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QJEEM22Macd2a: JEE Main 2022 Mathematics - Class 12 JEEM22Macd2a Solution: Let [t] denote the greatest integer less than or equal to t.…

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral ∫01[−8x2+6x−1]dx\int\limits_0^1 {[ - 8{x^2} + 6x - 1]dx} is equal to :

Multiple Choice Options

  • A −- 1
  • B −54{{ - 5} \over 4}
  • C 17−138{{\sqrt {17} - 13} \over 8}
  • D 17−168{{\sqrt {17} - 16} \over 8}

Step-by-Step Verified Solution

∫01[−8x2+6x−1]dx\int_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x

JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Definite Integration Question 135 English Explanation
=∫014(−1)dx+∫14340dx+∫1234−1dx+∫348−2dx+∫3+1781−3dx=\int_{0}^{\frac{1}{4}}(-1) d x+\int_{\frac{1}{4}}^{\frac{3}{4}} 0 d x+\int_{\frac{1}{2}}^{\frac{3}{4}}-1 d x+\int_{\frac{3}{4}}^{8}-2 d x+\int_{\frac{3+\sqrt{17}}{8}}^{1}-3 d x

=−14−14−2(3+178−34)−3(1−3+178)=-\frac{1}{4}-\frac{1}{4}-2\left(\frac{3+\sqrt{17}}{8}-\frac{3}{4}\right)-3\left(1-\frac{3+\sqrt{17}}{8}\right)

=17−138=\frac{\sqrt{17}-13}{8}

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