1. Problem 2.1: Find the constant $C$ such that $$\int_1^4 k(x)\,dx = f(4) + C$$ given that $k(x) = \frac{df}{dx}$. Step 1. Recognize that $k(x)$ is the derivative of $f(x)$, i.e. $k(x) = f'(x)$. Step 2. By the Fundamental Theorem of Calculus, $$\int_1^4 f'(x)\, dx = f(4) - f(1)$$. Step 3. Compare with the expression given: $$\int_1^4 k(x)\,dx = f(4) + C$$. Step 4. Equate: $$f(4) - f(1) = f(4) + C \implies C = -f(1)$$. Answer: $$C = -f(1)$$. 2. Problem 2.2: Given $$F(x) = \int_1^x \frac{t^2}{3+t} dt$$, find $F'(x)$. Step 1. By the Fundamental Theorem of Calculus, if $$F(x) = \int_1^x f(t) dt$$, then $$F'(x) = f(x)$$. Step 2. Identify $$f(t) = \frac{t^2}{3+t}$$ so Step 3. Therefore, $$F'(x) = \frac{x^2}{3+x}$$. 3. Problem 2.3: Given $$\int_0^8 f(x) dx = 7$$, $$\int_2^8 f(x) dx = 5$$, and $$\int_0^2 g(x) dx = 9$$, find $$\int_0^2 (3f(x) + g(x)) dx$$. Step 1. Use linearity of integrals: $$\int_0^2 (3f(x) + g(x)) dx = 3 \int_0^2 f(x) dx + \int_0^2 g(x) dx$$. Step 2. Find $$\int_0^2 f(x) dx$$ using given integrals: $$\int_0^8 f(x) dx = \int_0^2 f(x) dx + \int_2^8 f(x) dx \implies 7 = \int_0^2 f(x) dx + 5$$. Hence, $$\int_0^2 f(x) dx = 7 - 5 = 2$$. Step 3. Use known integral of $g(x)$: $$\int_0^2 g(x) dx = 9$$. Step 4. Calculate: $$3 \times 2 + 9 = 6 + 9 = 15$$. Answer: $$\int_0^2 (3f(x) + g(x)) dx = 15$$. 4. Problem 2.4: The average value of $$f(x) = \sqrt{x + 200}$$ on $$[-200, k]$$ equals 60. Find $k$. Step 1. The average value formula for $f(x)$ on $[a,b]$ is: $$\frac{1}{b - a} \int_a^b f(x) dx = \text{average}$$. Step 2. Given $a = -200$, $b = k$, and average value $= 60$: $$\frac{1}{k + 200} \int_{-200}^k \sqrt{x + 200} dx = 60$$. Step 3. Substitute $$u = x + 200$$, so when $x = -200$, $u=0$; when $x = k$, $u = k + 200$. Step 4. Then, $$\int_{-200}^k \sqrt{x + 200} dx = \int_0^{k+200} \sqrt{u} du$$. Step 5. Compute integral: $$\int_0^{k+200} u^{1/2} du = \left[ \frac{2}{3} u^{3/2} \right]_0^{k+200} = \frac{2}{3} (k+200)^{3/2}$$. Step 6. Average value equation becomes: $$\frac{1}{k+200} \times \frac{2}{3} (k+200)^{3/2} = 60$$ Simplify: $$\frac{2}{3} (k+200)^{1/2} = 60$$ Step 7. Multiply both sides by $\frac{3}{2}$: $$(k+200)^{1/2} = 60 \times \frac{3}{2} = 90$$ Step 8. Square both sides: $$k + 200 = 90^2 = 8100$$ Step 9. Solve for $k$: $$k = 8100 - 200 = 7900$$. Answer: $$k = 7900$$. 5. Problem 2.5: Given leak rate $$f'(t) = 400 e^{-0.01 t}$$ litres per hour, find total chemical spillage in first 24 hours. Step 1. Total spilled quantity is integral of leak rate over $t$ from 0 to 24: $$Q = \int_0^{24} 400 e^{-0.01 t} dt$$. Step 2. Factor out constant: $$Q = 400 \int_0^{24} e^{-0.01 t} dt$$. Step 3. Integral of exponential: $$\int e^{at} dt = \frac{1}{a} e^{at} + C$$. Here, $a = -0.01$, so $$\int_0^{24} e^{-0.01 t} dt = \left[ \frac{e^{-0.01 t}}{-0.01} \right]_0^{24} = -100 (e^{-0.24} - 1)$$. Step 4. Substitute back: $$Q = 400 \times -100 (e^{-0.24} - 1) = -40000 (e^{-0.24} - 1) = 40000 (1 - e^{-0.24})$$. Step 5. Calculate $e^{-0.24} \approx 0.7866$, so $$Q \approx 40000 (1 - 0.7866) = 40000 \times 0.2134 = 8536$$ litres. Answer: Approximately 8536 litres spill in first 24 hours.