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| | Hello from Germany. I'm glad to came here. My first name is Angus. <br>I live in a town called Nalbach in east Germany.<br>I was also born in Nalbach 21 years ago. Married in November year 2011. I'm working at the backery.<br><br>Visit my web-site; [http://reclaimstonehenge.co.uk/wiki/index.php?title=User:SueLowerwjnsbm Fifa 15 coin generator] |
| 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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| |-
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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>
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| ::::::<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> | |
| ::::::<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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Hello from Germany. I'm glad to came here. My first name is Angus.
I live in a town called Nalbach in east Germany.
I was also born in Nalbach 21 years ago. Married in November year 2011. I'm working at the backery.
Visit my web-site; Fifa 15 coin generator