compute\:the\:value\:of\:the\:definite\:integral\:\int _{-1}^1\:\left(x+5\right)dx\:as\:a\:limit\:of\:a\:Reimann\:sum

To compute the value of the definite integral \(\int_{-1}^1 (x+5) \, dx\) as a limit of a Riemann sum, we will follow the same procedure as before.

### Step 1: Partition the Interval
The interval of integration is \([-1, 1]\). We partition this interval into \(n\) subintervals, each of equal width \(\Delta x\).

\[
\Delta x = \frac{1 - (-1)}{n} = \frac{2}{n}
\]

### Step 2: Choose Sample Points
We choose the right endpoint as the sample point \(x_i^*\) for each subinterval:

\[
x_i^* = -1 + i\Delta x
\]

where \(i = 1, 2, \dots, n\).

### Step 3: Express the Function
The function we're integrating is \(f(x) = x + 5\). We evaluate this function at the sample points \(x_i^*\):

\[
f(x_i^*) = x_i^* + 5 = \left(-1 + \frac{2i}{n}\right) + 5 = \frac{2i}{n} + 4
\]

### Step 4: Write the Riemann Sum
The Riemann sum for this partition is:

\[
S_n = \sum_{i=1}^n f(x_i^*) \Delta x = \sum_{i=1}^n \left(\frac{2i}{n} + 4\right) \cdot \frac{2}{n}
\]

Expanding the sum:

\[
S_n = \frac{2}{n} \sum_{i=1}^n \left(\frac{2i}{n} + 4\right) = \frac{2}{n} \left(\frac{2}{n} \sum_{i=1}^n i + 4 \sum_{i=1}^n 1\right)
\]

We know the sum formulas:

\[
\sum_{i=1}^n i = \frac{n(n+1)}{2}, \quad \sum_{i=1}^n 1 = n
\]

Substituting these into the sum:

\[
S_n = \frac{2}{n} \left(\frac{2}{n} \cdot \frac{n(n+1)}{2} + 4n\right) = \frac{2}{n} \left(\frac{2n(n+1)}{2n} + 4n\right) = \frac{2}{n} \left(n+1 + 4n\right) = \frac{2}{n} \cdot (5n+1)
\]

Simplifying further:

\[
S_n = 10 + \frac{2}{n}
\]

### Step 5: Take the Limit
The definite integral is the limit of the Riemann sum as \(n \to \infty\):

\[
\int_{-1}^1 (x+5) \, dx = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(10 + \frac{2}{n}\right) = 10
\]

So, the value of the definite integral is \(10\).

Would you like any more details or have any questions about this process? Here are some related questions you might consider:

1. How does the integral \(\int_{-1}^1 (x+5) \, dx\) compare to \(\int_{-1}^1 (x+2) \, dx\) in terms of geometric interpretation?
2. What happens if you change the limits of integration? How does that affect the Riemann sum?
3. How would the result differ if you used the left endpoint or midpoint for the sample points?
4. Can you explain how the Riemann sum approximates the area under the curve?
5. How does the concept of Riemann sums extend to more complex functions?

**Tip:** When working with definite integrals, it's helpful to visualize the problem by sketching the function and the interval of integration. This helps in understanding the geometric interpretation of the integral as the area under the curve.

Learn how to compute the value of the definite integral ∫(-1)^(1) (x+5) dx as a limit of a Riemann sum. Follow a step-by-step solution that partitions the interval, chooses sample points, and evaluates the Riemann sum. Suitable for college-level mathematics students.

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