In this problem you will calculate $\displaystyle\int_{0}^{\,2} x^2 + 4 \;dx$ by using the formal definition of the definite integral: $$\int_{a}^{b} f(x) \;dx = \lim_{n \to \infty} \left[ \sum\limits_{k=1}^{n} f(x^{*}_{k}) \Delta x \right].$$ (a) The interval $[0,2]$ 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 $$\int_{0}^{\,2} x^2 + 4 \;dx = \lim_{n \to \infty} \left[ \sum\limits_{k=1}^{n} f(x^{*}_{k}) \Delta x \right].$$

What is $f(x^{*}_{k}) \Delta x$ (in terms of $k$ and $n$)?

$f(x^{*}_{k}) \Delta x$ =

(d) Express $\sum\limits_{k=1}^{n} f(x^{*}_{k}) \Delta x$ in closed form. (Your answer will be in terms of $n$.)

$\sum\limits_{k=1}^{n} f(x^{*}_{k}) \Delta x$ =

(e) Finally, complete the problem by taking the limit as $n \rightarrow \infty$ of the expression that you found in the previous part.

$\displaystyle\int_{0}^{\,2} x^2 + 4 \;dx = \lim_{n \to \infty} \left[ \sum\limits_{k=1}^{n} f(x^{*}_{k}) \Delta x \right]$ =