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Question

Mathematics Question on Definite Integral

∫02(∣2x2−3x∣+[x−12]) dx,\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

76\frac{7}{6}

B

1912\frac{19}{12}

C

3112\frac{31}{12}

D

32\frac{3}{2}

Answer

1912\frac{19}{12}

Explanation

Solution

The correct answer is (B) : 1912\frac{19}{12}
∫02∣2x2−3x∣ dx+∫02[x−12] dx\int_{0}^{2} |2x^2 - 3x| \, dx + \int_{0}^{2} \left[x - \frac{1}{2}\right] \, dx
=∫03/2(3x−2x2)dx+∫3/22(2x2−3x)dx+∫01/2−1dx+∫1/23/20dx+∫3/221dx= \int^{3/2}_{0}(3x-2x^2)dx+\int^{2}_{3/2}(2x^2-3x)dx+\int^{1/2}_{0}-1dx+\int^{3/2}_{1/2}0dx+\int^{2}_{3/2}1dx
=(3x22−2x33)∣03/2+(2x33−3x22)∣3/22−12+12= (\frac{3x^2}{2}-\frac{2x^3}{3})|^{3/2}_{0}+(\frac{2x^3}{3}-\frac{3x^2}{2})|^{2}_{3/2}-\frac{1}{2}+\frac{1}{2}
(278−2712)+(163−6−2712+278)\left(\frac{27}{8} - \frac{27}{12}\right) + \left(\frac{16}{3} - 6 - \frac{27}{12} + \frac{27}{8}\right)
=1912= \frac{19}{12}