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# Sum rule in integration

The sum rule in integration is very simple, and it helps you to integrate sums. It is derived from the sum rule in differentiation.

First note that from the definition of integration as the reverse process of differentiation:

$u = \int \frac{du}{dx} dx$
$v = \int \frac{dv}{dx} dx$

$u + v = \int \frac{du}{dx} dx + \int \frac{dv}{dx} dx \quad \mbox{(1)}$

Now take the sum rule in differentiation:

$\frac{d}{dx} \left(u+v\right) = \frac{du}{dx} + \frac{dv}{dx}$

Integrate both sides with respect to x:

$u + v = \int \left( \frac{du}{dx} + \frac{dv}{dx}\right) dx \quad \mbox{(2)}$

So we have, looking at (1) and (2):

$u+v = \int \frac{du}{dx} dx + \int \frac{dv}{dx}dx$
$u+v = \int \left(\frac{du}{dx} + \frac{dv}{dx}\right) dx$

Therefore:

$\int \left(\frac{du}{dx} + \frac{dv}{dx}\right) dx = \int \frac{du}{dx} dx + \int \frac{dv}{dx} dx$

Now subsitute:

$f = \frac{du}{dx}$
$g = \frac{dv}{dx}$

Then:

$\int \left(f + g\right) dx = \int f dx + \int g dx$

Obviously now we can re-subsitute u and v back as different thing to what they were before:

$u = f$
$v = g$

This gives the most common form for the sum rule in integration:

$\int \left(u + v \right) dx = \int u dx + \int v dx$

This proof seems somewhat long-winded, but is actually quite concise. Brackets are used to ensure clarity.

This is the most important rule in integration, fundamental to most integrals you will do. 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:

$\int \left(e^x + \cos{x}\right) dx = \int e^x dx + \int \cos{x}\ dx = e^x + \sin{x} + C$

It is quite easy to see some other results which come obviously as a result of this rule. For example:

$\int \left(u-v\right)dx$
$= \int \left(u+\left(-v\right)\right) dx$
$= \int u dx + \int \left(-v\right)dx$
$= \int u dx + \left(-\int vdx\right)$
$= \int u dx - \int v dx$

This rule is basically the same as the sum rule, but with pluses instead of minuses. Thus, the sum rule can be written as:

$\int u \pm v dx = \int u dx \pm \int v dx$

Note that the proof above relied on the special case of the constant factor rule in integration with k=-1.

Another thing is that the sigma and integral signs can be changed around. That is:

$\int \sum^b_{r=a} \left(f\left(r,x\right)\right) dx = \sum^b_{r=a} \left( \int f\left(r,x\right) dx\right)$

This is simply because:

$\int \sum^b_{r=a} \left(f\left(r,x\right)\right) dx$
$= \int \left( f\left(a,x\right) + f\left(\left(a+1\right),x\right) + f\left(\left(a+2\right),x\right) + \dots + f\left(\left(b-1\right),x\right) + f\left(b,x\right) \right) dx$
$= \int f\left(a,x\right)dx + \int f\left(\left(a+1\right),x\right) dx + \int f\left(\left(a+2\right),x\right) dx + \dots + \int f\left(\left(b-1\right),x\right) dx + \int f\left(b,x\right) dx$
$= \sum^b_{r=a} \left(\int f\left(r,x\right) dx\right)$

Since the integral is similar to a sum anyway, this is hardly surprising.

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