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+1 vote
9.8k views
in Mathematics by (43.5k points)
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Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral
\(\int\limits_0^1\) [-8x2 + 6x - 1] dx is equal to

(A) -1

(B) - 5/4

(C) \(\frac{\sqrt{17}-13}{8}\)

(D) \(\frac{\sqrt{17}-16}{8}\)

1 Answer

+2 votes
by (44.3k points)
selected by
 
Best answer

Correct option is (C) \(\frac{\sqrt{17}-13}{8}\)

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