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Mathematics Question on integral

If [x][x] denotes the greatest integer function, then ∫14([x]−1)([x]−2)([x]−3)([x]−4)dx=\int\limits_{1}^{4} \left(\left[x\right] -1\right)\left(\left[x\right] -2\right)\left(\left[x\right] -3\right)\left(\left[x\right] -4\right)dx =

A

00

B

33

C

66

D

18\frac{1}{8}

Answer

00

Explanation

Solution

∫14([x]−1)([x]−2)([x]−3)([x]−4)dx\int\limits_{1}^{4} \left(\left[x\right] -1\right)\left(\left[x\right] -2\right)\left(\left[x\right] -3\right)\left(\left[x\right] -4\right)dx
=∫12([x]−1)([x]−2)([x]−3)([x]−4)dx= \int\limits_{1}^{2} ([x]-1) ([x]- 2) ([x]- 3) ([x]- 4) dx
+∫23([x]−1)([x]−2)([x]−3)([x]−4)dx+ \int\limits_{2}^{3} ([x]-1) ([x]- 2) ([x]- 3) ([x]- 4) dx
+∫34([x]−1)([x]−2)([x]−3)([x]−4)dx=0+ \int\limits_{3}^{4} ([x]-1) ([x]- 2) ([x]- 3) ([x]-4) dx =0
[∵∫12([x]−1)dx=0 and∫23([x]−2)dx=0 and∫34([x]−3)dx=0]\begin{bmatrix} \because &\int_{1}^{2} \left(\left[x\right]-1\right)dx = 0 \\\ and&\int_{2}^{3} \left(\left[x\right]-2\right)dx = 0\\\ and &\int_{3}^{4} \left(\left[x\right]-3\right)dx = 0\end{bmatrix}