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Question

(A) First,evaluate the definite integral: $\int_{0}^{1} a^k x^k dx = a^k \int_{0}^{1} x^k dx = a^k [\frac{x^{k+1}}{k+1}]_{0}^{1} = \frac{a^k}{k+1}$. N

Vedclass · Accepted Answer

$\lim_{n 	o \infty} \frac{a^k (1^k + 2^k + 3^k + \dots + n^k)}{n^{k+1}}$

Vedclass · Answer

(A) First,evaluate the definite integral: $\int_{0}^{1} a^k x^k dx = a^k \int_{0}^{1} x^k dx = a^k [\frac{x^{k+1}}{k+1}]_{0}^{1} = \frac{a^k}{k+1}$. Now,consider option $A$: $\lim_{n 	o \infty} \frac{a^k (1^k + 2^k + 3^k + \dots + n^k)}{n^{k+1}} = a^k \lim_{n 	o \infty} \sum_{r=1}^{n} \frac{r^k}{n \cdot n^k} = a^k \lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^{n} (\frac{r}{n})^k$. By the definition of the definite integral as the limit of a sum,$\lim_{n 	o \infty} \frac{1}{n} \sum_{r=1}^{n} (\frac{r}{n})^k = \int_{0}^{1} x^k dx = \frac{1}{k+1}$. Therefore,the expression becomes $a^k \cdot \frac{1}{k+1} = \frac{a^k}{k+1}$. Thus,option $A$ is correct.

$\int_{0}^{1} a^k x^k dx =$

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