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Mathematics Question on Fundamental Theorem of Calculus

Let [t][t] denote the greatest integer less than or equal to tt. Then, the value of the integral ∫01[−8x2+6x−1]dx∫_0^1 [ -8x^2 + 6x - 1] dx is equal to

A

−1-1

B

−54-\frac 54

C

17−138\frac {\sqrt {17}-13}{8}

D

17−168\frac {\sqrt {17}-16}{8}

Answer

17−138\frac {\sqrt {17}-13}{8}

Explanation

Solution

Let t denote the greatest integer less than or equal to t

∫01[−8x2+6x−1]dx∫_0^1 [ -8x^2 + 6x - 1] dx

=∫014(−1)dx+∫14340dx+∫3412(−1)dx+∫3+17834(−1)dx+∫3+1781(−3)dx= ∫_0^\frac 14 (-1)dx + ∫_{\frac 14}^{\frac 34} 0dx + ∫_{\frac 34}^{\frac 12} (-1)dx + ∫_{\frac {3+\sqrt {17}}{8}}^{\frac 34} (-1)dx + ∫_{\frac {3+\sqrt {17}}{8}}^{1}(-3)dx

=−14−14−2(3+178−34)−3(1−3+178)= -\frac 14 -\frac 14 - 2 ( \frac {3+\sqrt {17}}{8} - \frac 34) - 3 (1 - \frac {3+\sqrt {17}}{8})

=17−138=\frac {\sqrt {17}-13}{8}

So, the correct option is (C): 17−138\frac {\sqrt {17}-13}{8}