There might be those of you who ask the question, “Why should I care about this debate about pants, because I am a girl?”, to which I reply, “Because you might want to sit on a penis and it might be encased in these pants.” Just because boys smell of meat, and don’t talk about feelings or like stroking kittens, it doesn’t mean that their undergarments should not be worshipped with the same reverence menfolk apply to the female knicker.

Paul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward Hawkwindses

or in an even shorter form, if we let u = f(x), v = g(x) and the differentials du = f ?(x) dx and dv = g?(x) dx, then it is in the form in which it is most often seen:

\int u\, dv=uv-\int v\, du.\!

Note that the original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ?g f? dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f and g, but the drawback of a nested integral:

\int f g\, dx = f \int g\, dx - \int \left ( f' \int g\, dx \right )\, dx.\!

This formula is valid whenever f is continuously differentiable and g is continuous.

More general formulations of integration by parts exist for the Riemann-Stieltjes integral and Lebesgue-Stieltjes integral.

Note: More complicated forms such as the one below are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

## (w)

## this is one messy period

.

## I'm so tempted to start that thread on the lipster.

## please do

## better than this surely?

http://www.thelipster.com/articles/2862777

## ...

There might be those of you who ask the question, “Why should I care about this debate about pants, because I am a girl?”, to which I reply, “Because you might want to sit on a penis and it might be encased in these pants.” Just because boys smell of meat, and don’t talk about feelings or like stroking kittens, it doesn’t mean that their undergarments should not be worshipped with the same reverence menfolk apply to the female knicker.

## I knew someone called fiona

## my sharona!

## THAT'S NOT THE RULES OF THIS THREAD

GOD YOU'RE SO HUNKY

## can everyone stop saying "eh"?

## ELLA ELLA ELLA

## damn

i was gonna put that

## you snooze you lose

## stupid english fucker.

:(

## Paul Silence and Thee Awkward Hawkwindses

Paul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward HawkwindsesPaul Silence and Thee Awkward Hawkwindses

## 10 votes

## .

) are two continuously differentiable functions. Then the integration by parts rule states that given an interval with endpoints a, b, one has

\int_a^b f(x) g'(x)\, dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\, dx\!

where we use the common notation

\left[ f(x) g(x) \right]_{a}^{b} = f(b) g(b) - f(a) g(a).\!

The rule is shown to be true by using the product rule for derivatives and the fundamental theorem of calculus. Thus

f(b)g(b) - f(a)g(a)\! = \int_a^b \frac{d}{dx} ( f(x) g(x) )\, dx\!

=\int_a^b f'(x) g(x) \, dx + \int_a^b f(x) g'(x)\, dx.\!

In the traditional calculus curriculum, this rule is often stated using indefinite integrals in the form

\int f(x) g'(x)\, dx = f(x) g(x) - \int f'(x) g(x)\, dx\!

or in an even shorter form, if we let u = f(x), v = g(x) and the differentials du = f ?(x) dx and dv = g?(x) dx, then it is in the form in which it is most often seen:

\int u\, dv=uv-\int v\, du.\!

Note that the original integral contains the derivative of g; in order to be able to apply the rule, the antiderivative g must be found, and then the resulting integral ?g f? dx must be evaluated.

One can also formulate a discrete analogue for sequences, called summation by parts.

An alternative notation has the advantage that the factors of the original expression are identified as f and g, but the drawback of a nested integral:

\int f g\, dx = f \int g\, dx - \int \left ( f' \int g\, dx \right )\, dx.\!

This formula is valid whenever f is continuously differentiable and g is continuous.

More general formulations of integration by parts exist for the Riemann-Stieltjes integral and Lebesgue-Stieltjes integral.

Note: More complicated forms such as the one below are also valid:

\int u v\, dw = u v w - \int u w\, dv - \int v w\, du.\!

## why did you block me on msn?

## I didnt

i just never use it anymore

## The End is near

but first this commercial

## _hi

## wow

they just came on the radio

## SPOOKY

## what was the point of this thread?

## eh Fiona?

## So we can all learn about integration and Muse?

## Can someone explain it to me?

## this is mildly funny now?

eh?

## I'm SO glad this thread was bumped today.

I forgot to bookmark it last night.

## I'm a Fiona

This is very confusing.