Let 0

Question

(B) The integral is given by $\int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x} = [\log(1+x)]_{1/(k+\beta)}^{1/(k+\alpha)} = \log(1+\frac{1}{k+\alpha})

Vedclass · Accepted Answer

$\log_{e} \frac{1+\beta}{1+\alpha}$

Vedclass · Answer

(B) The integral is given by $\int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x} = [\log(1+x)]_{1/(k+\beta)}^{1/(k+\alpha)} = \log(1+\frac{1}{k+\alpha}) - \log(1+\frac{1}{k+\beta})$.Summing from $k=1$ to $n$, we get $\sum_{k=1}^{n} (\log(\frac{k+\alpha+1}{k+\alpha}) - \log(\frac{k+\beta+1}{k+\beta}))$.This is a telescoping sum: $(\log(\frac{2+\alpha}{1+\alpha}) - \log(\frac{2+\beta}{1+\beta})) + (\log(\frac{3+\alpha}{2+\alpha}) - \log(\frac{3+\beta}{2+\beta})) + \dots + (\log(\frac{n+\alpha+1}{n+\alpha}) - \log(\frac{n+\beta+1}{n+\beta}))$.As $n \rightarrow \infty$, the terms $\log(\frac{n+\alpha+1}{n+\alpha}) \rightarrow \log(1) = 0$ and $\log(\frac{n+\beta+1}{n+\beta}) \rightarrow \log(1) = 0$.The remaining terms are $\log(\frac{1+\beta}{1+\alpha})$.

Let $0 < \alpha < \beta < 1$. Then $\lim_{n \rightarrow \infty} \sum_{k=1}^{n} \int_{1/(k+\beta)}^{1/(k+\alpha)} \frac{dx}{1+x}$ is

Similar Questions

$\int_{2}^{3} \frac{dx}{x^{2}+x} = $

If $[2,6]$ is divided into four intervals of equal length,then the approximate value of $\int_2^6 \frac{1}{x^2-x} dx$ using Simpson's rule is

If $\beta + 2 \int\limits_0^1 {{x^2}\,{e^{ - {x^2}}}}\, dx = \int\limits_0^1 {{e^{ - {x^2}}}}\, dx$,then the value of $\beta$ is

The set of values of $a$ which satisfy the equation $\int_{0}^{2} (t - \log_{2} a) \, dt = \log_{2} \left( \frac{4}{a^{2}} \right)$ is

The value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{1}{\sin 2 x(\tan ^5 x+\cot ^5 x)} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module