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6.6 Applying Properties of Definite Integrals

Syllabus
2020
Topic
6.6
Level
—

Learning objectives

—FUN-6.A—Calculate a definite integral using areas and properties of definite integrals• FUN-6.A Calculate a definite integral using areas and properties of definite integrals.• FUN-6.A.1 In some cases, a definite integral can be evaluated by using geometry and the connection between the definite integral and area.• FUN-6.A.2 Properties of definite integrals include the integral of a constant times a function, the integral of the sum of two functions, reversal of limits of integration, and the integral of a function over adjacent intervals.• FUN-6.A.3 The definition of the definite integral may be extended to functions with removable or jump discontinuities.• Enduring understanding FUN-6: Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.

Combine Definite-Integral Properties

Use definite-integral properties when a requested integral can be built from known values or familiar geometric regions. Keep two sources of sign visible: regions below the axis contribute negatively, and reversing the integration limits changes the sign.

Property Rule
Constant multiple ∫abcf(x) dx=c∫abf(x) dx\int_a^b c f(x)\,dx=c\int_a^b f(x)\,dx
Sum ∫ab(f+g) dx=∫abf dx+∫abg dx\int_a^b(f+g)\,dx=\int_a^b f\,dx+\int_a^b g\,dx
Reversed limits ∫baf(x) dx=−∫abf(x) dx\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx
Adjacent intervals ∫acf dx+∫cbf dx=∫abf dx\int_a^c f\,dx+\int_c^b f\,dx=\int_a^b f\,dx

If ∫02f(x) dx=5\int_0^2 f(x)\,dx=5 and ∫27f(x) dx=−1\int_2^7 f(x)\,dx=-1, then ∫07f(x) dx=5+(−1)=4\int_0^7 f(x)\,dx=5+(-1)=4. Therefore ∫703f(x) dx=3[−∫07f(x) dx]=3(−4)=−12\int_7^0 3f(x)\,dx=3[-\int_0^7 f(x)\,dx]=3(-4)=-12. The negative result comes from reversing the limits, not from discarding the given signed value.

If a graph on [−2,2][-2,2] is the upper semicircle of radius 22, its integral is the semicircle's area, 12π(2)2=2π\tfrac12\pi(2)^2=2\pi. The same semicircle below the axis would contribute −2π-2\pi.

A removable discontinuity or a jump discontinuity can still allow a definite integral; changing one isolated function value does not change accumulated area because a point has zero width. This extension does not mean every discontinuous or unbounded function is automatically integrable.