之前習慣在什麼章節就用甚麼方法...
到了現在總合在一起時卻不知道用甚麼樣的方法來解以下這一題
試過了分部積分法...但好像不是
請求各位幫忙解以下這題..
圖片參考：https://lh5.googleusercontent.com/68vB_8lcC_4X_JDZw84R8zx8t7H2gn2-zv3pbT_DZZubFf6nSGGM32fIkJeIgaGkOnYb7hOfVCmFiB9a-2bvXd1A0A=s512
關於微積分例題一問
2011-07-07 2:13 am
回答 (4)
2011-07-07 4:44 am
✔ 最佳答案
u=x^2+1----->x^2=u-11/2du=x dx
∫x^5*√(x^2+1) dx
=>∫x*[x^2]^2*√(x^2+1) dx
=(1/2)∫(u-1)^2*u^(1/2) du
=(1/2)∫(u^2-2u+1)*u^(1/2) du
=(1/2)∫u^(5/2)-2u^(3/2)+u^(1/2) du
=(1/2)[u^(7/2)-(4/5)u^(5/2)+(2/3)u^(3/2)]
=u^(7/2)/7-(2/5)u^(5/2)+u^(3/2)/3
=(x^2+1)^(7/2)/7-(2/5)(x^2+1)^(5/2)+(x^2+1)^(3/2)/3
範圍上線√3-下線0
=[128/7-64/5+8/3]-[1/7-2/5+1/3]
=127/7-62/5+7/3
=(1905-1302+245)/105
=848/105
2011-07-07 4:09 am
2011-07-07 2:40 am
另U=X平方+1 2Xdx=du
原式=X四次方*U1/2次方*1/2du
解出來會是(127*15-62*21+7*35)/105
原式=X四次方*U1/2次方*1/2du
解出來會是(127*15-62*21+7*35)/105
2011-07-07 2:33 am
∫(0 to √3)_x^5√(x^2+1)dx
Sol
Set y=√(x^2+1)
y^2=x^2+1
ydy=xdx
∫(0 to √3)_x^5√(x^2+1)dx
= ∫(0 to √3)_x^4√(x^2+1)(xdx)
= ∫(1 to 2)_(y^2－1)^2y(ydy)
= ∫(1 to 2)_y^6－2y^4+y^2dy
=y^7/7－2y^5/5+y^3/3｜(0 to 2)
=128/7－64/5+8/3
=856/105
Sol
Set y=√(x^2+1)
y^2=x^2+1
ydy=xdx
∫(0 to √3)_x^5√(x^2+1)dx
= ∫(0 to √3)_x^4√(x^2+1)(xdx)
= ∫(1 to 2)_(y^2－1)^2y(ydy)
= ∫(1 to 2)_y^6－2y^4+y^2dy
=y^7/7－2y^5/5+y^3/3｜(0 to 2)
=128/7－64/5+8/3
=856/105
收錄日期: 2021-05-02 10:44:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110706000010KK07653