#### Answer

$$\frac{{99}}{{10\ln 10}}$$

#### Work Step by Step

$$\eqalign{
& \int_{ - 1}^1 {{{10}^x}} dx \cr
& {\text{by the formula }}\int {{a^x}} dx = \frac{{{a^x}}}{{\ln a}} + C \cr
& {\text{letting }}a = 10 \cr
& \int_{ - 1}^1 {{{10}^x}} dx = \left. {\left( {\frac{{{{10}^x}}}{{\ln 10}}} \right)} \right|_{ - 1}^1 \cr
& = \left. {\frac{1}{{\ln 10}}\left( {{{10}^x}} \right)} \right|_{ - 1}^1 \cr
& {\text{use the fundamental theorem}} \cr
& = \frac{1}{{\ln 10}}\left( {{{10}^1} - {{10}^{ - 1}}} \right) \cr
& = \frac{1}{{\ln 10}}\left( {\frac{{99}}{{10}}} \right) \cr
& = \frac{{99}}{{10\ln 10}} \cr} $$