1. **Problem Statement:** Calculate the definite integral of the function $f(x) = x^2 + 3x + 2$ from 0 to 1. 2. **Formula Used:** The definite integral of a polynomial function $f(x)$ from $a$ to $b$ is given by $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. 3. **Find the antiderivative:** For $f(x) = x^2 + 3x + 2$, the antiderivative is $$F(x) = \frac{x^3}{3} + \frac{3x^2}{2} + 2x$$ 4. **Evaluate the definite integral:** $$\int_0^1 (x^2 + 3x + 2) \, dx = F(1) - F(0) = \left(\frac{1^3}{3} + \frac{3 \cdot 1^2}{2} + 2 \cdot 1\right) - 0 = \frac{1}{3} + \frac{3}{2} + 2$$ 5. **Simplify the sum:** Convert all terms to have a common denominator 6: $$\frac{1}{3} = \frac{2}{6}, \quad \frac{3}{2} = \frac{9}{6}, \quad 2 = \frac{12}{6}$$ So, $$\frac{2}{6} + \frac{9}{6} + \frac{12}{6} = \frac{23}{6}$$ 6. **Final answer:** $$\int_0^1 (x^2 + 3x + 2) \, dx = \frac{23}{6}$$ This means the area under the curve $f(x)$ from $x=0$ to $x=1$ is $\frac{23}{6}$. **Note:** The original transcription had a small error in the final sum; the correct sum is $\frac{23}{6}$, not $\frac{11}{3}$. --- **Additional info from the graph description:** - The parabola $f(x) = x^2 + 3x + 2$ opens upwards. - The sine wave $g(x) = \sin x$ oscillates between -1 and 1. - They intersect near $x = -2$ and $x = 0$. - The vertex of the parabola is near $x = -1.5$. These details help visualize the functions but are not needed for the integral calculation.