Answer
$\text {The area under the curve = $\frac{125}{6}$}$
Work Step by Step
$\text {It is given that}$
\begin{align}
f(x) = 5x-x^2; \ \ [a, b] = [0, 5]
\end{align}
$\text {Also, it is given that x$_k^*$ should be the left endpoint of each subinterval:}$
\begin{align}
x_k^* = a + (k-1) \varDelta x = (k-1) \varDelta x
\end{align}
$\text {According to the Definition 4.4.3:}$
\begin{align}
A = \lim\limits_{n \to \infty} \sum_{k=1}^n f(x_k^*) \varDelta x
\end{align}
$\text {Therefore:}$
\begin{align}
&A = \lim\limits_{n \to \infty} \sum_{k=1}^n (5(k-1) \varDelta x-(k-1)^2\varDelta x^2)\varDelta x = \\ & = \lim\limits_{n \to \infty} \sum_{k=1}^n (5k \varDelta x^2- 5 \varDelta x^2-k^2\varDelta x^3+2k\varDelta x^3 - \varDelta x^3) \\
& where \ \varDelta x = \frac{b-a}{n} = \frac{5}{n}
\end{align}
$\text {Thus,}$
\begin{align}
&A = \lim\limits_{n \to \infty} \sum_{k=1}^n \left(\frac{125k}{n^2} - \frac{125}{n^2} - \frac{125k^2}{n^3} + 2\frac{125k}{n^3} - \frac{125}{n^3}\right) = \\
& = \lim\limits_{n \to \infty} \left(\frac{125}{n^2}\sum_{k=1}^n k - \frac{125}{n^2}\sum_{k=1}^n 1- \frac{125}{n^3}\sum_{k=1}^n k^2 + \frac{125}{n^3}\sum_{k=1}^n 2k - \frac{125}{n^3}\sum_{k=1}^n 1\right) = \\
& = \lim\limits_{n \to \infty} \left(\frac{125}{n^2} \times \frac{n(n+1)}{2} - \frac{125}{n} -\frac{125}{n^3} \times \frac{n(n+1)(2n+1)}{6} \\ + \frac{125}{n^3} \times n(n+1) - \frac{125}{n^2}\right) = \\
& = \frac{125}{2} - \frac{125}{3} = \frac{125}{6}
\end{align}
