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6.7 The Fundamental Theorem of Calculus and Definite Integrals

Syllabus
2020
Topic
6.7
Level
—

Learning objectives

—FUN-6.B—Evaluate definite integrals analytically using the Fundamental Theorem of Calculus• FUN-6.B Evaluate definite integrals analytically using the Fundamental Theorem of Calculus.• FUN-6.B.1 An antiderivative of a function f is a function g whose derivative is f.• FUN-6.B.2 If a function f is continuous on an interval containing a, the function defined by F(x) = ∫ₐˣ f(t)dt is an antiderivative of f for x in the interval.• FUN-6.B.3 If f is continuous on the interval [a, b] and F is an antiderivative of f, then ∫ₐᵇ f(x)dx = F(b) − F(a).• Enduring understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.

Evaluate a Definite Integral with the FTC

An antiderivative of ff is any function FF satisfying F′(x)=f(x)F'(x)=f(x). If ff is continuous on [a,b][a,b], the Fundamental Theorem of Calculus evaluates its definite integral by the net change in an antiderivative.

\int_a^b f(x),dx=F(b)-F(a)\qquad\text{when }F'=f

First confirm that the integrand is continuous on the interval. Find one antiderivative FF, substitute the upper endpoint and lower endpoint separately, and compute upper minus lower. A quick derivative check of FF protects against power-rule and coefficient errors.

Evaluate ∫13(2x2−4x+1) dx\int_1^3(2x^2-4x+1)\,dx. A polynomial is continuous, and one antiderivative is F(x)=23x3−2x2+xF(x)=\tfrac23x^3-2x^2+x. Then F(3)=3F(3)=3 and F(1)=−13F(1)=-\tfrac13, so ∫13(2x2−4x+1) dx=3−(−13)=103\int_1^3(2x^2-4x+1)\,dx=3-(-\tfrac13)=\tfrac{10}{3}.

Use F(b)−F(a)F(b)-F(a), not the reverse. You may write +C+C while describing the family of antiderivatives, but it is unnecessary in a definite-integral evaluation because (F(b)+C)−(F(a)+C)(F(b)+C)-(F(a)+C) cancels the constant.