Question

Find the integral 2(e^(-.1t))dt

Answer 1

\[\int\limits{2e^{-t}dt}\]\[=\frac{2e^{-t}}{-1}+C\]

Answer 2

or is that 0.1t?

Answer 3

\[\int 2e^{-0.1t}dt\]
Substitute u = -0.1t and du = -0.1 dt.

Answer 4

yeah^^

Answer 5

Actually I'm not sure whether the asker meant -1 or -0.1.

Answer 6

-0.1

Answer 7

In this case, substitute u = -0.1t and du = -0.1 dt.
\[\frac{1}{-0.1}\int 2e^{u}du = -20\int e^{u}du\]

Answer 8

You should get:
\[-20e^{-0.1t}+\text{constant}\]

Answer 9

yes that is what I got thank you

Answer 10

You're welcome.