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| | Nice to satisfy you, I am Marvella Shryock. In her professional life she is a payroll clerk but she's usually needed her own business. California is our birth location. One of the things she loves most is to study comics and she'll be beginning something else alongside with it.<br><br>Check out my homepage - at home std testing, [http://gjycorp.com/Gcentre_Advisors/5743 see more], |
| In [[calculus]], the '''sum rule in integration''' states that the integral of a sum of two functions is equal to the sum of their integrals. It is of particular use for the [[integral|integration]] of [[sum]]s, and is one part of the [[linearity of integration]].
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| As with many properties of integrals in calculus, the sum rule applies both to [[definite integral]]s and [[indefinite integral]]s. For indefinite integrals, the sum rule states
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| :<math>\int \left(f + g\right) \,dx = \int f \,dx + \int g \,dx</math>
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| ==Application to indefinite integrals==
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| For example, if you know that the [[integral]] of exp(x) is exp(x) from [[calculus with exponentials]] and that the [[integral]] of cos(x) is sin(x) from [[calculus with trigonometry]] then:
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| :<math>\int \left(e^x + \cos{x}\right) \,dx = \int e^x \,dx + \int \cos{x}\ \,dx = e^x + \sin{x} + C</math>
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| Some other general results come from this rule. For example:
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| {|
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| |<math>\int \left(u-v\right)dx</math>
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| |<math>= \int u+\left(-v\right) \,dx</math>
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| |-
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| |<math>= \int u \,dx + \int \left(-v\right)\,dx</math>
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| |-
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| |<math>= \int u \,dx + \left(-\int v\,dx\right)</math>
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| |-
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| |<math>= \int u \,dx - \int v \,dx</math>
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| |}
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| The proof above relied on the special case of the [[constant factor rule in integration]] with k=-1.
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| Thus, the sum rule might be written as:
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| :<math>\int (u \pm v) \,dx = \int u\, dx \pm \int v\, dx</math>
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| Another basic application is that sigma and integral signs can be changed around. That is:
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| :<math>\int \sum^b_{r=a} f\left(r,x\right)\, dx = \sum^b_{r=a} \int f\left(r,x\right) \,dx</math>
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| This is simply because:
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| :<math>\int \sum^b_{r=a} f(r,x)\, dx</math>
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| :<math> = \int f\left(a,x\right) + f((a+1),x) + f((a+2),x) + \dots </math> | |
| ::::::<math>+ f((b-1),x) + f(b,x)\, dx</math>
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| :<math> = \int f(a,x)\,dx + \int f((a+1),x)\, dx + \int f((a+2),x) \,dx + \dots </math>
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| ::::::<math>+ \int f((b-1),x)\, dx + \int f(b,x)\, dx</math>
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| :<math> = \sum^b_{r=a} \int f(r,x)\, dx</math>
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| ==Application to definite integrals==
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| Passing from the case of indefinite integrals to the case of integrals over an interval [a,b], we get exactly the same form of rule (the [[arbitrary constant of integration]] disappears).
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| ==The proof of the rule==
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| First note that from the definition of [[integral|integration]] as the [[antiderivative]], the reverse process of [[derivative|differentiation]]:
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| :<math>u = \int \frac{du}{dx} \,dx</math>
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| :<math>v = \int \frac{dv}{dx} \,dx</math>
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| [[sum|Adding]] these,
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| :<math>u + v = \int \frac{du}{dx} \,dx + \int \frac{dv}{dx} \,dx \quad \mbox{(1)}</math>
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| Now take the [[sum rule in differentiation]]:
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| :<math>\frac{d}{dx} \left(u+v\right) = \frac{du}{dx} + \frac{dv}{dx}</math>
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| Integrate both sides with respect to x:
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| :<math>u + v = \int \left(\frac{du}{dx} + \frac{dv}{dx}\right) \,dx \quad \mbox{(2)}</math>
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| So we have, looking at (1) and (2):
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| :<math>u+v = \int \frac{du}{dx} \,dx + \int \frac{dv}{dx}\,dx</math>
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| :<math>u+v = \int \left(\frac{du}{dx} + \frac{dv}{dx}\right) \,dx</math>
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| Therefore:
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| :<math>\int \left(\frac{du}{dx} + \frac{dv}{dx}\right) \,dx = \int \frac{du}{dx} \,dx + \int \frac{dv}{dx} \,dx</math>
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| Now substitute:
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| :<math>f = \frac{du}{dx}</math>
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| :<math>g = \frac{dv}{dx}</math>
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| [[Category:Integral calculus]]
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Nice to satisfy you, I am Marvella Shryock. In her professional life she is a payroll clerk but she's usually needed her own business. California is our birth location. One of the things she loves most is to study comics and she'll be beginning something else alongside with it.
Check out my homepage - at home std testing, see more,