1. **Problem statement:** We have a storage tank with 10,000 litres of chemical leaking at a rate given by the derivative of the amount spilled, $$f'(t) = 400e^{-0.01t}$$ where $$t$$ is in hours. We want to find the total amount of chemical spilled in the first day (24 hours). 2. **Formula and explanation:** The rate of spill $$f'(t)$$ is the derivative of the amount spilled $$f(t)$$. To find the total amount spilled over a time interval, we integrate the rate over that interval: $$f(t) = \int f'(t) \, dt$$ The total amount spilled from $$t=0$$ to $$t=24$$ is: $$\int_0^{24} 400e^{-0.01t} \, dt$$ 3. **Integration:** We integrate: $$\int 400e^{-0.01t} \, dt = 400 \int e^{-0.01t} \, dt$$ Recall that: $$\int e^{kt} \, dt = \frac{1}{k} e^{kt} + C$$ Here, $$k = -0.01$$, so: $$\int e^{-0.01t} \, dt = \frac{1}{-0.01} e^{-0.01t} = -100 e^{-0.01t} + C$$ Therefore: $$\int 400 e^{-0.01t} \, dt = 400 (-100 e^{-0.01t}) + C = -40000 e^{-0.01t} + C$$ 4. **Evaluate definite integral:** Calculate: $$\int_0^{24} 400 e^{-0.01t} \, dt = [-40000 e^{-0.01t}]_0^{24} = -40000 e^{-0.01 \times 24} + 40000 e^{0}$$ Since $$e^0 = 1$$: $$= 40000 (1 - e^{-0.24})$$ 5. **Calculate numerical value:** Calculate $$e^{-0.24}$$: $$e^{-0.24} \approx 0.7866$$ So: $$40000 (1 - 0.7866) = 40000 \times 0.2134 = 8536$$ litres (approximately) 6. **Interpretation:** The total amount of chemical spilled in the first 24 hours is approximately 8536 litres. **Final answer:** $$\boxed{8536 \text{ litres}}$$