How do you find the antiderivative of $e^{- 2 x}$ $e^(-2x)$ ?

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Answer 1 · 2017-01-02T10:11:37

$\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x} + C$ $inte^(-2x)dx=-1/2e^(-2x)+C$

it is important to remember that

$\frac{d}{d x} (e^{x}) = e^{x}$ $d/dx(e^x)=e^x$

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

$y = e^{- 2 x}$ $y=e^(-2x)$

$u = - 2 x \Rightarrow \frac{d u}{d x} = - 2$ $u=-2x=>(du)/(dx)=-2$

$y = e^{u} \Rightarrow \frac{d y}{d u} = e^{u}$ $y=e^u=>(dy)/(du)=e^u$

by the chain rule we have:

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)xx(du)/(dx)$

giving us

$\frac{d y}{d x} = e^{u} \times - 2 e^{- 2 x} = - 2 e^{- 2 x}$ $(dy)/(dx)=e^uxx-2e^(-2x)=-2e^(-2x)$

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 $- 2$ $" -2$

$\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x} + C$ $inte^(-2x)dx=-1/2e^(-2x)+C$

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

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