How to integrate x e x without using antiderivatives or integration by parts.Yesterday, I sat for my Real Analysis II paper. There I found a question asking to integrate ∫ 0 1 x e x d x without using antiderivatives and integrating by parts.I tried it by choosing a partition P n = ( 0 , 1 n , 2 n , … , n − 1 n , 1 ) , but I was not able to show that lim n → ∞ U ( f , P n ) = lim n → ∞ L ( f , P n ) = 1 .

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How to integrate   x      e    x   without using antiderivatives or integration by parts.Yesterday, I sat for my Real Analysis II paper. There I found a question asking to integrate             ∫      0      1        x          e      x            d    x   without using antiderivatives and integrating by parts.I tried it by choosing a partition       P    n    =  (  0  ,      1    n    ,      2    n    ,  …  ,            n      −      1        n    ,  1  )  , but I was not able to show that             lim              n        →        ∞              U    (    f    ,          P      n        )    =          lim              n        →        ∞              L    (    f    ,          P      n        )    =    1  .

amanhantmk · Accepted Answer

Step 1Here is a rather different way of doing it. Let&#039;s assume we know the following:                    e        t            −      1        t    =      ∫    0    1        e          x      t           d  xStep 2Differentiate both sides (under the integral on the right, which may be proven easily)            (      t      −      1      )              e        t            +      1        t    =      ∫    0    1    x      e          x      t           d  xAnd with   t  =  1,  1  =      ∫    0    1    x      e    x       d  x

sebadillab0 · Accepted Answer

Step 1  U  (  f  ,      P    n    )  =      1          n      2            (          e                        1          n                      +    2          e                        2          n                      +    3          e                        3          n                      +    .    .    .    +    n          e      1        )    =      1          n      2                  ∑              k        =        1                    n              k          e                        k          n                      =          1              n        2                            ∑                  k          =          1                          n                    k              (        1        +                  k          n                +                              k            2                                2            !                          n              2                                      +                              k            3                                3            !                          n              3                                      +        .        .        )            =              1                  n          2                                      ∑                      k            =            1                                n                          k        +                  1                      n            2                                                ∑                          k              =              1                                      n                                k                      (                          k              n                        +                                          k                2                                            2                !                                  n                  2                                                      +                                          k                3                                            3                !                                  n                  3                                                      +            .            .            )                    =                                    n              (              n              +              1              )                                      2                              n                2                                              +                                    n              (              2              n              +              1              )              (              n              +              1              )                                      2              !              6                              n                3                                              +                                                    n                2                            (              n              +              1                              )                2                                                    3              !              4                              n                4                                              +          .          .          .          .                    Step 2Thus when   n  →  ∞,   U  (  f  ,      P    n    )  =      1    2    +      1          2      !      6        +      1          3      !      4        +      1          4      !      30        +  .  . should be 1.

How to integrate x e x </msup> without using antiderivatives or integration b

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New Questions in Calculus 1