马上考试高数一点不会啊。。。。极限与连续。。求大神解答! 2.3.4.5.6题 高数求大神解4,5,6要过程
\u9ad8\u6570\u6c42\u6781\u9650\uff0c\u7b2c\u4e00\u9898\u76842.3.4.5.62.-1/3
3.-2/3
4.-e/2
5.-1
6.e^((lna+lnb)/2)
4\u3001\u4ee4t=\u221ax\uff0c\u5219x=t^2\uff0cdx=2tdt
f(x)=e^x+2x\u222b(0,1)tf(t)dt
xf(x)=xe^x+2x^2*\u222b(0,1)tf(t)dt
\u222b(0,1)xf(x)dx=\u222b(0,1)xe^xdx+\u222b(0,1)2x^2dx*\u222b(0,1)tf(t)dt
=\u222b(0,1)xd(e^x)+(2/3)*x^3|(0,1)*\u222b(0,1)tf(t)dt
=xe^x|(0,1)-\u222b(0,1)e^xdx+(2/3)*\u222b(0,1)tf(t)dt
=e-e^x|(0,1)+(2/3)*\u222b(0,1)tf(t)dt
=1+(2/3)*\u222b(0,1)tf(t)dt
\u5373\u222b(0,1)xf(x)dx=\u222b(0,1)tf(t)dt=3
f(x)=e^x+6x
5\u3001\u4ee4t=\u221a(x-2)\uff0c\u5219x=t^2+2\uff0cdx=2tdt
\u539f\u5f0f=\u222b(0,+\u221e) 2/(t^2+9)dt
=(2/3)*arctan(t/3)|(0,+\u221e)
=(2/3)*(\u03c0/2)
=\u03c0/3
6\u3001\u222b(1/2,\u221a3/2) x^2/\u221a(1-x^2)dx\uff0c\u4ee4x=sint\uff0cdx=costdt
=\u222b(\u03c0/6,\u03c0/3) sin^2tdt
=(1/2)*\u222b(\u03c0/6,\u03c0/3) (1-cos2t)dt
=(1/2)*[t-(1/2)*sin2t]|(\u03c0/6,\u03c0/3)
=(1/2)*(\u03c0/3-\u221a3/4-\u03c0/6+\u221a3/4)
=\u03c0/12
\u6240\u4ee5\u5e73\u5747\u503c=(\u03c0/12)/(\u221a3/2-1/2)
=\u03c0/6(\u221a3-1)
=(\u221a3+1)\u03c0/12
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