Answer
$\approx 0.2932$
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 $\displaystyle \Delta x=\frac{b-a}{n}$.
-----------------
$f(x)=\displaystyle \frac{x}{1+x^{2}},\ \ [a,b]=[0,1],\ \ n=5$
$\displaystyle \Delta x=\frac{b-a}{n}=\frac{1-0}{5}=0.2$
$\left[\begin{array}{lllllll}
k & 0 & 1 & 2 & 3 & 4 & 5\\
x_{k} & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1\\
f(x_{k}) & 0 & \frac{0.2}{1.04} & \frac{0.4}{1.16} & \frac{0.6}{1.36} & \frac{0.8}{1.64} & ...
\end{array}\right]$
Left Riemann sum $=$
$=(0+\displaystyle \frac{0.2}{1.04}+\frac{0.4}{1.16}+\frac{0.6}{1.36}+\frac{0.8}{1.64})\cdot 0.2$
$\approx 0.2932$
