Question

高数求大神解4,5,6要过程

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Answer 1

4、令t=√x，则x=t^2，dx=2tdt
f(x)=e^x+2x∫(0,1)tf(t)dt
xf(x)=xe^x+2x^2*∫(0,1)tf(t)dt
∫(0,1)xf(x)dx=∫(0,1)xe^xdx+∫(0,1)2x^2dx*∫(0,1)tf(t)dt
=∫(0,1)xd(e^x)+(2/3)*x^3|(0,1)*∫(0,1)tf(t)dt
=xe^x|(0,1)-∫(0,1)e^xdx+(2/3)*∫(0,1)tf(t)dt
=e-e^x|(0,1)+(2/3)*∫(0,1)tf(t)dt
=1+(2/3)*∫(0,1)tf(t)dt
即∫(0,1)xf(x)dx=∫(0,1)tf(t)dt=3
f(x)=e^x+6x
5、令t=√(x-2)，则x=t^2+2，dx=2tdt
原式=∫(0,+∞) 2/(t^2+9)dt
=(2/3)*arctan(t/3)|(0,+∞)
=(2/3)*(π/2)
=π/3
6、∫(1/2,√3/2) x^2/√(1-x^2)dx，令x=sint，dx=costdt
=∫(π/6,π/3) sin^2tdt
=(1/2)*∫(π/6,π/3) (1-cos2t)dt
=(1/2)*[t-(1/2)*sin2t]|(π/6,π/3)
=(1/2)*(π/3-√3/4-π/6+√3/4)
=π/12
所以平均值=(π/12)/(√3/2-1/2)
=π/6(√3-1)
=(√3+1)π/12