Question

\(\int_{0}^{2} \left( |2x^2 - 3x| + \left[x - \frac{1}{2}\right] \right) \, dx,\)
where [t] is the greatest integer function, is equal to

• A. \(\frac{7}{6}\)
• B. \(\frac{19}{12}\)
• C. \(\frac{31}{12}\)
• D. \(\frac{3}{2}\)

Answer 1

The Correct Option is B

To solve the integral \(\int_{0}^{2} \left( |2x^2 - 3x| + \left[x - \frac{1}{2}\right] \right) \, dx,\) we need to analyze the expression under the integral sign.

1. Consider the expression \(|2x^2 - 3x|\). To evaluate this, we first find the point where \(2x^2 - 3x = 0\).

\(2x^2 - 3x = 0 \Rightarrow x(2x - 3) = 0 \Rightarrow x = 0 \text{ or } x = \frac{3}{2}\)

1. Next, examine the interval [0, 2] to determine the sign of \(2x^2 - 3x\).

• For \(x \in [0, \frac{3}{2}]\), \(2x^2 - 3x \leq 0\) therefore \(|2x^2 - 3x| = -(2x^2 - 3x) = -2x^2 + 3x\).
• For \(x \in [\frac{3}{2}, 2]\), \(2x^2 - 3x \geq 0\) therefore \(|2x^2 - 3x| = 2x^2 - 3x\).

1. Evaluate \(\left[x - \frac{1}{2}\right]\).

• For \(x \in [0, \frac{1}{2})\), \(\left[x - \frac{1}{2}\right] = -1\).
• For \(x \in [\frac{1}{2}, \frac{3}{2})\), \(\left[x - \frac{1}{2}\right] = 0\).
• For \(x \in [\frac{3}{2}, 2]\), \(\left[x - \frac{1}{2}\right] = 1\).

1. Split the integration into intervals as indicated by the expression’s changes:

• Integral from 0 to \(\frac{1}{2}\): \(\int_{0}^{\frac{1}{2}} \left( -2x^2 + 3x - 1 \right) \, dx\)
• Integral from \(\frac{1}{2}\) to \(\frac{3}{2}\): \(\int_{\frac{1}{2}}^{\frac{3}{2}} \left( -2x^2 + 3x \right) \, dx\)
• Integral from \(\frac{3}{2}\) to 2: \(\int_{\frac{3}{2}}^{2} \left( 2x^2 - 3x + 1 \right) \, dx\)

1. Calculate each integral separately:

• First Interval: \([0, \frac{1}{2}] \) 

\[\int_{0}^{\frac{1}{2}} (-2x^2 + 3x - 1) \, dx = \left[-\frac{2}{3}x^3 + \frac{3}{2}x^2 - x\right]_{0}^{\frac{1}{2}} = \left(-\frac{1}{12} + \frac{3}{8} - \frac{1}{2}\right) = -\frac{1}{24}\]

• Second Interval: \([\frac{1}{2}, \frac{3}{2}] \) 

\[\int_{\frac{1}{2}}^{\frac{3}{2}} (-2x^2 + 3x) \, dx = \left[-\frac{2}{3}x^3 + \frac{3}{2}x^2\right]_{\frac{1}{2}}^{\frac{3}{2}} = \left[ -\frac{27}{12} + \frac{27}{4} \right] - \left[ -\frac{1}{24} + \frac{3}{8} \right] = \frac{5}{6}\]

• Third Interval: \([\frac{3}{2}, 2] \) 

\[\int_{\frac{3}{2}}^{2} (2x^2 - 3x + 1) \, dx = \left[\frac{2}{3}x^3 - \frac{3}{2}x^2 + x\right]_{\frac{3}{2}}^{2} = \left[\frac{16}{3} - 6 + 2\right] - \left[\frac{27}{12} - \frac{27}{4} + \frac{3}{2}\right] = \frac{5}{8}\]

1. Add up values from all intervals: \(-\frac{1}{24} + \frac{5}{6} + \frac{5}{8} = \frac{19}{12}\)

Thus, the value of the integral \(\int_{0}^{2} \left( |2x^2 - 3x| + \left[x - \frac{1}{2}\right] \right) \, dx\) is \(\frac{19}{12}\), which matches the correct option.