To solve for $R_5$ of $f(x) = 3x^2 + 8x$ over the interval $[-1, 1]$, we need to compute the Right Riemann Sum with $n = 5$. Here’s a step-by-step outline:

Step 1: Define the Interval

The interval $[-1, 1]$ has a length $L = 1 - (-1) = 2$.

Step 2: Calculate the Width of Each Subinterval

Since we are using 5 subintervals, each subinterval width $\Delta x$ is: $\Delta x = \frac{2}{5} = \frac{2}{5}$

Step 3: Determine the Right Endpoints of Each Subinterval

The right endpoints $x_i$ for each subinterval $i = 1, 2, 3, 4, 5$ are calculated as follows: $x_i = -1 + i \cdot \Delta x = -1 + i \cdot \frac{2}{5}$ Thus, the right endpoints are:

1. $x_1 = -1 + \frac{2}{5} = -\frac{3}{5}$
2. $x_2 = -1 + 2 \cdot \frac{2}{5} = -\frac{1}{5}$
3. $x_3 = -1 + 3 \cdot \frac{2}{5} = \frac{1}{5}$
4. $x_4 = -1 + 4 \cdot \frac{2}{5} = \frac{3}{5}$
5. $x_5 = -1 + 5 \cdot \frac{2}{5} = 1$

Step 4: Evaluate $f(x)$ at Each Right Endpoint

Now we plug each $x_i$ into $f(x) = 3x^2 + 8x$:

1. $f\left(-\frac{3}{5}\right) = 3\left(-\frac{3}{5}\right)^2 + 8\left(-\frac{3}{5}\right)$
2. $f\left(-\frac{1}{5}\right) = 3\left(-\frac{1}{5}\right)^2 + 8\left(-\frac{1}{5}\right)$
3. $f\left(\frac{1}{5}\right) = 3\left(\frac{1}{5}\right)^2 + 8\left(\frac{1}{5}\right)$
4. $f\left(\frac{3}{5}\right) = 3\left(\frac{3}{5}\right)^2 + 8\left(\frac{3}{5}\right)$
5. $f(1) = 3(1)^2 + 8(1)$

Let's evaluate each:

1. $f\left(-\frac{3}{5}\right) = 3 \cdot \frac{9}{25} - \frac{24}{5} = \frac{27}{25} - \frac{120}{25} = -\frac{93}{25}$
2. $f\left(-\frac{1}{5}\right) = 3 \cdot \frac{1}{25} - \frac{8}{5} = \frac{3}{25} - \frac{40}{25} = -\frac{37}{25}$
3. $f\left(\frac{1}{5}\right) = 3 \cdot \frac{1}{25} + \frac{8}{5} = \frac{3}{25} + \frac{40}{25} = \frac{43}{25}$
4. $f\left(\frac{3}{5}\right) = 3 \cdot \frac{9}{25} + \frac{24}{5} = \frac{27}{25} + \frac{120}{25} = \frac{147}{25}$
5. $f(1) = 3 \cdot 1 + 8 = 3 + 8 = 11$

Step 5: Calculate the Right Riemann Sum $R_5$

The Right Riemann Sum $R_5$ is given by: $R_5 = \Delta x \cdot \sum_{i=1}^{5} f(x_i)$ Substitute $\Delta x = \frac{2}{5}$ and the calculated $f(x_i)$ values: $R_5 = \frac{2}{5} \left( -\frac{93}{25} - \frac{37}{25} + \frac{43}{25} + \frac{147}{25} + 11 \right)$

Convert 11 to $\frac{275}{25}$: $R_5 = \frac{2}{5} \left( -\frac{93}{25} - \frac{37}{25} + \frac{43}{25} + \frac{147}{25} + \frac{275}{25} \right)$ $R_5 = \frac{2}{5} \cdot \frac{335}{25} = \frac{2 \cdot 335}{5 \cdot 25} = \frac{670}{125} = \frac{134}{25}$

Final Answer

$R_5 = \frac{134}{25}$

Would you like more details or have any questions?

Relative Questions:

1. How would the answer change if we used a Left Riemann Sum instead?
2. What is the effect of increasing $n$ on the accuracy of the Riemann sum?
3. How can we compute the Midpoint Riemann Sum for this function over the same interval?
4. How does using Riemann sums approximate the area under the curve?
5. What is the exact integral value of $f(x) = 3x^2 + 8x$ over $[-1, 1]$?

Tip: For symmetric functions, consider whether you can simplify calculations by observing symmetry over intervals like $[-a, a]$.