f（x）=e^2x,则∫（0,1）xf′（x）dx答案是1／2（e^2x+1)
问题描述：
f（x）=e^2x,则∫（0,1）xf′（x）dx
答案是1／2（e^2x+1)
答
= (1/2)∫xe^(2x)d(2x)
= (1/2)∫xd(e^(2x))
= (1/2)[xe^(2x) - ∫e^(2x)dx]
= (1/2)[xe^(2x) - (1/2)∫e^(2x)d(2x)]
= (1/2)e^(2x)[x - (1/2)] (0 -> 1)
= (1/2)e^2(1 - 1/2) - (1/2)(0 - 1/2)
= (e^2 + 1)/4
答
解:
∫[0,1]xf'(x)dx=xf(x)|[0,1]-∫[0,1]f'(x)dx
=xf(x)|[0,1]-f(x)|[0,1]
=e^2-0-(e^2-1)=1
答
∫[0,1]xf'(x)dx
=∫[0,1]2xe^2x dx
=∫[0,1]x de^2x
=x e^2x[0,1]-∫[0,1]e^2x dx
=e^2-1/2 e^2x [0,1]
=e^2-1/2( e^2-1)
=1/2( e^2+1)
我做的是正确的!