Riemann Sums

Let’s consider two simple problems, one for Left Riemann Sum and one for Right Riemann Sum:

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1. Left Riemann Sum

We approximate the integral $\int_a^b f(x) \, dx$ using the left endpoints of subintervals.

Example Problem: Evaluate the left Riemann sum for $\int_0^2 (x^2) \, dx$ using $n = 4$ subintervals.

Solution:

1. Interval length: $b - a = 2 - 0 = 2$, and subinterval width: $\Delta x = \frac{2}{4} = 0.5$.
2. Subintervals are $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$.
3. Use the left endpoints: $x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5$.
4. Compute the sum: [ L_4 = \Delta x \sum_{i=0}^{3} f(x_i) = 0.5 \big[f(0) + f(0.5) + f(1) + f(1.5)\big] = 0.5 \big[0^2 + 0.5^2 + 1^2 + 1.5^2\big] = 0.5 \big[0 + 0.25 + 1 + 2.25\big] = 0.5 \times 3.5 = 1.75 ]

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2. Right Riemann Sum

We approximate the integral $\int_a^b f(x) \, dx$ using the right endpoints of subintervals.

Example Problem: Evaluate the right Riemann sum for $\int_0^2 (x^2) \, dx$ using $n = 4$ subintervals.

Solution:

1. Interval length: $b - a = 2 - 0 = 2$, and subinterval width: $\Delta x = \frac{2}{4} = 0.5$.
2. Subintervals are $[0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]$.
3. Use the right endpoints: $x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2$.
4. Compute the sum: [ R_4 = \Delta x \sum_{i=1}^{4} f(x_i) = 0.5 \big[f(0.5) + f(1) + f(1.5) + f(2)\big] = 0.5 \big[0.5^2 + 1^2 + 1.5^2 + 2^2\big] = 0.5 \big[0.25 + 1 + 2.25 + 4\big] = 0.5 \times 7.5 = 3.75 ]

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Final Results:

1. Left Riemann Sum: $L_4 = 1.75$
2. Right Riemann Sum: $R_4 = 3.75$

Would you like further details or clarifications? Here are some related questions for you:

1. How do these approximations compare to the actual integral value?
2. Can we visualize this as rectangles under the curve?
3. What happens if we increase the number of subintervals $n$?
4. How do we compute the error bounds for Riemann sums?
5. What is the formula for a midpoint Riemann sum, and how does it differ?

Tip: Increasing the number of subintervals ($n$) improves the accuracy of the Riemann sum approximation.