Let's break down the problem step by step: ### Problem Statement Find the area of the region bounded by \[ y = \frac{x}{x^2 + 1} \], the x-axis, \( x = 0 \), and \( x = 3 \). --- ## Step 1: **Set Up the Definite Integral** The area \(A\) under the curve from \(x = 0\) to \(x = 3\) above the x-axis is: \[ A = \int_{0}^{3} \frac{x}{x^2 + 1} \, dx \] --- ## Step 2: **Integrate** Let’s solve the integral: \[ \int \frac{x}{x^2 + 1} \, dx \] Let \( u = x^2 + 1 \implies du = 2x \, dx \implies \frac{du}{2} = x \, dx \) Substitute: \[ \int \frac{x}{x^2 + 1} \, dx = \int \frac{1}{u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C \] Back substitute \( u = x^2 + 1 \): \[ = \frac{1}{2} \ln|x^2 + 1| + C \] --- ## Step 3: **Evaluate the Definite Integral** \[ A = \left[ \frac{1}{2} \ln(x^2 + 1) \right]_{0}^{3} \] Calculate at upper and lower limits: - When \( x = 3 \): \( x^2 + 1 = 9 + 1 = 10 \) - When \( x = 0 \): \( x^2 + 1 = 0 + 1 = 1 \) \[ A = \frac{1}{2} \ln(10) - \frac{1}{2} \ln(1) \] But \( \ln(1) = 0 \): \[ A = \frac{1}{2} \ln(10) - 0 = \frac{1}{2} \ln(10) \] --- ## Step 4: **Final Answer** \[ \boxed{A = \frac{1}{2} \ln(10)} \] Or, as a decimal (using \( \ln(10) \approx 2.3026 \)): \[ A \approx \frac{1}{2} \times 2.3026 \approx 1.1513 \] --- ## Step 5: **Graph the Function (Sketch)** - The function \( y = \frac{x}{x^2 + 1} \) is 0 at \( x = 0 \). - It rises, reaches a maximum, and then decreases toward 0 as \( x \) increases. - The area under the curve from \( x = 0 \) to \( x = 3 \) and above the x-axis is shaded. #### Quick Sketch ```plaintext y ↑ / | / | / | / |___/__________→ x 0 3 ``` --- ### **Summary** - **Integral:** \( \int_{0}^{3} \frac{x}{x^2 + 1} dx \) - **Area:** \( \boxed{\frac{1}{2}\ln(10) \approx 1.1513} \) - **Graph:** See sketch above.