Definite Integration Question 1
Question 1
- $\lim _{n \rightarrow \infty} \frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots .+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}$ is equal to (a) $\frac{4}{3}(2)^{4 / 3}$ (b) $\frac{3}{4}(2)^{4 / 3}-\frac{4}{3}$ (c) $\frac{3}{4}(2)^{4 / 3}-\frac{3}{4}$ (d) $\frac{4}{3}(2)^{3 / 4}$
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Answer:
Correct Answer: 1. (c)
Solution:
- Let $p=\lim _{n \rightarrow \infty} \frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}$
$$ =\lim {n \rightarrow \infty} \sum{r=1}^{n} \frac{(n+r)^{1 / 3}}{n^{4 / 3}} $$
$$ \begin{aligned} & =\lim {n \rightarrow \infty} \sum{r=1}^{n} \frac{1+\frac{r}{n}^{1 / 3} n^{1 / 3}}{n^{4 / 3}} \ & =\lim {n \rightarrow \infty} \frac{1}{n} \sum{r=1}^{n} 1+\frac{r}{n}^{1 / 3} \end{aligned} $$
Now, as per integration as limit of sum.
$$ \text { Let } \frac{r}{n}=x \text { and } \frac{1}{n}=d x $$
$[\because n \rightarrow \infty]$
Then, upper limit of integral is 1 and lower limit of integral is 0.
$$ \text { So, } \begin{aligned} p & =\int_{0}^{1}(1+x)^{1 / 3} d x \quad \because \lim {n \rightarrow \infty} \frac{1}{n} \sum{r=1}^{n} f \frac{r}{n}=\int_{0}^{1} f(x) d x \ & =\frac{3}{4}(1+x)^{4 / 3}=\frac{3}{4}\left(2^{4 / 3}-1\right)=\frac{3}{4}(2)^{4 / 3}-\frac{3}{4} \end{aligned} $$
