## Thomas' Calculus 13th Edition

$$\frac{{{2^{11/2}}}}{3}$$
\eqalign{ & \int_0^{\ln 9} {{e^\theta }{{\left( {{e^\theta } - 1} \right)}^{1/2}}} d\theta \cr & {\text{use substitution}}{\text{: }} \cr & {\text{ }}u = {e^\theta } - 1,{\text{ so that }}\frac{{du}}{{d\theta }} = {e^\theta },\,\,\,dr = \frac{{du}}{{{e^\theta }}} \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}\theta = \ln 9,{\text{ }}u = {e^{\ln 9}} - 1 = 8 \cr & \,\,\,\,\,\,{\text{If }}\theta = 0,{\text{ }}u = {e^0} - 1 = 0 \cr & {\text{then}} \cr & \int_0^{\ln 9} {{e^\theta }{{\left( {{e^\theta } - 1} \right)}^{1/2}}} d\theta = \int_0^8 {{e^\theta }{u^{1/2}}} \left( {\frac{{du}}{{{e^\theta }}}} \right) \cr & = \int_0^8 {{u^{1/2}}} du \cr & {\text{integrating, we get:}} \cr & = \left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)_0^8 \cr & = \frac{2}{3}\left( {{u^{3/2}}} \right)_0^8 \cr & {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = \frac{2}{3}\left( {{8^{3/2}} - {0^{3/2}}} \right) \cr & {\text{simplifying, we get:}} \cr & = \frac{2}{3}\left( {{8^{3/2}} - {0^{3/2}}} \right) \cr & = \frac{{2{{\left( {{2^3}} \right)}^{3/2}}}}{3} \cr & = \frac{{{2^{11/2}}}}{3} \cr}