Re: [積分] 後醫92微積分
※ 引述《johnnyjaiu (芭樂)》之銘言：
: The area between the curve y=x(3x+1)^1/2 and the lines y=0,
: x=0 and x=1 is_____
: 這題是要代公式去算嗎？還是用分部積分就行了？
: 我的算式：S 1到-1/3 x(3x+1)^1/2 dx
: 請高手幫忙解答，小弟感謝m<_ _>m
分段積分(不是分部積分)
0 1
∫ -x(3x+1)^1/2 dx + ∫ x(3x+1)^1/2 dx
-1/3 0
1 4
=∫-[(u-1)/3]*u^1/2*(1/3)du + ∫[(u-1)/3]*u^1/2*(1/3)du (令u=3x+1)
0 1
1 4
=∫-[(u^3/2 - u^1/2)/9]du + ∫[(u^3/2 - u^1/2)/9]du
0 1
=-(1/9)[(2/5)u^5/2-(2/3)u^3/2]|1 + (1/9)[(2/5)u^5/2-(2/3)u^3/2]|4
| |
|0 |1
=-(1/9)[2/5-2/3] + (1/9)[(2/5)(32)-(2/3)(8)-2/5+2/3]
=4/135+116/135=120/135=8/9
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