How to determine if the improper integral converges or diverges ∫ x 2 e - x d x from 0 to infinity?

Question

How to determine if the improper integral converges or diverges       ∫          x      2              e              -        x                    d      x       from 0 to infinity?

Geovanni Marquez · Accepted Answer

To solve the indefinite integral, we will use integration by parts. Next, we will use limits to assess the incorrect definite integral.      ∫          x      2              e              -        x                    d      x      Integration by parts 1:Let       u    =          x      2       and       d    v    =          e              -        x                    d      x      Then       d    u    =          d      x       and       v    =    -          e              -        x            Applying the integration by parts formula       ∫    u    d    v    =    u    v    -    ∫    v    d    u  :      ∫          x      2              e              -        x                    d      x        =    -          x      2              e              -        x              +    2    ∫    x          e              -        x                    d      x      Integration by parts 2:Let       u    =    x   and       d    v    =          e              -        x                    d      x      Then       d    u    =    2    x          d      x       and       v    =    -          e              -        x            Applying the formula to the remaining integral:      ∫    x          e              -        x                    d      x        =    -    x          e              -        x              +    ∫          e              -        x                    d      x            =    -    x          e              -        x              -          e              -        x              +    C  Substituting this back in, we have      ∫          x      2              e              -        x                    d      x        =    -          x      2              e              -        x              +    2          (      -      x              e                  -          x                    -              e                  -          x                    )        +    C        =    -          e              -        x                    (              x        2            +      2      x      +      2      )        +    C  Now we can check the definite integral:                    ∫        0        ∞                    x      2              e              -        x                    d      x        =          lim              M        →        ∞                            ∫        0        M                    x      2              e              -        x                    d      x            =          lim              M        →        ∞                                      [          -                      e                          -              x                                            (                          x              2                        +            2            x            +            2            )                    ]                0        M                  =          lim              M        →        ∞              -                            M          2                +        2        M        +        2                    e        M              +    2  Intuitively we can say at this point that the first term will go to       0  , as the exponential             e      x       in the denominator will grow much faster than the polynomial in the numerator. This will give us an answer of       2  . Still, we can prove this using L&#039;Hopital&#039;s rule  to show that the initial term does converge to       0  .            lim              x        →        ∞                                      x          2                +        2        x        +        2                    e        x              =                            d                      d            x                                    (                      x            2                    +          2          x          +          2          )                                      d                      d            x                                    e          x                          =          lim              x        →        ∞                            2        x        +        2                    e        x                  =          lim              x        →        ∞                                      d                      d            x                                    (          2          x          +          2          )                                      d                      d            x                                    e          x                          =          lim              x        →        ∞                                      d                      d            x                                    (          2          x          +          2          )                                      d                      d            x                                    e          x                                2      ∞            =    0  With that, using that if two functions converge at a limit, then the limit of their sum is equal to the sum of their limits, we have:            lim              M        →        ∞              2    -                            M          2                +        2        M        +        2                    e        M              =          lim              M        →        ∞              2    -          lim              M        →        ∞                                      M          2                +        2        M        +        2                    e        M                  =    2    +    0        +    2  Therefore the integral converges to       2  .

How to determine if the improper integral converges or diverges intx^2 e^-x dx from 0 to infinity?

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