微積分「積分」

2011-05-16 7:02 pm
f(x) = ∫ [0 to 1] (y^x-1) / lny dy

求f(e-1) = ?

回答 (2)

2011-05-17 6:21 am
✔ 最佳答案
f(x)=∫[0~1] (y^x -1)/lny dy
=∫[0~1] ∫[0~x] y^t dt dy
=∫[0~x] ∫[0~1] y^t dy dt
=∫[0~x] 1/(t+1) dt
=ln| x+1 |
so, f(e -1)= ln|e|=1

Note: ∫[0~1] y^t dy = y^(t+1)/(t+1) sub. y=0~1
= 1/(t+1) - 0 = 1/(t+1)

2011-05-18 01:50:13 補充:
直接積分可能要用複變積分,沒有簡易作法,改為重積分較易
另法: 先求f'(x)得∫[0,1] y^x dy=1/(x+1), 再積分得ln(1+x), 最後代x=e-1
2011-05-17 11:51 pm
請問不行直接積分嗎

∫ [0 to 1] (y^(e-1)-1) / lny dy


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