JEE Math Practice Question
Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $$\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} $$ is equal to :
- A.$$-$$1
- B.$${{ - 5} \over 4}$$
- C.$${{\sqrt {17} - 13} \over 8}$$
- D.$${{\sqrt {17} - 16} \over 8}$$
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