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JEE Main 202228 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Definite Integration

Let [t] denote the greatest integer less than or equal to t . Then,...

Let [t]\left[t\right] denote the greatest integer less than or equal to tt. Then, the value of the integral ${∫}_{0}^{1}\left[- 8{x}^{2} + 6x - 1\right]dx is equal toLet [t] denote the greatest integer less than or equal to t . Then, the value of the integral ∫ _ 0 ^ 1 [- 8 x ^ 2 + 6x

Options

  1. A−1- 1
  2. B−54- \dfrac{5}{4}
  3. C17−138\dfrac{\sqrt{17} - 13}{8}
  4. D17−168\dfrac{\sqrt{17} - 16}{8}

Correct answer

C. 17−138\dfrac{\sqrt{17} - 13}{8}

Step-by-step solution

The graph of y=−8x2+6x−1y = - 8{x}^{2} + 6x - 1 will be as shown

Let [t] denote the greatest integer less than or equal to t . Then, the value of the integral ∫ _ 0 ^ 1 [- 8 x ^ 2 + 6x

Now ∫01[−8x2+6x−1]dx{∫}_{0}^{1}\left[- 8{x}^{2} + 6x - 1\right]dx=∫014(−1)dx+∫1412(0)dx+∫1234(−1)dx= {∫}_{0}^{\dfrac{1}{4}}\left(- 1\right)dx + {∫}_{\dfrac{1}{4}}^{\dfrac{1}{2}}\left(0\right)dx + {∫}_{\dfrac{1}{2}}^{\dfrac{3}{4}}\left(- 1\right)dx

+∫343+178(−2)dx+∫3+1781(−3)dx+ {∫}_{\dfrac{3}{4}}^{\dfrac{3 + \sqrt{17}}{8}}\left(- 2\right)dx + {∫}_{\dfrac{3 + \sqrt{17}}{8}}^{1}\left(- 3\right)dx

=−[x]014+0−[x]1234+−2[x]343+178−3[x]3+1781= - {\left[x\right]}_{0}^{\dfrac{1}{4}} + 0 - {\left[x\right]}_{\dfrac{1}{2}}^{\dfrac{3}{4}} + - 2{\left[x\right]}_{\dfrac{3}{4}}^{\dfrac{3 + \sqrt{17}}{8}} - 3{\left[x\right]}_{\dfrac{3 + \sqrt{17}}{8}}^{1}

=−(14−0)−(34−12)−2(3+178−34)−3(1−3+178)= - \left(\dfrac{1}{4} - 0\right) - \left(\dfrac{3}{4} - \dfrac{1}{2}\right) - 2\left(\dfrac{3 + \sqrt{17}}{8} - \dfrac{3}{4}\right) - 3\left(1 - \dfrac{3 + \sqrt{17}}{8}\right)

=−14−14+−6−2178+32−3+9+3178= - \dfrac{1}{4} - \dfrac{1}{4} + \dfrac{- 6 - 2\sqrt{17}}{8} + \dfrac{3}{2} - 3 + \dfrac{9 + 3\sqrt{17}}{8}

=17−138= \dfrac{\sqrt{17} - 13}{8}

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