By the definition of the definite integral,th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given limit can be expressed as a Riemann sum: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \in

Vedclass · Accepted Answer

$\frac{1}{5} \log 2$

Vedclass · Answer

(B) The given limit can be expressed as a Riemann sum: $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{r^5+n^5} = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r^4}{n^5( (r/n)^5 + 1 )} = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{(r/n)^4}{(r/n)^5 + 1}$. By the definition of the definite integral,this is equal to $\int_0^1 \frac{x^4}{x^5+1} dx$. Let $t = x^5 + 1$,then $dt = 5x^4 dx$,or $x^4 dx = \frac{dt}{5}$. When $x=0, t=1$. When $x=1, t=2$. Thus,the integral becomes $\int_1^2 \frac{1}{t} \cdot \frac{dt}{5} = \frac{1}{5} [\log |t|]_1^2 = \frac{1}{5} (\log 2 - \log 1) = \frac{1}{5} \log 2$.

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ is

Similar Questions

$\int_0^{1.5} [x^2] dx$ is equal to

$\int_0^a {x^4 \sqrt{a^2 - x^2}} \,dx = $

$\int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx = } $

Evaluate the definite integral $\int_{-1}^{2} \left[ \frac{[x]}{1 + x^2} \right] dx$,where $[\cdot]$ denotes the Greatest Integer Function $(GIF)$.

The value of the definite integral,$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin 5x}}{{\sin x}}\,dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module