Answer
The left endpoint approximation for the area $A=6.75$.
Work Step by Step
We have $x^{2}=f(x),$ and the interval [1,3]
\[
\Delta x=\frac{-a+b}{n}=\frac{-1+3}{4}=\frac{1}{2}
\]
Left endpoints $x_{k}^{*}=x_{k-1}=a+(-1+k) \Delta=1+(-1+k) \frac{1}{2}$
$f\left(x_{k}^{*}\right)=\left(1+\frac{1}{2}(-1+k)\right)^{2}$
\[
\begin{aligned}
A=\sum_{k=1}^{4} f\left(x_{k}^{*}\right) \Delta &=\sum_{k=1}^{4}\left(1+\frac{1}{2}(-1+k)\right)^{2} \cdot \frac{1}{2} \\
&=\frac{1}{2}\left[4+\frac{25}{4}+1+\frac{9}{4}\right] \\
&=\frac{54}{4} \cdot \frac{1}{2} \\
&=6.75
\end{aligned}
\]
And therefore, the left endpoint approximation for the area is $A=6.75$
