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

$\int_0^\pi \frac{1}{4+3 \cos x} d x=$

If $I=\int_1^3 \sqrt{3+x+x^2} dx$,then $I$ lies in the interval

Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3 \left( [x^2] + [\frac{x^2}{2}] \right) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$,where $a, b, c \in \mathbb{Z}$,then $a + b + c$ is equal to:

Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t - a \sin 4 t}{t^2} d t = \frac{a \sin 4 x}{x} - 1$. Find the values of $a$ and $b$.

Let $I = \int_{10}^{19} \frac{\sin x}{1+x^{6}} dx$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module