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)

To find the area bounded by the curve \( y = x^2 + x + 4 \), the \( x \)-axis, and the lines \( x = 1 \) and \( x = 3 \), we need to compute the definite integral of the function from \( 1 \) to \( 3 \). First, we can set up the integral: \[ \text{Area} = \int_{1}^{3} (x^2 + x + 4) \, dx \] Next, we'll calculate the integral. The indefinite integral of \( x^2 + x + 4 \) can be determined as follows: \[ \int (x^2 + x + 4) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + 4x + C \] Now, we evaluate this from \( 1 \) to \( 3 \): \[ \left[ \frac{x^3}{3} + \frac{x^2}{2} + 4x \right]_{1}^{3} \] Calculating at the upper limit \( x = 3 \): \[ \frac{3^3}{3} + \frac{3^2}{2} + 4(3) = \frac{27}{3} + \frac{9}{2} + 12 = 9 + 4.5 + 12 = 25.5 \] Now, calculating at the lower limit \( x = 1 \): \[ \frac{1^3}{3} + \frac{1^2}{2} + 4(1) = \frac{1}{3} + \frac{1}{2} + 4 = \frac{1}{3} + \frac{3}{6} + \frac{24}{6} = \frac{1}{3} + \frac{27}{6} = \frac{1}{3} + 4.5 = \frac{1}{3} + 4.5 = \frac{1 + 13.5}{3} = \frac{14.5}{3} \approx 4.8333 \] Subtract the value at the lower limit from the upper limit: \[ 25.5 - 4.8333 = 20.6667 \] Thus, the area bounded by the curve, the \( x \)-axis, and the lines \( x=1 \) and \( x=3 \) is: \[ \text{Area} \approx 20.67 \] Therefore, the final answer for the area is approximately \( 20.67 \) square units.