Integrate the following function with respect to t
f(t) = 0,2e^−0.2t
a.− e^−0.2t + c
b. e^−0.2t + c
c.−o.4e^−0.2t
d.−o.4e^−0.2t + c
∫0.2 e−0.2 tdt=−∫ e−0.2 td(−0.2t)=−e−0.2 t+C,\int0.2\,e^{-0.2\,t}dt=-\int\,e^{-0.2\,t}d(-0.2t)=-e^{-0.2\,t}+C,∫0.2e−0.2tdt=−∫e−0.2td(−0.2t)=−e−0.2t+C, C∈RC\in{\mathbb{R}}C∈R. We used a formula for a well-known integral ∫exdx\int e^x dx∫exdx and a substitution: z=−0.2 tz=-0.2\,tz=−0.2t.
Answer: a. −e−0.2 t+C-e^{-0.2\,t}+C−e−0.2t+C.
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