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Definite Integral

jacks

Well-known member
Apr 5, 2012
226
Calculation of : $\displaystyle \int_{0}^{1991}\{ \frac{2x+5}{x+1}\}[ x]dx$, where $[ x]$ and $\{ x \}$ denote the integral and fractional part of $x$

My Trial :: $\displaystyle \int_{0}^{1991}\left\{\frac{(2x+2)+3}{x+1}\right\}\cdot [x]dx$

$\displaystyle \int_{0}^{1991}\left\{2+\frac{3}{x+1}\right\}[x]dx = \int_{0}^{1991}\left\{\frac{3}{x+1}\right\}[x]dx$

Using the formula $\left\{z+x\right\} = \left\{x\right\}$,where $z$ is an Integer

Now How can i slve after that

Help me

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Calculation of : $\displaystyle \int_{0}^{1991}\{ \frac{2x+5}{x+1}\}[ x]dx$, where $[ x]$ and $\{ x \}$ denote the integral and fractional part of $x$

My Trial :: $\displaystyle \int_{0}^{1991}\left\{\frac{(2x+2)+3}{x+1}\right\}\cdot [x]dx$

$\displaystyle \int_{0}^{1991}\left\{2+\frac{3}{x+1}\right\}[x]dx = \int_{0}^{1991}\left\{\frac{3}{x+1}\right\}[x]dx$

Using the formula $\left\{z+x\right\} = \left\{x\right\}$,where $z$ is an Integer

Now How can i slve after that
Split the interval from $0$ to $1991$ into subintervals of length $1$: $$ \int_{0}^{1991}\left\{\frac{3}{x+1}\right\}\cdot [x]dx = \sum_{n=0}^{1990} \int_n^{n+1} \left\{\frac{3}{x+1}\right\}\cdot [x]dx.$$ In the first interval, from $0$ to $1$, $[x]=0$, so you can ignore the integral over that interval. In the next interval, from $1$ to $2$, $\frac{3}{x+1}$ lies between $1$ and $2$ and $[x]=1$. After that, in each of the remaining intervals, $\frac{3}{x+1}$ is less than $1$, and $[x]=n$. So you can find \(\displaystyle \int_n^{n+1} \left\{\frac{3}{x+1}\right\}\cdot [x]dx\) for each $n$. That will give the answer in the form of a series. But it might not be easy to find the sum of that series.
 

johng

Well-known member
MHB Math Helper
Jan 25, 2013
236
Hi,
Opalg certainly pointed you in the right direction. The following should help you to arrive at a relatively simple formula.

MHBcalculus2.png