1. **Problem Statement:** We are given the function $f(x) = \frac{2x}{x^2 + 1}$, representing velocity over time $t \geq 0$. We want to find its antiderivative $F(x)$, graph it, and interpret its y-intercept and end behavior. 2. **Formula and Rules:** The antiderivative (indefinite integral) of a function $f(x)$ is given by: $$F(x) = \int f(x) \, dx + C$$ where $C$ is the constant of integration. 3. **Find the antiderivative:** Start with $$F(x) = \int \frac{2x}{x^2 + 1} \, dx$$ Use substitution: let $u = x^2 + 1$, then $du = 2x \, dx$. Rewrite the integral: $$F(x) = \int \frac{1}{u} \, du = \ln|u| + C = \ln(x^2 + 1) + C$$ 4. **Interpretation of constant $C$ and initial condition:** Since $t \geq 0$ and the problem context is physical (position from velocity), we can set $F(0) = 0$ for convenience (starting position zero). Calculate $F(0)$: $$F(0) = \ln(0^2 + 1) + C = \ln(1) + C = 0 + C = C$$ Set $C = 0$. So, $$F(x) = \ln(x^2 + 1)$$ 5. **Maximum value scaling:** The maximum value of $f(x)$ occurs near $x=1$: $$f(1) = \frac{2(1)}{1^2 + 1} = \frac{2}{2} = 1$$ We want the maximum of $F(x)$ to be twice the maximum of $f(x)$, i.e., $2 \times 1 = 2$. Calculate $F(1)$: $$F(1) = \ln(1^2 + 1) = \ln(2) \approx 0.693$$ To scale $F(x)$ so that $F(1) = 2$, multiply by $\frac{2}{0.693} \approx 2.886$: $$F_{scaled}(x) = 2.886 \times \ln(x^2 + 1)$$ 6. **Estimate y-intercept and end behavior:** - At $x=0$: $$F_{scaled}(0) = 2.886 \times \ln(1) = 0$$ - As $x \to +\infty$: $$F_{scaled}(x) \approx 2.886 \times \ln(x^2) = 2.886 \times 2 \ln x = 5.772 \ln x \to +\infty$$ 7. **Interpretation:** - The y-intercept $F(0) = 0$ represents the initial position of the particle at time zero. - The end behavior $F(x) \to +\infty$ as $x \to +\infty$ means the position increases without bound over time, consistent with continuous positive displacement.