JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \([t]\) denote the greatest integer less than or equal to \(t.\) Then, the value of the integral \(\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x\) is equal to
- A \(-1\)
- B \(-\frac{5}{4}\)
- C \(\frac{\sqrt{17}-13}{8}\)
- D \(\frac{\sqrt{17}-16}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{17}-13}{8}\)
Step-by-step Solution
Detailed explanation
\(\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x\) \(=\int\limits_{0}^{1 / 4}-1 dx +\int\limits_{1 / 4}^{1 / 2} 0 dx +\int\limits_{1 / 2}^{3 / 4}-1 dx\) \(+\int\limits_{3 / 4}^{\frac{3+\sqrt{17}}{8}}-2 dx +\int\limits_{\frac{3+\sqrt{17}}{8}}^{1}-3 dx\)…
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