Chapter 4: Integration
Section 4.5: Improper Integrals
Does the improper integral ∫1∞1x+ex ⅆx converge or diverge?
Note that an antiderivative for 1x+ⅇx is not available in Maple, so an appeal to Theorem 4.5.1 will be essential.
∫1∞1x+ⅇx ⅆx = ∫1∞1x+ⅇxⅆx
For x≥1, fx=1ex=e−x>gx=1x+ex because making the value of a denominator smaller makes the value of the fraction larger.
Since ∫1∞ⅇ−x ⅆx = ⅇ−1, by conclusion (1) in Theorem 4.5.1, ∫1∞1x+ex ⅆx must likewise converge.
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