5.5 The Substitution Rule/59

$\int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx$

${\begin{aligned}u&={\frac {1}{x}}\\[2ex]du&=-{\frac {1}{x^{2}}}dx\\[2ex]-du&={\frac {1}{x^{2}}}dx\\[2ex]\end{aligned}}$

New upper limit:${\frac {1}{2}}={\frac {1}{(2)}}$
New lower limit: $1={\frac {1}{(1)}}$

${\begin{aligned}\int _{1}^{2}{\frac {e^{\frac {1}{x}}}{x^{2}}}\,dx&=\int _{1}^{2}e^{\frac {1}{x}}({\frac {1}{x^{2}}}\,dx)&=\int _{1}^{\frac {1}{2}}e^{u}\,(-du)\\[2ex]&=-\int _{1}^{\frac {1}{2}}e^{u}\,du\\[2ex]&=-e^{u}{\bigg |}_{1}^{\frac {1}{2}}\\[2ex]&=-{\sqrt {e}}-(-e^{1})\\[2ex]&=e-{\sqrt {e}}\end{aligned}}$