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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\left[ {{1^2}\sin \frac{1}{n} + {2^2}\sin \frac{2}{n} + {3^2}\sin \frac{3}{n} + ....+{n^2}\sin \frac{n}{n}} \right]$ equals
  • A
    $cos1 + 2sin1$
  • B
    $2sin1 -2$
  • C
    $cos1 -2sin1 -2$
  • ✓
    $cos1 + 2sin1 -2$
✓

Answer

Correct option: D.
$cos1 + 2sin1 -2$
d
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum {\frac{{{r^2}}}{{{n^2}}}} \sin \frac{r}{n} = \int_0^1 {{x^2}} \sin xdx$

$\left[x^{2}(-\cos x)-2 x(-\sin x)+2 \cos x\right]_{0}^{1}$

$=(-\cos 1+2 \sin 1+2 \cos 1)-2$

$=\cos 1+2 \sin 1-2$

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