Find the area bounded by the curve $$y=x^2+x+4$$​, the x-axis and the ordinates x = 1 and x = 3.

Correct option is A

​$$\begin{aligned}&\int (x^2 + x + 4)\,dx = \int x^2\,dx + \int x\,dx + \int 4\,dx = \frac{x^3}{3} + \frac{x^2}{2} + 4x \\[10pt]&\text{Now apply the limits from 1 to 3:} \\[10pt]&{\text{\checkmark\ Step 3: Evaluate the definite integral}} \\[5pt]&A = \left[ \frac{x^3}{3} + \frac{x^2}{2} + 4x \right]_1^3 \\[8pt]&\text{At } x = 3: \\&\quad \frac{3^3}{3} + \frac{3^2}{2} + 4(3) = \frac{27}{3} + \frac{9}{2} + 12 = 9 + 4.5 + 12 = 25.5 \\[10pt]&\text{At } x = 1: \\&\quad \frac{1^3}{3} + \frac{1^2}{2} + 4(1) = \frac{1}{3} + \frac{1}{2} + 4 = \frac{2 + 3}{6} + 4 = \frac{5}{6} + 4 = \frac{29}{6} \approx 4.8333 \\[10pt]&{\text{\checkmark\ Final Step: Subtract}} \\[5pt]&A = 25.5 - \frac{29}{6} = \frac{153}{6} - \frac{29}{6} = \frac{124}{6} = {\frac{62}{3}} \text{ square units}\end{aligned}$$​​