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6、∫(0,1)xf"(2x)dx
=(1/2)∫(0,1)xdf'(2x)
=(1/2)xf'(2x)|(0,1)-(1/2)∫(0,1)f'(2x)dx
=(1/2)xf'(2x)|(0,1)-(1/4)∫(0,1)df(2x)
=(1/2)xf'(2x)|(0,1)-(1/4)f(2x)|(0,1)
=(1/2)f'(2)-(1/4)[f(2)-f(0)]
=2
7、∫(0,x)tf(t)dt=√(x²+9)-3
∫(0,3)x³f(x²)dx
=(1/3)∫(0,9)x²f(x²)dx²
=(1/3)[√(9²+9)-3]
=√10-1
8、f(x)=e^x+x∫(0,1)f(√x)dx=e^x+x∫(0,1)f(√t)dt
设∫(0,1)f(√t)dt=a,则f(x)=e^x+ax
a=∫(0,1)f(√t)dt=∫(0,1)f(u)du²=2∫(0,1)u(e^u+au)du
=2∫(0,1)(ue^u+au²)du
=2∫(0,1)(ue^u+au²)du
=2(1+a/3)
a=6
=(1/2)∫(0,1)xdf'(2x)
=(1/2)xf'(2x)|(0,1)-(1/2)∫(0,1)f'(2x)dx
=(1/2)xf'(2x)|(0,1)-(1/4)∫(0,1)df(2x)
=(1/2)xf'(2x)|(0,1)-(1/4)f(2x)|(0,1)
=(1/2)f'(2)-(1/4)[f(2)-f(0)]
=2
7、∫(0,x)tf(t)dt=√(x²+9)-3
∫(0,3)x³f(x²)dx
=(1/3)∫(0,9)x²f(x²)dx²
=(1/3)[√(9²+9)-3]
=√10-1
8、f(x)=e^x+x∫(0,1)f(√x)dx=e^x+x∫(0,1)f(√t)dt
设∫(0,1)f(√t)dt=a,则f(x)=e^x+ax
a=∫(0,1)f(√t)dt=∫(0,1)f(u)du²=2∫(0,1)u(e^u+au)du
=2∫(0,1)(ue^u+au²)du
=2∫(0,1)(ue^u+au²)du
=2(1+a/3)
a=6
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