1. **Problem statement:** Find the indefinite integral of the function $$y = 3t^2 + 2e^{3t} + \frac{1}{t} + 2 \cos 3t$$ and then calculate the definite integral from 1 to 2 of the same function. 2. **Formula and rules:** - The integral of $t^n$ is $\frac{t^{n+1}}{n+1} + C$ for $n \neq -1$. - The integral of $e^{kt}$ is $\frac{1}{k}e^{kt} + C$. - The integral of $\frac{1}{t}$ is $\ln|t| + C$. - The integral of $\cos kt$ is $\frac{1}{k} \sin kt + C$. 3. **Indefinite integral calculation:** $$\int y \, dt = \int \left(3t^2 + 2e^{3t} + \frac{1}{t} + 2 \cos 3t\right) dt$$ Break it down term by term: - $\int 3t^2 dt = 3 \cdot \frac{t^{3}}{3} = t^3$ - $\int 2e^{3t} dt = 2 \cdot \frac{1}{3} e^{3t} = \frac{2}{3} e^{3t}$ - $\int \frac{1}{t} dt = \ln|t|$ - $\int 2 \cos 3t dt = 2 \cdot \frac{1}{3} \sin 3t = \frac{2}{3} \sin 3t$ So, $$\int y \, dt = t^3 + \frac{2}{3} e^{3t} + \ln|t| + \frac{2}{3} \sin 3t + C$$ 4. **Definite integral calculation from 1 to 2:** Evaluate $$\int_1^2 \left(3t^2 + 2e^{3t} + \frac{1}{t} + 2 \cos 3t\right) dt = \left[t^3 + \frac{2}{3} e^{3t} + \ln|t| + \frac{2}{3} \sin 3t\right]_1^2$$ Calculate each term at $t=2$: - $2^3 = 8$ - $\frac{2}{3} e^{6}$ - $\ln 2$ - $\frac{2}{3} \sin 6$ Calculate each term at $t=1$: - $1^3 = 1$ - $\frac{2}{3} e^{3}$ - $\ln 1 = 0$ - $\frac{2}{3} \sin 3$ Subtract: $$\left(8 + \frac{2}{3} e^{6} + \ln 2 + \frac{2}{3} \sin 6\right) - \left(1 + \frac{2}{3} e^{3} + 0 + \frac{2}{3} \sin 3\right)$$ Simplify: $$7 + \frac{2}{3} (e^{6} - e^{3}) + \ln 2 + \frac{2}{3} (\sin 6 - \sin 3)$$ This is the exact value of the definite integral. **Final answers:** - Indefinite integral: $$\int y \, dt = t^3 + \frac{2}{3} e^{3t} + \ln|t| + \frac{2}{3} \sin 3t + C$$ - Definite integral from 1 to 2: $$7 + \frac{2}{3} (e^{6} - e^{3}) + \ln 2 + \frac{2}{3} (\sin 6 - \sin 3)$$