Answer
$\text {Area under the graph = $\frac{32}{3}$}$
Work Step by Step
$\text {It is given that}$
\begin{align}
f(x) = 4x-x^2; \ \ [a, b] = [0, 4]
\end{align}
$\text {Also, it is given that x$_k^*$ should be the left endpoint of each subinterval:}$
\begin{align}
x_k^* = a + k \varDelta x = k \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 (4k \varDelta x-k^2\varDelta x^2)\varDelta x = \lim\limits_{n \to \infty} \sum_{k=1}^n (4k \varDelta x^2-k^2\varDelta x^3) \\
& where \ \varDelta x = \frac{b-a}{n} = \frac{4}{n}
\end{align}
$\text {Thus,}$
\begin{align}
&A = \lim\limits_{n \to \infty} \sum_{k=1}^n \left(\frac{64k}{n^2} - \frac{64k^2}{n^3}\right) = \lim\limits_{n \to \infty} \left(\frac{64}{n^2}\sum_{k=1}^n k - \frac{64}{n^3}\sum_{k=1}^n k^2\right) = \\
& = \lim\limits_{n \to \infty} \left(\frac{64}{n^2} \times \frac{n(n+1)}{2} - \frac{64}{n^3} \times \frac{n(n+1)(2n+1)}{6}\right) = \\
& = \lim\limits_{n \to \infty} \left(32 \left(1+\frac{1}{n} \right) - \frac{32}{3} \left(1+\frac{1}{n} \right) \left(2+\frac{1}{n} \right)\right) = 32 - \frac{64}{3} =\frac{32}{3}
\end{align}
