数学简单的微积分题目求解 一道数学题不过 关于微积分的 很简单
\u9ad8\u7b49\u6570\u5b66 \u4e00\u5230\u7b80\u5355\u7684\u79ef\u5206\u9898\uff1f\u222b(0->\u03c0/2) dt/(1+tant)
=\u222b(0->\u03c0/2) cost/(sint+cost) dt
=(1/2)\u222b(0->\u03c0/2) [ (sint+cost) + ( cost -sint ) ]/(sint+cost) dt
=(1/2)[ t +ln|sint+cost|]| (0->\u03c0/2)
=\u03c0/4
\u7403\u7684\u4f53\u79ef\uff1aV=4/3\u03c0r³\uff0c\u5219
dV=4\u03c0r²dr\u3002\u5f53r\u589e\u52a0\u8f83\u5c0f\u65f6\uff0c
DetV=4\u03c0r²Detr\uff0cDet\u6307\u589e\u52a0\u91cf
=4*3.14*5²*1/16=19.62 \u7acb\u65b9\u5398\u7c73
\u7b54\u6848\u53c2\u8003\u4e8e\u767e\u5ea6
给个好评
(8)
f(x)=4e^(-4x) .ln(x)
f'(x)=4( 1/x - 4lnx) e^(-4x)
f'(3) = 4( 1/3 - 4ln3) e^(-12)
(9)
f(x)=x^3.arctan(7x)
f'(x)
= x^3. { 7/[ 1+ (7x)^2]} + 3x^2.arctan(7x)
=[7x^3/(1+49x^2)] + 3x^2.arctan(7x)
(10)
(10)
G(u) = ln√[(2u+7)/(2u-7)]
=(1/2)ln(2u+7) -(1/2)ln(2u-7)
dG(u) =[1/(2u+7) - 1/(2u-7)] du
希望有所帮助
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