Answer
$$10,000$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{{x^{1.0001}}}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_1^\infty {\frac{1}{{{x^{1.0001}}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{{x^{1.0001}}}}} dx \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_1^b {{x^{ - 1.0001}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{x^{ - 1.0001 + 1}}}}{{ - 1.0001 + 1}}} \right]_1^b \cr
& = - 10,000\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{x^{0.0001}}}}} \right]_1^b \cr
& {\text{ using ftc}} \cr
& = - 10,000\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{b^{0.0001}}}} - \frac{1}{{{1^{0.0001}}}}} \right] \cr
& = - 10,000\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{{b^{0.0001}}}} - 1} \right] \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = - 10,000\left( {\frac{1}{{{\infty ^{0.0001}}}} - 1} \right) \cr
& = - 10,000\left( { - 1} \right) \cr
& = 10,000 \cr} $$
