What is the integral of $e^{- 0.2 t} d t$ $e^(-0.2t) dt$ ?

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Answer 1 · 2015-10-19T16:00:57

$\int e^{- 0.2 t} d t = 5 e^{- 0.2 t} + C$ $int e^(-0.2t) dt = 5e^(-0.2t) +C$

Two methods:

Substitution

Let $u = 0.2 t$ $u = 0.2t$ so $d u = - 0.2 d t$ $du = -0.2 dt$

$\int e^{- 0.2 t} d t$ $int e^(-0.2t) dt$ becomes

$- \frac{1}{0.2} \int e^{u} d u = - \frac{1}{\frac{2}{10}} e^{u} + C$ $-1/0.2 int e^u du = -1/(2/10) e^u +C$

$= - \frac{10}{2} e^{- 0.2 t} + C$ $= -10/2 e^(-0.2t) +C$

$= - 5 e^{- 0.2 t} + C$ $= -5 e^(-0.2t) +C$

Check the answer by differentiating.

Learn the rule

$\int e^{k u} d u = \frac{1}{k} e^{k u} + C$ $int e^(ku) du = 1/k e^(ku) +C$

(Verify be differentiating.)

So

$\int e^{- 0.2 t} d t = \frac{1}{- 0.2} e^{- 0.2 t} + C$ $int e^(-0.2t) dt = 1/(-0.2)e^(-0.2t) +C$

$= - 5 e^{- 0.2 t} + C$ $= -5 e^(-0.2t) +C$

What is the integral of e−0.2tdte^(-0.2t) dt?