Re: [討論] 微積分~
※ 引述《greg741010 (昇哥)》之銘言：
: 一定很多人很好奇
: 我出的最後一題
: ∫e^(X^2)dX
: 這答案
: 到底可不可以用大一所學的微積分得到不定積分的答案
: (幾乎所有的回答都是:
: 1.拉普拉斯
: 2.極座標(你們應該還沒有學到) )
: 而以上兩種都可以在:物理數學中學到
: (或許極座標這個方法可以在學微積分的大一就會用)
: 我在這裡以另一種方式將這不定積分推導出來:
: =============================開始=====================================
: ∫exp(X^2)dX
: First of all, we need to use a really useful skill, partial fractions.
: It is a technique that we often used to solve an integration which we
: can't solve by using substitution or some other method.
: We let u = exp(X^2) and dv = dX, after computing, we get:
: du = 2Xexp(X^2)dX , v = X
: ∴∫exp(X^2)dX = Xexp(X^2) - ∫2(X^2(exp(X^2))dX
: After that, we can use substitution:
: let X = sinY , dX = cosYdY
: Take our assume into our function:
: ∫exp(X^2)dX = Xexp(X^2) - (∫2((sinY)^2)(exp((sinY)^2))cosYdY)
: Now we need to solve a new intergral:∫2((sinY)^2)(exp((sinY)^2))cosYdY
: We can use a equation: the sum of sine's square and cosine's square is 1,
: and get:
: ∫exp(X^2)dX = Xexp(X^2) - (∫2((sinY)^2)(exp((sinY)^2))cosYdY)
: = Xexp(X^2) - (∫2(1-(cosY)^2)(exp((sinY)^2))cosYdY)
: after that, we can get:
: 3∫exp(X^2)dX = Xexp(X^2) - (∫2((cosY)^3)(exp((sinY)^2))dY)
: And now we need to use the same method that we used at first: partial
: fractions. We let u = (cosY)^2 , dv = (exp((sinY)^2))cosYdY
: So, we can also know that: du = -2sinYcosYdY , v = exp((sinY)^2)
: Take our result into our function:
: 3∫exp(X^2)dX = Xexp(X^2) - (∫2((cosY)^3)(exp((sinY)^2))dY))
: = Xexp(X^2) - ((cosY)^2)exp((sinY)^2) + ~
: ~ = ∫exp((sinY)^2)*(-2sinYcosYdY)
: According to equation:sin(2α) = 2sinαcosα
: cos(2β) = 1 - 2(sinβ)^2
: ∴ ~ = ∫exp((sinY)^2)*(-2sinYcosYdY)
: = -∫exp((1/2)(1 - cos(2Y)))*(sin(2Y))dY
: = -(1/2)∫exp((1/2)(1 - cos(2Y)))*(sin(2Y))d2Y
: = -(1/2)(exp(1/2))∫exp((1/2)(- cos(2Y)))*(sin(2Y))d2Y
: Let S = -cos(2Y), dS = sin(2Y)d2Y
: ~ = -(1/2)(exp(1/2))∫exp((1/2)(- cos(2Y)))*(sin(2Y))d2Y
: = -(1/2)(exp(1/2))∫exp((1/2)(S)*dS
: = -(exp(1/2))∫exp((1/2)(S)*d(1/2)S
: = -(exp(1/2))*exp((1/2)(S)
: = -exp((1/2)(1-cos2Y)) = exp((sinY)^2)
: Take it back to our function:
: 3∫exp(X^2)dX = Xexp(X^2) - (∫2((cosY)^3)(exp((sinY)^2))dY))
: = Xexp(X^2) - ((cosY)^2)*exp((sinY)^2) - exp((sinY)^2)
: = Xexp(X^2) - [exp((sinY)^2]*[((cosY)^2) + 1]
: = Xexp(X^2) - [exp((sinY)^2]*[2 - ((sinY)^2)]
: Take our substitution away, that is:
: 3∫exp(X^2)dX = Xexp(X^2) - [exp((sinY)^2]*[2 - ((sinY)^2)]
: = Xexp(X^2) - [exp(X^2)]*[2 - X^2]
: = exp(X^2) * [X - 2 + X^2]
: =========================================================================
: 不知道這樣對不對~~~
: 打的好累~~~
: 跟電動王一樣
: 解15分鐘打1個小時
: XD...
http://tw.knowledge.yahoo.com/question/question?qid=1405101500701
http://www.physicsforums.com/showthread.php?t=30928
http://www.physicsforums.com/showthread.php?t=25798&page=2
如果補習班老師沒晃點我的話
這個不定積分應該不存在 只有在有上下限時才是高斯積分
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