Find the area bounded by the curve \( y=x^{2}+x+4 \), the \( x \)-axis and the lines \( x=1 \) and \( x=3 \). (15 pts)

We wish to find the area bounded by the curve \[ y = x^2 + x + 4, \] the \(x\)-axis, and the vertical lines \(x=1\) and \(x=3\). Since the function is always positive (its discriminant is negative), the area is given by the integral of the function from \(x=1\) to \(x=3\). Thus, we set up the integral: \[ \text{Area} = \int_{1}^{3}(x^2 + x + 4) \, dx. \] **Step 1. Compute the indefinite integral** Find the antiderivative of \(x^2 + x + 4\): \[ \int (x^2 + x + 4) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + 4x + C. \] **Step 2. Evaluate the definite integral from \(x=1\) to \(x=3\)** Substitute the limits into the antiderivative: \[ \text{Area} = \left[ \frac{x^3}{3} + \frac{x^2}{2} + 4x \right]_{1}^{3}. \] First, evaluate at \(x=3\): \[ \frac{3^3}{3} + \frac{3^2}{2} + 4(3) = \frac{27}{3} + \frac{9}{2} + 12 = 9 + \frac{9}{2} + 12. \] Simplify: \[ 9 + 12 = 21, \quad\text{so} \quad 21 + \frac{9}{2} = \frac{42}{2} + \frac{9}{2} = \frac{51}{2}. \] Next, evaluate at \(x=1\): \[ \frac{1^3}{3} + \frac{1^2}{2} + 4(1) = \frac{1}{3} + \frac{1}{2} + 4. \] Simplify by writing \(4\) as a fraction with denominator \(6\): \[ \frac{1}{3} = \frac{2}{6},\quad \frac{1}{2} = \frac{3}{6},\quad 4 = \frac{24}{6}. \] Thus, \[ \frac{2}{6} + \frac{3}{6} + \frac{24}{6} = \frac{29}{6}. \] **Step 3. Subtract the evaluation at \(x=1\) from the evaluation at \(x=3\)** \[ \text{Area} = \frac{51}{2} - \frac{29}{6}. \] To combine these fractions, convert \(\frac{51}{2}\) to sixths: \[ \frac{51}{2} = \frac{51 \times 3}{6} = \frac{153}{6}. \] Thus, \[ \text{Area} = \frac{153}{6} - \frac{29}{6} = \frac{153 - 29}{6} = \frac{124}{6} = \frac{62}{3}. \] The area bounded by the curve, the \(x\)-axis, and the vertical lines \(x=1\) and \(x=3\) is \[ \boxed{\frac{62}{3}}. \]