1. **Problem Statement:** (a) Find the indefinite integral $\int (1 - t)^2 \, dt$. (b) Evaluate the definite integrals: (i) $\int_1^3 (x^2 - 4x + 3) \, dx$ (ii) $\int_0^2 3 \sin t \, dt$ (c) Sketch the curves $y = x^2 + 3$ and $y = 7 - 3x$ and find the area enclosed by them using integration. 2. **Formulas and Rules:** - Indefinite integral: $\int f(x) \, dx = F(x) + C$, where $F'(x) = f(x)$. - Definite integral: $\int_a^b f(x) \, dx = F(b) - F(a)$. - Area between curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is $\int_a^b |f(x) - g(x)| \, dx$. --- ### (a) Indefinite integral $\int (1 - t)^2 \, dt$ 1. Expand the integrand: $$ (1 - t)^2 = 1 - 2t + t^2 $$ 2. Integrate term-by-term: $$ \int (1 - 2t + t^2) \, dt = \int 1 \, dt - 2 \int t \, dt + \int t^2 \, dt $$ 3. Use power rule $\int t^n dt = \frac{t^{n+1}}{n+1} + C$: $$ = t - 2 \cdot \frac{t^2}{2} + \frac{t^3}{3} + C = t - t^2 + \frac{t^3}{3} + C $$ --- ### (b)(i) Evaluate $\int_1^3 (x^2 - 4x + 3) \, dx$ 1. Find the antiderivative: $$ \int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C $$ 2. Evaluate at limits: $$ F(3) = \frac{3^3}{3} - 2 \cdot 3^2 + 3 \cdot 3 = 9 - 18 + 9 = 0 $$ $$ F(1) = \frac{1^3}{3} - 2 \cdot 1^2 + 3 \cdot 1 = \frac{1}{3} - 2 + 3 = \frac{1}{3} + 1 = \frac{4}{3} $$ 3. Compute definite integral: $$ F(3) - F(1) = 0 - \frac{4}{3} = -\frac{4}{3} $$ --- ### (b)(ii) Evaluate $\int_0^2 3 \sin t \, dt$ 1. Factor out constant: $$ 3 \int_0^2 \sin t \, dt $$ 2. Antiderivative of $\sin t$ is $-\cos t$: $$ 3 [-\cos t]_0^2 = 3 (-\cos 2 + \cos 0) $$ 3. Calculate values: $$ \cos 0 = 1, \quad \cos 2 \approx -0.4161 $$ 4. Substitute: $$ 3 ( -(-0.4161) + 1 ) = 3 (0.4161 + 1) = 3 \times 1.4161 = 4.2483 $$ --- ### (c) Area enclosed by $y = x^2 + 3$ and $y = 7 - 3x$ 1. Find intersection points by solving: $$ x^2 + 3 = 7 - 3x $$ $$ x^2 + 3x - 4 = 0 $$ 2. Factor or use quadratic formula: $$ (x + 4)(x - 1) = 0 \implies x = -4, 1 $$ 3. The area is: $$ \int_{-4}^1 \big[(7 - 3x) - (x^2 + 3)\big] \, dx = \int_{-4}^1 (4 - 3x - x^2) \, dx $$ 4. Integrate: $$ \int (4 - 3x - x^2) \, dx = 4x - \frac{3x^2}{2} - \frac{x^3}{3} + C $$ 5. Evaluate at limits: $$ F(1) = 4(1) - \frac{3(1)^2}{2} - \frac{1^3}{3} = 4 - 1.5 - 0.3333 = 2.1667 $$ $$ F(-4) = 4(-4) - \frac{3(-4)^2}{2} - \frac{(-4)^3}{3} = -16 - 24 + \frac{64}{3} = -40 + 21.3333 = -18.6667 $$ 6. Area: $$ F(1) - F(-4) = 2.1667 - (-18.6667) = 20.8334 $$ --- **Final answers:** (a) $t - t^2 + \frac{t^3}{3} + C$ (b)(i) $-\frac{4}{3}$ (b)(ii) Approximately $4.2483$ (c) Area enclosed is approximately $20.8334$