Answer
The given integral is
$$
\begin{aligned}
\int_{0}^{1}\left(x^{3}-3 x^{2}\right) d x&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=-\frac{3}{4}
\end{aligned}
$$
Work Step by Step
$$
\int_{0}^{1}\left(x^{3}-3 x^{2}\right) d x
$$
The width of the subintervals is
$$ \Delta x=\frac{1-0}{n}=\frac{1}{n} ,
$$
and
$$
x_{i}=0+i \Delta x=\frac{i}{n}$$
The given integral is
$$
\begin{aligned}
\int_{0}^{1}\left(x^{3}-3 x^{2}\right) d x&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(x_{i}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n} f\left(\frac{i}{n}\right) \Delta x \\
&=\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[\left(\frac{i}{n}\right)^{3}-3\left(\frac{i}{n}\right)^{2}\right] \frac{1}{n}\\
&=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n}\left[\frac{i^{3}}{n^{3}}-\frac{3 i^{2}}{n^{2}}\right]\\
&=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{n^{3}} \sum_{i=1}^{n} i^{3}-\frac{3}{n^{2}} \sum_{i=1}^{n} i^{2}\right]\\
&=\lim _{n \rightarrow \infty}\left\{\frac{1}{n^{4}}\left[\frac{n(n+1)}{2}\right]^{2}-\frac{3}{n^{3}} \frac{n(n+1)(2 n+1)}{6}\right\}\\
&=\lim _{n \rightarrow \infty}\left[\frac{1}{4} \frac{n+1}{n} \frac{n+1}{n}-\frac{1}{2} \frac{n+1}{n} \frac{2 n+1}{n}\right] \\
&=\lim _{n \rightarrow \infty}\left[\frac{1}{4}\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)-\frac{1}{2}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right]\\
&=\frac{1}{4}(1)(1)-\frac{1}{2}(1)(2)\\
&=-\frac{3}{4}
\end{aligned}
$$
