# 緊急!!微積分的逐次積分 !!20點!!

20點立刻送上!!

### 1 個解答

• Tony
Lv 4
7 年前
最佳解答

這是我的標記, 這是不正式, 請不要在其他地方使用

(2∫1)(x)dx=積分x由1到2

2[x]1=(2-1)

解釋: (2∫1)(xy)dx

因為dx是對象x, 所以可以先把y當成常數,

(2∫1)(xy)dx = 2[(1/2)(x^2)(y)]1

然後2和1是作用x身上,

所以[(1/2)(2^2)(y) - (1/2)(1^2)(y)]

1. (ln2∫0)[(y∫0)(e)^(3x+y)dx]dy

=(ln2∫0) {y[(1/3)(e)^(3x+y)]0} dy

=(ln2∫0) [(1/3)(e)^(3y+y) - (1/3)(e)^(3(0)+y)] dy

=(ln2∫0) [(1/3)(e)^(4y) - (1/3)(e)^(y)] dy

=ln2[(1/12)(e)^(4y)]0 - ln2[(1/3)(e)^(y)]0

=[(1/12)(e)^(4ln2) - (1/12)(e)^(4(0))]

- [(1/3)(e)^(ln2) - [(1/3)(e)^(0)]

=[(1/12)(e)^(ln2^4) - (1/12)] - [(1/3)(e)^(ln2) - (1/3)]

=[(1/12)(2^4) - (1/12)] - [(1/3)(2) - (1/3)]

=16/12 - 1/12 - 2/3 + 1/3

=11/12

2. (1∫0)[(x^2∫x)(2x+xy)dy]dx

=(1∫0) {x^2[(2xy+(1/2)(xy^2)]x} dx

=(1∫0) { [(2x)(x^2)+(1/2)(x)(x^2)^2] - [(2x)(x)+(1/2)(x)(x)^2] } dx

=(1∫0) { [(2x^3)+(1/2)(x^5)] - [(2x^2)+(1/2)(x^3)] } dx

=(1∫0) [ (2x^3)+(1/2)(x^5) - (2x^2) - (1/2)(x^3) ] dx

=(1∫0) [ (1/2)(x^5) + (3/2)(x^3) - (2x^2) ] dx

=1[(1/12)(x^6) + (3/8)(x^4) - (2/3)(x^3)]0 dx

=[(1/12)(1^6) + (3/8)(1^4) - (2/3)(1^3)]

- [(1/12)(0^6) + (3/8)(0^4) - (2/3)(0^3)]

=(1/12) + (3/8) - (2/3)

= -5/24

3. (e∫1)[(x∫0)(lnx)dy]dx

=(e∫1) {x[(lnx)(y)]0} dx

=(e∫1) [(lnx)(x) - (lnx)(0)] dx

=(e∫1)[(lnx)(x)]dx

(dx^2)/dx = 2x

=> 2xdx=dx^2

=> xdx=(1/2)dx^2

∫ [(lnx)(x)]dx

=(1/2) ∫ [(lnx)]d(x^2)

=(1/2)(x^2)(lnx) - (1/2) ∫ [(x^2)]d(lnx)

=(1/2)(x^2)(lnx) - (1/2) ∫ [(x^2)(1/x)]dx

=(1/2)(x^2)(lnx) - (1/2) ∫ [(x)]dx

=(1/2)(x^2)(lnx) - (1/4)(x^2)

所以,(e∫1)[(x∫0)(lnx)dy]dx

=(e∫1)[(lnx)(x)]dx

=e[(1/2)(x^2)(lnx) - (1/4)(x^2)]1

=[(1/2)(e^2)(lne) - (1/4)(e^2)] - [(1/2)(e^2)(ln1) - (1/4)(1^2)]

=[(1/2)(e^2) - (1/4)(e^2)] - [0 - (1/4)]

=(1/4)(e^2) + (1/4)

=(1/4)(e^2 + 1)

積分的步驟好容易出錯, 但現在你應該知道怎樣計算吧

因為我是香港人, 所以可能打的中文數學用語有點奇怪(我用英文學的)

希望你看得懂.

希望幫到你!

參考資料： 自己