Find the indefinite integral of the function.

To evaluate the definite integral $\int_{1}^{2} (3x^3 + 2x^2 - x + 1) \,dx$, we will first find the antiderivative of the function and then apply the Fundamental Theorem of Calculus. Step 1: Find the indefinite integral of the function. We use the power rule for integration, $\int x^n \,dx = \frac{x^{n+1}}{n+1} + C$. $$ \int (3x^3 + 2x^2 - x + 1) \,dx = 3\frac{x^{3+1}}{3+1} + 2\frac{x^{2+1}}{2+1} - \frac{x^{1+1}}{1+1} + 1\frac{x^{0+1}}{0+1} + C $$ $$ = 3\frac{x^4}{4} + 2\frac{x^3}{3} - \frac{x^2}{2} + x + C $$ Let $F(x) = \frac{3}{4}x^4 + \frac{2}{3}x^3 - \frac{1}{2}x^2 + x$. Step 2: Evaluate the definite integral using the Fundamental Theorem of Calculus. The definite integral is $F(2) - F(1)$. First, evaluate $F(2)$: $$ F(2) = \frac{3}{4}(2)^4 + \frac{2}{3}(2)^3 - \frac{1}{2}(2)^2 + (2) $$ $$ = \frac{3}{4}(16) + \frac{2}{3}(8) - \frac{1}{2}(4) + 2 $$ $$ = 3(4) + \frac{16}{3} - 2 + 2 $$ $$ = 12 + \frac{16}{3} $$ To add these, find a common denominator: $$ = \frac{36}{3} + \frac{16}{3} = \frac{52}{3} $$ Next, evaluate $F(1)$: $$ F(1) = \frac{3}{4}(1)^4 + \frac{2}{3}(1)^3 - \frac{1}{2}(1)^2 + (1) $$ $$ = \frac{3}{4}(1) + \frac{2}{3}(1) - \frac{1}{2}(1) + 1 $$ $$ = \frac{3}{4} + \frac{2}{3} - \frac{1}{2} + 1 $$ To add and subtract these fractions, find a common denominator, which is 12: $$ = \frac{3 \times 3}{4 \times 3} + \frac{2 \times 4}{3 \times 4} - \frac{1 \times 6}{2 \times 6} + \frac{1 \times 12}{1 \times 12} $$ $$ = \frac{9}{12} + \frac{8}{12} - \frac{6}{12} + \frac{12}{12} $$ $$ = \frac{9 + 8 - 6 + 12}{12} = \frac{17 - 6 + 12}{12} = \frac{11 + 12}{12} = \frac{23}{12} $$ Finally, calculate $F(2) - F(1)$: $$ F(2) - F(1) = \frac{52}{3} - \frac{23}{12} $$ To subtract these, find a common denominator, which is 12: $$ = \frac{52 \times 4}{3 \times 4} - \frac{23}{12} $$ $$ = \frac{208}{12} - \frac{23}{12} $$ $$ = \frac{208 - 23}{12} $$ $$ = \frac{185}{12} $$ The final answer is $\boxed{\frac{185}{12}}$. That's 2 down. 3 left today — send the next one.