Answer
$\approx 0.7456$
Work Step by Step
The interval [a,b] is subdivided into n intervals,
f is a continuous function over [a,b],
Left Riemann sum $=\displaystyle \sum_{k=0}^{n-1}f(x_{k})\Delta x$
$=[f(x_{0})+f(x_{1})+\cdots+f(x_{n-1})]\Delta x$,
where $a=x_{0} < x_{1} < \cdots < x_{n}=b$
are the endpoints of the subdivisions,
and $\displaystyle \Delta x=\frac{b-a}{n}$.
-----------------
$f(x)=\displaystyle \frac{1}{1+x},\ \ [a,b]=[0,1],\ \ n=5$
$\displaystyle \Delta x=\frac{b-a}{n}=\frac{1-0}{5}=0.2$
$\left[\begin{array}{lllllll}
k & 0 & 1 & 2 & 3 & 4 & 5\\
x_{k} & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1\\
f(x_{k}) & 1 & 5/6 & 5/7 & 5/8 & 5/9 & ...
\end{array}\right]$
$ f(0)=\displaystyle \frac{1}{1}\quad$
$f(0.2)=\displaystyle \frac{1}{1.2}=\frac{10}{12}=\frac{5}{6}$
$f(0.4)=\displaystyle \frac{1}{1.4}=\frac{10}{14}=\frac{5}{7}$
$f(0.6)=\displaystyle \frac{1}{1.6}=\frac{10}{16}=\frac{5}{8}$
$f(0.8)=\displaystyle \frac{1}{1.8}=\frac{10}{18}=\frac{5}{9}$
Left Riemann sum
$=(1+\displaystyle \frac{5}{6}+\frac{5}{7}+\frac{5}{8}+\frac{5}{9})\cdot 0.2\approx 0.7456$
