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Let's go through the steps outlined in the problem for \( f(x) = x + 1 \) on the interval \([0, 4]\) with \( n = 4 \).

### a. Sketch the graph of the function
The function \( f(x) = x + 1 \) is a straight line with a slope of 1 and a y-intercept of 1. On the interval \([0, 4]\), the graph is a straight line passing through the points:
- \( (0, 1) \)
- \( (4, 5) \)

### b. Calculate \( \Delta x \) and the grid points \( x_0, x_1, \dots, x_n \)
To compute \( \Delta x \), use the formula:
\[
\Delta x = \frac{b - a}{n}
\]
where \( a = 0 \), \( b = 4 \), and \( n = 4 \). Therefore,
\[
\Delta x = \frac{4 - 0}{4} = 1
\]

The grid points are calculated as follows:
\[
x_0 = 0, \quad x_1 = 1, \quad x_2 = 2, \quad x_3 = 3, \quad x_4 = 4
\]

### c. Illustrate the left and right Riemann sums
- **Left Riemann Sum**: The left Riemann sum uses the left endpoint of each subinterval. The function values at the left endpoints \( x_0, x_1, x_2, x_3 \) are:
\[
f(x_0) = f(0) = 1, \quad f(x_1) = f(1) = 2, \quad f(x_2) = f(2) = 3, \quad f(x_3) = f(3) = 4
\]
So, the left Riemann sum is:
\[
L_4 = \Delta x [f(x_0) + f(x_1) + f(x_2) + f(x_3)] = 1 [1 + 2 + 3 + 4] = 10
\]

- **Right Riemann Sum**: The right Riemann sum uses the right endpoint of each subinterval. The function values at the right endpoints \( x_1, x_2, x_3, x_4 \) are:
\[
f(x_1) = f(1) = 2, \quad f(x_2) = f(2) = 3, \quad f(x_3) = f(3) = 4, \quad f(x_4) = f(4) = 5
\]
So, the right Riemann sum is:
\[
R_4 = \Delta x [f(x_1) + f(x_2) + f(x_3) + f(x_4)] = 1 [2 + 3 + 4 + 5] = 14
\]

- **Which underestimates and overestimates?**  
Since the function \( f(x) = x + 1 \) is increasing, the left Riemann sum underestimates the area under the curve, while the right Riemann sum overestimates it.

### d. Calculate the left and right Riemann sums
- **Left Riemann Sum**: \( L_4 = 10 \)
- **Right Riemann Sum**: \( R_4 = 14 \)

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Would you like more details or have any questions about the steps?  
Here are some additional related questions:

1. How do we interpret the difference between the left and right Riemann sums?
2. How can the Riemann sum change if we increase the number of subintervals \( n \)?
3. How do we compute the actual area under the curve using integration?
4. How does the shape of the function affect whether the Riemann sum overestimates or underestimates?
5. How can midpoint Riemann sums provide a better approximation?

**Tip**: Increasing the number of subintervals (increasing \( n \)) reduces the difference between the left and right Riemann sums, leading to a more accurate approximation of the area under the curve.

For the function f(x) = x + 1 on [0, 4] with n = 4, calculate the left and right Riemann sums, and determine which sum underestimates or overestimates the area under the curve.

Calculate the left and right Riemann sums for the linear function f(x) = x + 1 on the interval [0, 4] using n = 4 subintervals. Learn how Riemann sums approximate the area under the curve and determine whether the left or right sum underestimates or overestimates the area.

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