※ 引述《kohana (little flower)》之铭言： : 1 : (1) 求定积分 ∫ xe^√x dx (请问这题的算式) : 0 : (2) d cosx 1 : ----∫ ----- dt (想确定一下答案) : dx 0 1+t^4 : 请大家帮忙 谢谢 (1) 令√x = t , 则 x = t^2 => dx = 2tdt x = 0 , x = 1 => t = 0 , t = 1 1 所以 ∫ xe^√x dx 0 1 = ∫ (t^2)*(e^(t))*(2t) dt 0 1 = 2∫ (t^3)*(e^(t)) dt 0 |1 1 = 2( (t^3)*(e^(t)) | - ∫ (e^(t))*(3t^2) dt ) |0 0 1 = 2( e - 3∫ (t^2)*(e^(t)) dt ) 0 |1 1 = 2( e - 3( (t^2)*(e^(t)) | - ∫ (e^(t))*(2t)dt )) |0 0 1 = 2( e - 3e + 6∫ (t)*(e^(t)) dt ) 0 |1 1 = 2( -2e + 6( (t)*(e^(t)) | - ∫ (e^(t)) dt ) |0 0 |1 = 2( -2e + 6e - 6( e^t | ) ) |0 = 2( 4e - 6(e - 1)) = 2( -2e + 6) = 12 - 4e (2) d cosx 1 -----∫ ------- dt dx 0 1 + t^4 1 d = ------------ -----(cosx) 1 + (cosx)^4 dx -sinx = ------------ 1 + (cosx)^4 --

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