i*b
发帖数: 11 | 1 e(y2)dydx，
其中2是y的指数，括号整体是e的指数，y从2x积到2，x从0积到1
抱歉不能输入公式原样。 | s***e
发帖数: 911 | 2
我来提供一个做法(分部积分):
定义f[x]=Integate[Exp[y^2],{y,2*x,2}]
你的积分就是:
I=Integrate[f[x],{x,0,1}]=Intergrate[x*(f[x]/x),{x,0,1}]
=(1/2)*x^2*f[x]_{0,1}-Integrate[(1/2)*x^2*(-f/x^2+f'/x),{x,0,1}]
=A+(1/2)*I-(1/2)*Integrate[x*f'[x],{x,0,1}
=>
(3/2)*I=A-(1/2)*Integrate[x*f'[x],{x,0,1}
其中A=(1/2)*x^2*f[x]_{0,1}, f'[x]可以根据f[x]=Integate[Exp[y^2],{y,2*x,2}]
求出. 下面的计算看起来很简单,你自己作作看看吧.
【在 i*b 的大作中提到】
: e(y2)dydx，
: 其中2是y的指数，括号整体是e的指数，y从2x积到2，x从0积到1
: 抱歉不能输入公式原样。
| s*****g
发帖数: 352 | 3 Another solution, easy for hand-calculations:
If you note the area of integration: x=0..1 and y=2x..2,
it is a triangle on the x-y plane, with its three corners
at (0,0) (1,2) and (0,2). Like this:
|
-+---------------> y
|\**********|
| \********|
| \******|
| \****|
| \**|
| \|
| (x=1,y=2)
|
V x
We wish to integrate the function of exp(y^2) in this region.
The original integral integrate y first, and then x. We can
also change the order of integration, i.e., i | p*****e
发帖数: 58 | 4 Another solution, if not better than shaking's.
Stocks theorem:
Int{(curl V)z}ds over area A is equal to Int V.dl along the
counterclockwise boundary of A, namely (0, 0) -> (1, 2) ->
(0, 2) -> (0, 0). V is a vector field.
If we can find a V satisfying (curl V) = (dVy/dx - dVx/dy) =
exp(y^2), we can use that V to convert the area integral to
an line integral. Just look at the function exp(y^2) we can
get one such vector V = {0, x * exp(y^2), 0}, and the line
integral of this vector is 0 along (1
【在 s*****g 的大作中提到】
: Another solution, easy for hand-calculations:
: If you note the area of integration: x=0..1 and y=2x..2,
: it is a triangle on the x-y plane, with its three corners
: at (0,0) (1,2) and (0,2). Like this:
: |
: -+---------------> y
: |\**********|
: | \********|
: | \******|
: | \****|
|
|
