In this problem you will calculate $\int_{3}^{5} 5 x d x$ by using the formal definition of the definite integral:
\[
\int_{a}^{b} f(x) d x=\lim _{n \rightarrow \infty}\left[\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x\right] .
\]
(a) The interval $[3,5]$ is divided into $n$ equal subintervals of length $\Delta x$. What is $\Delta x$ (in terms of $n$ )?
\[
\Delta x=
\]
(b) The right-hand endpoint of the $k$ th subinterval is denoted $x_{k}^{*}$. What is $x_{k}^{*}$ (in terms of $k$ and $n$ )?
\[
x_{k}^{*}=
\]
(c) Using these choices for $x_{k}^{*}$ and $\Delta x$, the definition tells us that
\[
\int_{3}^{5} 5 x d x=\lim _{n \rightarrow \infty}\left[\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x\right]
\]
What is $f\left(x_{k}^{*}\right) \Delta x$ (in terms of $k$ and $n$ )?
\[
f\left(x_{k}^{*}\right) \Delta x=
\]
(d) Express $\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x$ in closed form. (Your answer will be in terms of $n$.)
\[
\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x=\square
\]
(e) Finally, complete the problem by taking the limit as $n \rightarrow \infty$ of the expression that you found in the
\[
\int_{3}^{5} 5 x d x=\lim _{n \rightarrow \infty}\left[\sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x\right]=\square
\]
Answer
\( \boxed{\int_{3}^{5} 5 x d x = 35} \)
Step 1 ：\( \Delta x = \frac{2}{n} \)
Step 2 ：\( x_{k}^{*} = 3 + k \left( \frac{2}{n} \right) \)
Step 3 ：\( f\left(x_{k}^{*}\right) \Delta x = 5 \left(3 + k \left(\frac{2}{n}\right)\right) \left(\frac{2}{n}\right) \)
Step 4 ：\( \sum_{k=1}^{n} f\left(x_{k}^{*}\right) \Delta x = \sum_{k=1}^{n} 5 \left(\frac{6}{n} + \frac{2k}{n^2}\right) = \frac{30}{n} \sum_{k=1}^{n} 1 + \frac{10}{n^2} \sum_{k=1}^{n} k \)
Step 5 ：\( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
Step 6 ：\( \int_{3}^{5} 5 x d x = \lim _{n \rightarrow \infty}\left[30 + \frac{10(n+1)}{2n}\right] = 30 + \lim _{n \rightarrow \infty}\frac{10(n+1)}{2n} \)
Step 7 ：\( \boxed{\int_{3}^{5} 5 x d x = 35} \)
