In this problem you will calculate ${\int }_3^{5}5xdx$ by using the formal definition of the definite integral:
$${\int }_a^{b}f(x)dx=\underset{n\to \mathrm{\infty }}{lim}\left[\sum _{k=1}^{n}f\left(x_k^{\ast }\right)\mathrm{\Delta }x\right].$$
(a) The interval $[3,5]$ is divided into $n$ equal subintervals of length $\mathrm{\Delta }x$. What is $\mathrm{\Delta }x$ (in terms of $n$ )?
$$\mathrm{\Delta }x=$$
(b) The right-hand endpoint of the $k$ th subinterval is denoted $x_k^{\ast }$. What is $x_k^{\ast }$ (in terms of $k$ and $n$ )?
$$x_k^{\ast }=$$
(c) Using these choices for $x_k^{\ast }$ and $\mathrm{\Delta }x$, the definition tells us that
$${\int }_3^{5}5xdx=\underset{n\to \mathrm{\infty }}{lim}\left[\sum _{k=1}^{n}f\left(x_k^{\ast }\right)\mathrm{\Delta }x\right]$$

What is $f\left(x_k^{\ast }\right)\mathrm{\Delta }x$ (in terms of $k$ and $n$ )?
$$f\left(x_k^{\ast }\right)\mathrm{\Delta }x=$$
(d) Express $\sum _{k=1}^{n}f\left(x_k^{\ast }\right)\mathrm{\Delta }x$ in closed form. (Your answer will be in terms of $n$.)
$$\sum _{k=1}^{n}f\left(x_k^{\ast }\right)\mathrm{\Delta }x=◻$$
(e) Finally, complete the problem by taking the limit as $n\to \mathrm{\infty }$ of the expression that you found in the
$${\int }_3^{5}5xdx=\underset{n\to \mathrm{\infty }}{lim}\left[\sum _{k=1}^{n}f\left(x_k^{\ast }\right)\mathrm{\Delta }x\right]=◻$$