How do you find the antiderivative of #e^(-2x)#?

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Answer 1 · 2017-01-02T10:11:37

#inte^(-2x)dx=-1/2e^(-2x)+C#

it is important to remember that

#d/dx(e^x)=e^x#

so let us see what happens if we differentiate the given function

#y=e^(-2x)#

#u=-2x=>(du)/(dx)=-2#

#y=e^u=>(dy)/(du)=e^u#

by the chain rule we have:

#(dy)/(dx)=(dy)/(du)xx(du)/(dx)#

giving us

#(dy)/(dx)=e^uxx-2e^(-2x)=-2e^(-2x)#

now integration is the reverse of differentiation, so comparing what we have after differentiating and the function we are given to integrate.

we have to adjust the function by a suitable constant to cancel the #" -2#

#inte^(-2x)dx=-1/2e^(-2x)+C#

if you now differentiate the resulting function you will see it gives the original function.