How do you determine if the improper integral converges or diverges e^(-2t) dt  from negative infinity to -1?

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Explanation:

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Jan 4, 2018

The integral is divergent

Explanation:

The improper ntegral is

$I = {\int}_{x}^{-} 1 {e}^{- 2 t} \mathrm{dt}$

$= {\left[- \frac{1}{2} {e}^{-} \left(2 t\right)\right]}_{x}^{-} 1$

$= {\left[\frac{1}{2} {e}^{- 2 t}\right]}_{-} {1}^{x}$

$= \frac{1}{2} {e}^{- 2 x} - \frac{1}{2} {e}^{2}$

Therefore,

${\lim}_{x \to - \infty} I = {\lim}_{x \to - \infty} \left(\frac{1}{2} {e}^{- 2 x} - \frac{1}{2} {e}^{2}\right) = + \infty$

The integral is divergent

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