To solve this problem, we are considering the function $f(t) = 2t^2$ on the interval $[1, 4]$, and dividing this interval into $n$ subintervals.

Step 1: Identify $a$, $b$, and $\Delta t$

• $a$ is the left endpoint of the interval, so $a = 1$.
• $b$ is the right endpoint of the interval, so $b = 4$.
• The width of each subinterval is $\Delta t$, which is given by: $\Delta t = \frac{b - a}{n} = \frac{4 - 1}{n} = \frac{3}{n}$

Step 2: Set up the upper sum using right endpoints

The right endpoint of the $i$-th subinterval is given by: $t_i = a + i \cdot \Delta t = 1 + i \cdot \frac{3}{n}$ where $i = 1, 2, 3, \dots, n$.

To find the upper sum using the right endpoints, we sum up the function evaluated at each right endpoint. The upper sum $U_n$ is given by: $U_n = \sum_{i=1}^{n} f(t_i) \cdot \Delta t$ Substitute $f(t) = 2t^2$ and $t_i = 1 + \frac{3i}{n}$: $U_n = \sum_{i=1}^{n} 2\left( 1 + \frac{3i}{n} \right)^2 \cdot \frac{3}{n}$ Now, we can expand the term $\left( 1 + \frac{3i}{n} \right)^2$: $\left( 1 + \frac{3i}{n} \right)^2 = 1 + 2\cdot \frac{3i}{n} + \left(\frac{3i}{n}\right)^2 = 1 + \frac{6i}{n} + \frac{9i^2}{n^2}$ Thus, the upper sum becomes: $U_n = \sum_{i=1}^{n} 2\left( 1 + \frac{6i}{n} + \frac{9i^2}{n^2} \right) \cdot \frac{3}{n}$ Distribute the $2$ and $\frac{3}{n}$: $U_n = \frac{6}{n} \sum_{i=1}^{n} \left( 1 + \frac{6i}{n} + \frac{9i^2}{n^2} \right)$ Now, separate the sum: $U_n = \frac{6}{n} \left( \sum_{i=1}^{n} 1 + \sum_{i=1}^{n} \frac{6i}{n} + \sum_{i=1}^{n} \frac{9i^2}{n^2} \right)$

Step 3: Use known summation formulas

The summation formulas for $i$ and $i^2$ are: $\sum_{i=1}^{n} 1 = n$ $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

Now, substitute these into the upper sum expression: $U_n = \frac{6}{n} \left( n + \frac{6}{n} \cdot \frac{n(n+1)}{2} + \frac{9}{n^2} \cdot \frac{n(n+1)(2n+1)}{6} \right)$

Step 4: Simplify the upper sum expression

Simplifying each term:

1. The first term is just: $\frac{6}{n} \cdot n = 6$
2. The second term simplifies to: $\frac{6}{n} \cdot \frac{6n(n+1)}{2n} = 18 \cdot \frac{n+1}{n}$
3. The third term simplifies to: $\frac{6}{n} \cdot \frac{9n(n+1)(2n+1)}{6n^2} = 9 \cdot \frac{(n+1)(2n+1)}{n^2}$

So the upper sum $U_n$ becomes: $U_n = 6 + 18 \cdot \frac{n+1}{n} + 9 \cdot \frac{(n+1)(2n+1)}{n^2}$

This is the expression for the upper sum in terms of $n$.

Would you like more details on the simplification or any further clarifications?

---

Here are 5 related questions you might want to consider:

1. What is the lower sum for the same interval and subintervals?
2. How does the upper sum change as $n \to \infty$?
3. What would the midpoint sum look like for this function and interval?
4. How can we interpret the upper sum in terms of the Riemann sum and integral approximation?
5. How does this process change if we divide the interval into unequal subintervals?

Tip: When calculating sums involving powers of $i$, using summation formulas helps simplify complex expressions efficiently.