∫02 2x2−3x+x−12dx, where [t] is the greatest integer function, is equal to :

∫02 2x2−3xdx =∫032 3x−2x2dx+∫322 2x2−3xdx=1912∫02 x−12dx=∫−1232 [t] dt =∫−120 (−1)dt+∫01 0⋅dt+∫132 1⋅dt=0.

∫02 2x2−3x+x−12dx, where [t] is the greatest integer function, is equal to :

∫02 2x2−3xdx =∫032 3x−2x2dx+∫322 2x2−3xdx=1912∫02 x−12dx=∫−1232 [t] dt =∫−120 (−1)dt+∫01 0⋅dt+∫132 1⋅dt=0.

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

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$\frac{7}{6}$

$\frac{19}{12}$

$\frac{31}{12}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \int_{0}^{2} |2 x^{2} - 3 x| dx \\ = \int_{0}^{\frac{3}{2}} (3 x - 2 x^{2}) dx + \int_{\frac{3}{2}}^{2} (2 x^{2} - 3 x) dx = \frac{19}{12} \\ \int_{0}^{2} [x - \frac{1}{2}] dx = \int_{\frac{- 1}{2}}^{\frac{3}{2}} [t] dt \\ = \int_{- \frac{1}{2}}^{0} (- 1) dt + \int_{0}^{1} 0 \cdot dt + \int_{1}^{\frac{3}{2}} 1 \cdot dt = 0 . \end{array}$