Answer
$\approx 25$
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 $\displaystyle \Delta x=\frac{b-a}{n}$.
-----------------
$f(x)=e^{-x^{2}},\ \ [a,b]=[0,100],\ \ n=4$
$\displaystyle \Delta x=\frac{b-a}{n}=\frac{100-0}{4}=25$
$\left[\begin{array}{llllll}
k & 0 & 1 & 2 & 3 & 4\\
x_{k} & 0 & 25 & 50 & 75 & 100\\
f(x_{k}) & 1 & e^{-625} & e^{-2500} & e^{-5625} & ...
\end{array}\right]$
Left Riemann sum $=$
$=(1+e^{-625}+e^{-2500}+e^{-5625})\cdot 25$
$\approx$25.0000000003
$\approx 25$
