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Mathematics · Ch 11 — Integral Calculus

Properties of Integrals

11.5

Properties of Integrals

Integration behaves linearly with respect to constant multiples and sums/differences of functions — this is what lets a complicated-looking integrand be broken into manageable pieces and integrated term by term.

Note

Property 1. If kk is any constant, then

∫kf(x) dx=k∫f(x) dx.\int k f(x)\,dx = k\int f(x)\,dx.

Property 2. For two functions f1,f2f_1,f_2,

∫(f1(x)±f2(x)) dx=∫f1(x) dx±∫f2(x) dx.\int \big(f_1(x)\pm f_2(x)\big)\,dx = \int f_1(x)\,dx \pm \int f_2(x)\,dx.

Note 11.1 — combining and extending both properties. The two properties above combine, and extend to any finite number of terms, as

∫(k1f1(x)±k2f2(x)±k3f3(x)±⋯±knfn(x)) dx=k1 ⁣∫ ⁣f1(x)dx±k2 ⁣∫ ⁣f2(x)dx±k3 ⁣∫ ⁣f3(x)dx±⋯±kn ⁣∫ ⁣fn(x)dx.\int \big(k_1f_1(x)\pm k_2f_2(x)\pm k_3f_3(x)\pm\cdots\pm k_nf_n(x)\big)\,dx = k_1\!\int\! f_1(x)dx \pm k_2\!\int\! f_2(x)dx \pm k_3\!\int\! f_3(x)dx \pm\cdots\pm k_n\!\int\! f_n(x)dx.

In plain words: the integral of a linear combination of finitely many functions equals the same linear combination of their individual integrals. This is the justification for the routine step, used constantly from here on, of integrating a multi-term expression one term at a time and simply adding up the results (each carrying its own working constant, which are then merged into one overall arbitrary constant cc).

Worked illustration (pure decomposition). To integrate 2cos⁡x−4sin⁡x+5sec⁡2x+cosec2x2\cos x - 4\sin x + 5\sec^2x + \text{cosec}^2x, split into four separate standard integrals and combine:

∫(2cos⁡x−4sin⁡x+5sec⁡2x+cosec2x) dx=2sin⁡x+4cos⁡x+5tan⁡x−cot⁡x+c.\int(2\cos x-4\sin x+5\sec^2x+\text{cosec}^2x)\,dx = 2\sin x + 4\cos x + 5\tan x - \cot x + c.

(Each coefficient is pulled out by Property 1, and each piece read off directly from the standard-integrals table of §11.3.)

Worked illustration (mixing decomposition with the linear-argument rule of §11.4). A realistic problem typically needs both properties of this section and the ∫f(ax+b)dx\int f(ax+b)dx shortcut together. For instance,

∫ ⁣(12(4x−5)3+63x+2+16e4x+3) ⁣dx=12 ⁣∫ ⁣(4x−5)−3dx+6 ⁣∫ ⁣dx3x+2+16 ⁣∫ ⁣e4x+3 dx,\int\!\left(\frac{12}{(4x-5)^3}+\frac{6}{3x+2}+16e^{4x+3}\right)\!dx = 12\!\int\!(4x-5)^{-3}dx + 6\!\int\!\frac{dx}{3x+2} + 16\!\int\! e^{4x+3}\,dx,

and each of the three pieces is then evaluated by §11.4's rule before being recombined:

=12(14) ⁣(−12(4x−5)2)+6(13)log⁡∣3x+2∣+16(14)e4x+3+c=−32(4x−5)2+2log⁡∣3x+2∣+4e4x+3+c.= 12\left(\frac14\right)\!\left(-\frac{1}{2(4x-5)^2}\right) + 6\left(\frac13\right)\log|3x+2| + 16\left(\frac14\right)e^{4x+3}+c = -\frac{3}{2(4x-5)^2}+2\log|3x+2|+4e^{4x+3}+c.

Similarly, a mixture of a reciprocal-square-root term (§11.4 inverse-trig case) with a cosec-cot product term decomposes the same way — each term integrated on its own by whichever standard/linear-argument rule applies, then the results simply added with their original signs: …