∫(0→1) xe^x dx = ∫(0→1) x d(e^x)
= xe^x - ∫(0→1) e^x dx
= [(1)e^(1) - (0)e^(0)] - e^x
= e - [e^(1) - e^(0)]
= e - e + 1
= 1
∫(0→e) xlnx dx = ∫(0→e) lnx d(x²/2)
= (1/2)x²lnx - (1/2)∫(0→e) x² d(lnx)
= [(1/2)(e²)ln(e) - (1/2)(0)] - (1/2)∫(0→e) x dx
= (1/2)e² - (1/2)(x²/2)
= (1/2)e² - (1/4)(e² - 0)
= (1/4)e²第一個算對了！第二個我書上的答案是e²/4+1/4=1/4(1+e²)可能你上限打錯了吧，1e什麼意思？好意思！上限是個e，沒有1！不好意思??！∫ xlnx dx = ∫ lnx d(x²/2) = (1/2)∫ lnx d(x²) = (1/2)x²lnx - (1/2)∫ x² d(lnx) = (1/2)x²lnx - (1/2)∫ x²(1/x) dx = (1/2)x²lnx - (1/2)∫ x dx = (1/2)x²lnx - (1/2)(x²/2) + C = (1/2)x²lnx - (1/4)x² + C = (1/4)x²(2lnx - 1) + C 代入上下限得： = (1/4)(e²)[2ln(e) - 1] - (1/4)(0) = (1/4)(e²)(2 - 1) = e²/4 還有意見不？你要有鑒別能力，不只是光信答案就可以。謝謝了??！一會給你發(fā)過去幾道題！幫忙解答一下！