Question

Please tell me how to solve this $\int\limits^0_1 {e^{-4x} } \, dx$

Answer 1

$\frac{1}{4}( \frac{1}{ {e}^{4}} - 1 )$

Step-by-step explanation:

$Given, \: \: \: \int\limits^0_1 {e^{-4x} } \, dx \\ let, \: \: y = - 4x \\ Differentiating \: \: with \: \: respect \: \: to \: \: x, \\ dy = - 4dx \\ or, \: \: \: dx = \frac{ - 1}{4} dy \\ and \: \: y = 0 \: \: at \: \: x = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: y = - 4 \: \: at \: \: x = 1 \\ now, \: \: \int\limits^0_1 {e^{-4x} } \, dx = \frac{ -1 }{4} \int\limits^0_ {- 4} {e^{y} }dy \\ = \frac{ - 1}{4} [ {e}^{y} ]^0_ {- 4} \\ = \frac{ - 1}{4} [ {e}^{0} - {e}^{ - 4} ] \\ = \frac{ - 1}{4}(1 - \frac{1}{ {e}^{4} } ) \\ = \frac{1}{4}( \frac{1}{ {e}^{4}} - 1 )$