Answer
$$c = \root 3 \of {\frac{{18}}{4}} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{9}{{{x^3}}},{\text{ interval }}\left[ {1,3} \right] \cr
& {\text{By the Mean Value Theorem for integrals}}\left( {{\text{Theorem 4}}{\text{.10}}} \right) \cr
& \int_a^b {f\left( x \right)} dx = f\left( c \right)\left( {b - a} \right) \cr
& \int_a^b {f\left( x \right)} dx = \frac{9}{{{c^3}}}\left( {3 - 1} \right) \cr
& \int_a^b {f\left( x \right)} dx = \frac{{18}}{{{c^3}}} \cr
& {\text{Then,}} \cr
& \int_1^3 {\frac{9}{{{x^3}}}} dx = \frac{{18}}{{{c^3}}} \cr
& \left[ {\frac{{9{x^{ - 2}}}}{{ - 2}}} \right]_1^3 = \frac{{18}}{{{c^3}}} \cr
& - \frac{9}{2}\left[ {\frac{1}{{{x^2}}}} \right]_1^3 = \frac{{18}}{{{c^3}}} \cr
& - \frac{9}{2}\left[ {\frac{1}{{{{\left( 3 \right)}^2}}} - \frac{1}{{{{\left( 1 \right)}^2}}}} \right] = \frac{{18}}{{{c^3}}} \cr
& 4 = \frac{{18}}{{{c^3}}} \cr
& {c^3} = \frac{{18}}{4} \cr
& c = \root 3 \of {\frac{{18}}{4}} \cr
& c \approx 1.6509 \cr
& c{\text{ is in the interval }}\left[ {1,3} \right] \cr} $$
