Answer
a) $f(t)=-e^{-0.01t}+C$
b) $f(10)=-e^{-0.01 \cdot 10}+1\approx 0.10$
Work Step by Step
a)
The equation of $f$ can be obtained as:
$$f(t)=\int f'(x)dt$$
$$f(t)=\int 0.01e^{-0.01t}dt$$
$$f(t)=-e^{-0.01t}+C$$
b)
We have $f(0)=0$ so:
$$f(0)=-e^{-0.01\cdot 0}+C$$
$$0=-e^{-0.01\cdot 0}+C$$
$$0=-1+C$$
$$1=C$$
So the equation of $f$ is:
$$f(t)=-e^{-0.01t}+1$$
$$f(10)=-e^{-0.01 \cdot 10}+1\approx 0.10$$
