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

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(D) We use the definition of the definite integral as the limit of a sum: $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} f\left(\f

Vedclass · Accepted Answer

$\frac{1}{5}$

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(D) We use the definition of the definite integral as the limit of a sum: $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight) = \int_0^1 f(x) dx$. For the first term: $\mathop {\lim }\limits_{n 	o \infty } \frac{\sum_{r=1}^{n} r^4}{n^5} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}ight)^4 = \int_0^1 x^4 dx = \left[ \frac{x^5}{5} ight]_0^1 = \frac{1}{5}$. For the second term: $\mathop {\lim }\limits_{n 	o \infty } \frac{\sum_{r=1}^{n} r^3}{n^5} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n^2} \left( \frac{1}{n} \sum_{r=1}^{n} \left(\frac{r}{n}ight)^3 ight) = 0 	imes \int_0^1 x^3 dx = 0 	imes \frac{1}{4} = 0$. Thus,the result is $\frac{1}{5} - 0 = \frac{1}{5}$.

$\mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^4} + {3^4} + .... + {n^4}}}{{{n^5}}} - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^3} + {3^3} + .... + {n^3}}}{{{n^5}}} = $

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Evaluate the following definite integral as a limit of sums: $\int_{a}^{b} x \, dx$

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots \left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=$

For a sufficiently large value of $n$,the sum of the square roots of the first $n$ positive integers,i.e.,$\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n}$,is approximately equal to:

$\mathop {\lim }\limits_{n \to \infty } \left( \frac{1^2}{1^3 + n^3} + \frac{2^2}{2^3 + n^3} + \dots + \frac{n^2}{n^3 + n^3} \right)$ is equal to

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