variable is a routine and costly error.

Appears in: Unit 6: Integration and Accumulation