Respostas

2014-05-05T02:11:35-03:00
 \int\limits^a_b { \frac{1}{e^x} -2 \, dx

 \frac{1}{e^x} =e^{-x}

a integral fica
 \int\limits^a_b {e^{-x}-2.} \, dx

a integral de e^x= \frac{e^x}{1} \\\\e^{-x}= \frac{e^{-x}}{-1} =-e^{-x}
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\int\limits^a_b {e^{-x}-2.} \, dx= \ [-e^{-x}-2x   \left\begin{array}{ccc}\end{array}\right] \limts^a_b\\\\\\(-e^{-a} -2a)-(-e^{-b} -2b)