mathop {lim }limits_{n to infty } frac{1}{{{n | vedclass

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Question

(D) The given limit can be expressed as a Riemann sum: $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} \left( \frac{r}{n} ight)^2

Vedclass · Accepted Answer

$\cos1 + 2\sin1 - 2$

Vedclass · Answer

(D) The given limit can be expressed as a Riemann sum: $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} \left( \frac{r}{n} ight)^2 \sin \left( \frac{r}{n} ight) = \int_0^1 x^2 \sin x \, dx$ Using integration by parts,let $u = x^2$ and $dv = \sin x \, dx$. Then $du = 2x \, dx$ and $v = -\cos x$. $\int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx$ Applying integration by parts again for $\int 2x \cos x \, dx$ with $u = 2x$ and $dv = \cos x \, dx$: $\int 2x \cos x \, dx = 2x \sin x - \int 2 \sin x \, dx = 2x \sin x + 2 \cos x$ Combining these,we get: $\left[ -x^2 \cos x + 2x \sin x + 2 \cos x ight]_0^1$ Evaluating at the limits $0$ and $1$: $= (-1^2 \cos 1 + 2(1) \sin 1 + 2 \cos 1) - (-0^2 \cos 0 + 2(0) \sin 0 + 2 \cos 0)$ $= (-\cos 1 + 2 \sin 1 + 2 \cos 1) - (0 + 0 + 2)$ $= \cos 1 + 2 \sin 1 - 2$

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\left[ {{1^2}\sin \frac{1}{n} + {2^2}\sin \frac{2}{n} + {3^2}\sin \frac{3}{n} + ....+{n^2}\sin \frac{n}{n}} \right]$ equals

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