Question

The area bounded by the parabola \(y = x^2 + 2\) and the lines \(y = x\), \(x = 1\) and \(x = 2\) (in square units) is:

• A. \(\frac{31}{6}\)
• B. \(\frac{29}{6}\)
• C. \(\frac{25}{6}\)
• D. \(\frac{17}{6}\)
• E. \(\frac{13}{6}\)

Hint:

Always check which function is on top when setting up the integral for the area between curves to ensure correct calculation of the area.

Answer 1

The Correct Option is D

To find the area, integrate the difference between the upper function and the lower function from \(x = 1\) to \(x = 2\).
The upper function in this case is the parabola \(y = x^2 + 2\), and the lower function is the line \(y = x\).
Calculate the integral: \[ {Area} = \int_{1}^{2} ((x^2 + 2) - x) \, dx \] \[ = \int_{1}^{2} (x^2 - x + 2) \, dx \] \[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_1^2 \] \[ = \left( \frac{2^3}{3} - \frac{2^2}{2} + 2 \times 2 \right) - \left( \frac{1^3}{3} - \frac{1^2}{2} + 2 \times 1 \right) \] \[ = \left( \frac{8}{3} - 2 + 4 \right) - \left( \frac{1}{3} - \frac{1}{2} + 2 \right) \] \[ = \left( \frac{8}{3} + 2 \right) - \left( \frac{1}{3} + \frac{3}{2} \right) \] \[ = \left( \frac{8}{3} + 2 \right) - \left( \frac{2}{6} + \frac{9}{6} \right) \] \[ = \left( \frac{8}{3} + \frac{6}{3} \right) - \left( \frac{11}{6} \right) \] \[ = \frac{14}{3} - \frac{11}{6} \] \[ = \frac{28}{6} - \frac{11}{6} \] \[ = \frac{17}{6} \] Thus, the area bounded by the given curves is \(\frac{17}{6}\) square units.