The summation in the definition of the definite integral is then,

\[\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \sum\limits_{i = 1}^n {\left[ {\left( {\frac{{6i}}{n}} \right)\left( {\frac{i}{n} - 1} \right)} \right]\left[ {\frac{1}{n}} \right]} = \sum\limits_{i = 1}^n {\left[ {\frac{{6{i^2}}}{{{n^3}}} - \frac{{6i}}{{{n^2}}}} \right]} \] Show Step 3

Now we need to use the formulas from the Summation Notation section in the Extras chapter to “evaluate” the summation.

\[\begin{align*}\sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} & = \sum\limits_{i = 1}^n {\left[ {\frac{{6{i^2}}}{{{n^3}}}} \right]} - \sum\limits_{i = 1}^n {\left[ {\frac{{6i}}{{{n^2}}}} \right]} = \frac{6}{{{n^3}}}\sum\limits_{i = 1}^n {{i^2}} - \frac{6}{{{n^2}}}\sum\limits_{i = 1}^n i \\ & = \frac{6}{{{n^3}}}\left( {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right) - \frac{6}{{{n^2}}}\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right) = \frac{{2{n^2} + 3n + 1}}{{{n^2}}} - \frac{{3n + 3}}{n}\end{align*}\] Show Step 4

Finally, we can use the definition of the definite integral to determine the value of the integral.

\[\int_{0}^{1}{{6x\left( {x - 1} \right)\,dx}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {f\left( {x_i^*} \right)\Delta x} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{2{n^2} + 3n + 1}}{{{n^2}}} - \frac{{3n + 3}}{n}} \right] = 2 - 3 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 1}}\]