f(2x+1)=xe^x,求∫(上限5,下限3)f(t)dt答案是2e^2,我算的不对,求f(2x+1)=xe^x,求∫(上限5,下限3)f(t)dt答案是2e^2,我算的不对,求详解,谢那x=(t-1)/2,f(t)=[(t-1)/2]e^(t-1)/2为什么不对?

Question

f(2x+1)=xe^x,求∫(上限5,下限3)f(t)dt 答案是2e^2,我算的不对,求
f(2x+1)=xe^x,求∫(上限5,下限3)f(t)dt 答案是2e^2,我算的不对,求详解,谢
那x=(t-1)/2,f(t)=[(t-1)/2]e^(t-1)/2为什么不对?

Answer 1

令t = 2x + 1,dt = 2 dx
t = 3 => x = 1
t = 5 => x = 2
∫(3→5) ƒ(t) dt = ∫(1→2) ƒ(2x + 1) * 2 dx
= 2∫(1→2) xe^x dx
= 2∫(1→2) x de^x
= 2[xe^x] |(1→2) - 2∫(1→2) e^x dx
= 2[2e² - e] - 2[e^x] |(1→2)
= 4e² - 2e - 2[e² - e]
= 2e²
你那个换元也不是错误的,可能过程会比较复杂,所以容易做错吧~
ƒ(2x + 1) = xe^x
t = 2x + 1,x = (t - 1)/2
ƒ(t) = [(t - 1)/2]e^[(t - 1)/2]
∫(3→5) ƒ(t) dt
= ∫(3→5) [(t - 1)/2]e^[(t - 1)/2] dt
= ∫(3→5) (t - 1)e^[(t - 1)/2] d[(t - 1)/2]
= ∫(3→5) (t - 1) de^[(t - 1)/2]
= (t - 1)e^[(t - 1)/2] |(3→5) - ∫(3→5) e^[(t - 1)/2] d(t - 1)
= [4e² - 2e] - 2∫(3→5) e^[(t - 1)/2] d[(t - 1)/2]
= [4e² - 2e] - 2e^[(t - 1)/2] |(3→5)
= [4e² - 2e] - 2[e² - e]
= 2e²