1. Problem 2.1: Given $k(x) = \frac{df}{dx}$, find the constant $C$ such that $\int_4^1 k(x) \, dx = f(4) + C$. Step 1: Recall the Fundamental Theorem of Calculus: If $k(x) = f'(x)$, then $\int_a^b k(x) \, dx = f(b) - f(a)$. Step 2: Here, $\int_4^1 k(x) \, dx = f(1) - f(4)$. Step 3: The problem states $\int_4^1 k(x) \, dx = f(4) + C$. Step 4: Equate the two expressions: $f(1) - f(4) = f(4) + C$. Step 5: Solve for $C$: $C = f(1) - 2f(4)$. 2. Problem 2.2: Given $F(x) = \int_2^x \frac{1}{t^3 + t} \, dt$, find $F'(x)$. Step 1: By the Fundamental Theorem of Calculus, if $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$. Step 2: Therefore, $F'(x) = \frac{1}{x^3 + x}$. 3. Problem 2.3: Given $\int_8^0 f(x) \, dx = 7$, $\int_8^2 f(x) \, dx = 5$, and $\int_0^2 g(x) \, dx = 9$, find $\int_2^0 (3f(x) + g(x)) \, dx$. Step 1: Use properties of integrals: $\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$. Step 2: Find $\int_0^2 f(x) \, dx$ using $\int_8^0 f(x) \, dx = 7$ and $\int_8^2 f(x) \, dx = 5$. Step 3: Note $\int_8^0 f(x) \, dx = \int_8^2 f(x) \, dx + \int_2^0 f(x) \, dx$. Step 4: Substitute values: $7 = 5 + \int_2^0 f(x) \, dx$ so $\int_2^0 f(x) \, dx = 2$. Step 5: Calculate $\int_2^0 (3f(x) + g(x)) \, dx = 3 \int_2^0 f(x) \, dx + \int_2^0 g(x) \, dx$. Step 6: Use $\int_0^2 g(x) \, dx = 9$ so $\int_2^0 g(x) \, dx = -9$. Step 7: Substitute: $3 \times 2 + (-9) = 6 - 9 = -3$. 4. Problem 2.4: Find $k$ such that the average value of $f(x) = x + \sqrt{200}$ on $[-200, k]$ is 60. Step 1: Average value formula: $\frac{1}{k - (-200)} \int_{-200}^k f(x) \, dx = 60$. Step 2: Simplify denominator: $\frac{1}{k + 200} \int_{-200}^k (x + \sqrt{200}) \, dx = 60$. Step 3: Compute integral: $\int_{-200}^k x \, dx = \frac{k^2}{2} - \frac{(-200)^2}{2} = \frac{k^2}{2} - 20000$. Step 4: Compute integral: $\int_{-200}^k \sqrt{200} \, dx = \sqrt{200} (k + 200)$. Step 5: Sum integrals: $\frac{k^2}{2} - 20000 + \sqrt{200} (k + 200)$. Step 6: Set average value equation: $$\frac{\frac{k^2}{2} - 20000 + \sqrt{200} (k + 200)}{k + 200} = 60$$ Step 7: Multiply both sides by $k + 200$: $$\frac{k^2}{2} - 20000 + \sqrt{200} (k + 200) = 60(k + 200)$$ Step 8: Rearrange and solve quadratic for $k$. 5. Problem 2.5: Given $f'(t) = 400 e^{-0.01 t}$, find total chemical spilled in first day ($t$ in hours, 24 hours). Step 1: Total spilled is $\int_0^{24} f'(t) \, dt$. Step 2: Compute integral: $$\int_0^{24} 400 e^{-0.01 t} \, dt = 400 \int_0^{24} e^{-0.01 t} \, dt$$ Step 3: Integral of $e^{at}$ is $\frac{1}{a} e^{at}$, so: $$400 \times \left[ \frac{e^{-0.01 t}}{-0.01} \right]_0^{24} = -40000 \left( e^{-0.24} - 1 \right)$$ Step 4: Simplify: $$40000 (1 - e^{-0.24})$$ Step 5: Calculate numeric value: $$40000 (1 - e^{-0.24}) \approx 40000 (1 - 0.7866) = 40000 \times 0.2134 = 8536$$ Step 6: Total chemical spilled is approximately 8536 litres.