intergration

白羊座aries

Find the value of a when
intergrate 0 to infinity  [1/(x^2+a^)] = 1


做到1/[a (arc tan x/a)]infinity to 1 = 1 . 要怎么接下去？

puangenlun

1/(x^2+a^?)
integrate 1/(x^2+a^2) = (tan^(-1)(x/a))/a+constant
integrate 1/(x^2+a^2) from 0 to inf = pi/2a

白羊座aries

integrate 0 to 1 x ln x dx

puangenlun

integration by part
integral x log(x) dx from 0 to 1= -1/4

白羊座aries

integration by part
integral x log(x) dx from 0 to 1= -1/4
puangenlun 发表于 26-12-2011 02:35 AM

然后会拿到 1/2x^2 ln x -1/4x^2

当sub 0 ->ln x要怎样？

puangenlun

l'hospital rule

lim_(x->0) x^2 log(x) = 0

白羊座aries

l'hospital rule

lim_(x->0) x^2 log(x) = 0
puangenlun 发表于 27-12-2011 03:00 PM

是不是 x^2 / 1/ln x ? 然后 differentiate = 2x / 1/1/x  ,怎样拿到1/4?

另外， (3xy-2py^2)dx + ( x^2 - 2pxy )dy = 0  is not an exact equation .


Multiply the equation by f( x ) and determine the function f( x ) such that the equation become exact , 请问要怎样知道f ( x ) ?

puangenlun

lim_(x->0) x^2 ln(x)
=lim_(x->0) log(x) / (1 / x^2)
=lim_(x->0) (1 / x) / (-2 / x^3)
=lim_(x->0) -x^2 / 2
=0

puangenlun

"怎样拿到1/4?"

1/4是 integration by parts 出来的

puangenlun

(3xy-2py^2)dx + (x^2-2pxy)dy = 0
dg/dx=3xy-2py^2
dg/dy=x^2-2pxy
d2g/dxdy=x^2y-pxy^2
d2g/dydx=3x-4py

let f(x)=ax+b
since (3xy-2py^2)fdx + (x^2-2pxy)fdy is exact equation
d(3xy-2py^2)f/dy=d(x^2-2pxy)f/dx
(3x-4py)f=(2x-2py)f+ (x^2-2pxy)f'
(x-2py)(ax+b)=(x^2-2pxy)a
a=any constant and b=0
f=ax

白羊座aries

(3xy-2py^2)dx + (x^2-2pxy)dy = 0
dg/dx=3xy-2py^2
dg/dy=x^2-2pxy
d2g/dxdy=x^2y-pxy^2
d2g/dydx=3x- ...
puangenlun 发表于 7-1-2012 02:58 PM

  为什么你会intergrate  w.r.t y ？ d2g/dxdy 不是differentiate w.r.t x ? 2x -2py?

puangenlun

exact equation
f = 0
df/dx dx + df/dy dy = 0
d2f/dxdy = d2f/dydx

http://www.efunda.com/math/ode/ode1_exact.cfm