1. **State the problem:** 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 spilled in the first day (24 hours). 2. **Understand the rate function:** The function $$f'(t)$$ represents the rate of chemical spillage in litres per hour at time $$t$$. 3. **Find the total amount spilled:** The total amount spilled in the first 24 hours is the integral of the rate from $$t=0$$ to $$t=24$$: $$ \text{Total spilled} = \int_0^{24} 400e^{-0.01t} \, dt $$ 4. **Calculate the integral:** $$ \int 400e^{-0.01t} dt = 400 \int e^{-0.01t} dt $$ Using the substitution $$u = -0.01t$$, $$du = -0.01 dt$$, so $$dt = \frac{du}{-0.01}$$: $$ 400 \int e^u \frac{du}{-0.01} = -40000 \int e^u du = -40000 e^u + C = -40000 e^{-0.01t} + C $$ 5. **Evaluate definite integral:** $$ \int_0^{24} 400e^{-0.01t} dt = \left[-40000 e^{-0.01t}\right]_0^{24} = -40000 e^{-0.24} + 40000 e^{0} $$ Since $$e^0 = 1$$: $$ = 40000 (1 - e^{-0.24}) $$ 6. **Calculate numerical value:** $$ e^{-0.24} \approx 0.7866 $$ So, $$ \text{Total spilled} \approx 40000 (1 - 0.7866) = 40000 \times 0.2134 = 8536 $$ 7. **Interpretation:** Approximately 8536 litres of chemical spill in the first 24 hours. **Final answer:** $$\boxed{8536 \text{ litres}}$$