SATHEE: Indefinite Integration 4 Question 1

Indefinite Integration 4 Question 1

1. $lim_{n \to \infty} \frac{(n + 1)^{1 / 3}}{n^{4 / 3}} + \frac{(n + 2)^{1 / 3}}{n^{4 / 3}} + \dots . + \frac{(2 n)^{1 / 3}}{n^{4 / 3}}$ is equal to

(a) $\frac{4}{3} (2)^{4 / 3}$

(b) $\frac{3}{4} (2)^{4 / 3} - \frac{4}{3}$

(d) $\frac{4}{3} (2)^{3 / 4}$

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Answer:

Correct Answer: 1. (c)

Solution:

Let $p = lim_{n \to \infty} \frac{(n + 1)^{1 / 3}}{n^{4 / 3}} + \frac{(n + 2)^{1 / 3}}{n^{4 / 3}} + \dots + \frac{(2 n)^{1 / 3}}{n^{4 / 3}}$

$= lim_{n \to \infty} \sum_{r = 1}^{n} \frac{(n + r)^{1 / 3}}{n^{4 / 3}}$

$\begin{aligned} = lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1 + {\frac{r}{n}}^{1 / 3} n^{1 / 3}}{n^{4 / 3}} \\ = lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} 1 + {\frac{r}{n}}^{1 / 3} \end{aligned}$

Now, as per integration as limit of sum.

$Let \frac{r}{n} = x and \frac{1}{n} = d x$

$[∵ n \to \infty]$

Then, upper limit of integral is 1 and lower limit of integral is 0.

$So, \begin{aligned} p & = \int_{0}^{1} (1 + x)^{1 / 3} d x ∵ lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f \frac{r}{n} = \int_{0}^{1} f (x) d x \\ = \frac{3}{4} (1 + x)^{4 / 3} = \frac{3}{4} (2^{4 / 3} - 1) = \frac{3}{4} (2)^{4 / 3} - \frac{3}{4} \end{aligned}$

Indefinite Integration 4 Question 1

1. $lim_{n \to \infty} \frac{(n + 1)^{1 / 3}}{n^{4 / 3}} + \frac{(n + 2)^{1 / 3}}{n^{4 / 3}} + \dots . + \frac{(2 n)^{1 / 3}}{n^{4 / 3}}$ is equal to

Answer:

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Indefinite Integration 4 Question 1

1. limn→∞(n+1)1/3n4/3+(n+2)1/3n4/3+….+(2n)1/3n4/3 is equal to

Answer:

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1. $lim_{n \to \infty} \frac{(n + 1)^{1 / 3}}{n^{4 / 3}} + \frac{(n + 2)^{1 / 3}}{n^{4 / 3}} + \dots . + \frac{(2 n)^{1 / 3}}{n^{4 / 3}}$ is equal to