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On this page
  • Rules
  • Sum/Difference
  • Constant multiple
  • Reverse interval
  • Zero-length interval
  • Adding intervals
  • Example

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  1. High Level Math
  2. Integration
  3. Definite integral

Properties of definite integral

PreviousDefinite integralNextSecond fundamental theorem of calculus

Last updated 6 years ago

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Rules

Sum/Difference

∫ab[f(x)±g(x)]dx=∫abf(x)dx±∫abg(x)dx\int_a^b [f(x) \pm g(x)] dx = \int_a^b f(x) dx \pm \int_a^b g(x) dx∫ab​[f(x)±g(x)]dx=∫ab​f(x)dx±∫ab​g(x)dx

Constant multiple

∫abk⋅f(x)dx=k∫abf(x)dx\int_a^b k \cdot f(x) dx = k \int_a^b f(x) dx∫ab​k⋅f(x)dx=k∫ab​f(x)dx

Reverse interval

∫abf(x)dx=−∫baf(x)dx\int_a^b f(x) dx = - \int_b^a f(x) dx∫ab​f(x)dx=−∫ba​f(x)dx

Zero-length interval

Adding intervals

Example

∫aaf(x)dx=0\int_a^a f(x) dx = 0∫aa​f(x)dx=0
∫abf(x)dx+∫bcf(x)dx=∫acf(x)dx\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx∫ab​f(x)dx+∫bc​f(x)dx=∫ac​f(x)dx

Given ∫−31g(x)dx=6\int_{-3}^{1} g(x) dx = 6∫−31​g(x)dx=6 and ∫15g(x)dx=8\int_{1}^{5} g(x) dx = 8∫15​g(x)dx=8

∫−35g(x)dx=6+8=14\int_{-3}^5 g(x) dx = 6 + 8 = 14∫−35​g(x)dx=6+8=14

Given ∫−13f(x)dx=−2\int_{-1}^{3} f(x) dx = -2∫−13​f(x)dx=−2 and ∫−13g(x)dx=5\int_{-1}^{3} g(x) dx = 5∫−13​g(x)dx=5

∫−13(3f(x)−2g(x))dx=−6−10=−16\int_{-1}^{3} (3f(x) - 2g(x)) dx = -6 - 10 = -16∫−13​(3f(x)−2g(x))dx=−6−10=−16

Given ∫−51g(x)dx=3\int_{-5}^{1} g(x) dx = 3∫−51​g(x)dx=3 and ∫−31g(x)dx=7\int_{-3}^{1} g(x) dx = 7∫−31​g(x)dx=7

∫−5−3g(x)dx=3−7=−4\int_{-5}^{-3} g(x) dx = 3 - 7 = -4∫−5−3​g(x)dx=3−7=−4