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General
Easy

Question

integral subscript 0 superscript 1   open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals

  1. -1
  2. 2
  3. 1 plus e to the power of negative 1 end exponent
  4. 2 minus 1 over straight e

hintHint:

Integrate the each term from the given range.

The correct answer is: 2 minus 1 over straight e


    Given That:
    integral subscript 0 superscript 1   open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals
    >>> Integrate each term:
    integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x
    >>>Integration of 1 becomes x and integration of e-x becomes -e-x.
    >>>              = left parenthesis 1 minus 1 over e plus 1 right parenthesis
    = (2-1 over e)
    >>> Therefore, the integration of integral subscript 0 superscript 1   open parentheses 1 plus e to the power of negative x end exponent close parentheses d x equals becomes  (2-1 over e).

    integral subscript 0 superscript 1 left parenthesis 1 plus e to the power of negative x end exponent right parenthesis d x
    >>>Integration of 1 becomes x and integration of e-x becomes -e-x.
    >>>              = left parenthesis 1 minus 1 over e plus 1 right parenthesis
    = (2-1 over e)

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