Re: [微積] 變數代換後的上下限疑問
※ 引述《znmkhxrw (QQ)》之銘言:
: ※ 引述《YmemY (**米)》之銘言:
: : 有個積分:
: : 1 ___
: : ∫ √2-x^2 dx 令x=√2cosθ 上下限要改成怎樣才對呢?
: : 0
: : 是π/2 ~ π/4 還是 -π/2 ~ -π/4 ??
: : 我認為dx是正的,如果上下限用第一種改法,dθ會是負的,所以
: : dx= √2(-sinθ)(-dθ) 這樣對嗎?
: : 反之,第二種上下限因為dx和dθ都是正的,所以不用加負號.
: : (如果我的想法正確的話,書上的解答就錯了~)
我做一次喔~~
1. θ:π/2 ~ π/4
1 ___
∫ √2-x^2 dx , x=√2cosθ
0
π/4 ______
=∫ √2-2(cosθ)^2 √2(-sinθ)dθ
π/2
π/4
=∫ √2│sinθ│√2(-sinθ)dθ
π/2
因為在此區間 sinθ is positive
π/4
= 2∫ (sinθ)(-sinθ)dθ
π/2
π/4
= 2∫ -(sinθ)^2 dθ
π/2
2. θ:-π/2 ~ -π/4
1 ___
∫ √2-x^2 dx , x=√2cosθ
0
-π/4 ______
=∫ √2-2(cosθ)^2 √2(-sinθ)dθ
-π/2
-π/4
=∫ √2│sinθ│√2(-sinθ)dθ
-π/2
因為在此區間 sinθ is negative
-π/4
= 2∫ (-sinθ)(-sinθ)dθ
-π/2
-π/4
= 2∫ (sinθ)^2 dθ
-π/2
算一算答案是一樣的
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在Change of variables 的證明:
假設 g : [p,q] → R is differentiable
f : [a,b] → R is continuous
g([p,q]) 包含於 [a,b]
we want to prove
b q
∫ f(x)dx = ∫ f(g(t))g'(t)dt , where g(p)=a , g(q)=b
a p
Since f is cont.
By F.C.T. there exists F€C^1([a,b]) s.t. F'(x) = f(x)
so Left hand side say:
b b
∫ f(x)dx = ∫ F'(x)dx = F(b) - F(a)
a a
and Right hand side say:
q q q
∫ f(g(t))g'(t)dt = ∫ F'(g(t))g'(t)dt = ∫ d(F(g(t)))/dt dt
p p p
q
= F(g(t))│
p
= F(b) - F(a)
看證明過程
g(p) = a , g(q) = b
沒有規定不能 g(p) = g(p') = a
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