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Code Scholars Tutoring · AP Calculus AB

Integral properties and the evaluation theorem

Unit 6 · Topics 6.6, 6.7. Original AP Calculus AB notes, worked example, and self-check.

Understand the idea

Definite integrals add across adjacent intervals and reverse sign when bounds are exchanged. Linearity allows constants to factor out and sums to split. Evaluate using an antiderivative when available. Symmetry can simplify work: odd integrands have zero signed integral on symmetric bounds, while even integrands give twice the positive-half integral. State continuity or other sufficient conditions when using the FTC.

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

Worked example

Evaluate ∫02(3x2−1) dx\displaystyle \int_0^2(3x^2-1)\,dx.

An antiderivative is F(x)=x3−x\displaystyle F(x)=x^3-x. Therefore the integral is F(2)−F(0)=(8−2)−0=6\displaystyle F(2)-F(0)=(8-2)-0=6. The definite integral is a number, not a family with a constant.

Try it yourself

If ∫14f=7\displaystyle \int_1^4f=7 and ∫12f=3\displaystyle \int_1^2f=3, find ∫42f\displaystyle \int_4^2f.

Show solution and reasoning

First ∫24f=7−3=4\displaystyle \int_2^4f=7-3=4, then reverse the bounds to get ∫42f=−4\displaystyle \int_4^2f=-4.

Unit 6 MCQNext: Basic antiderivatives and substitution →