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Riemann Sums and Area Approximation

Riemann Sums and Area Approximation

Finding the exact area under a curve can be tricky. Riemann sums provide a practical way to approximate this area by breaking the region down into simpler geometric shapes, like rectangles or trapezoids.

What is a Riemann Sum?

A Riemann sum approximates the area under a curve y=f(x)y = f(x) on an interval [a,b][a, b] by dividing the interval into nn smaller subintervals and building a shape (usually a rectangle) on each one.

If we divide the interval into nn subintervals of equal width, the width of each subinterval is: Δx=b−an\Delta x = \frac{b - a}{n}

Types of Rectangular Approximations

Depending on where you evaluate the function to determine the height of each rectangle, you get different types of sums:

  • Left Riemann Sum: Uses the left endpoint of each subinterval for the height.
  • Right Riemann Sum: Uses the right endpoint of each subinterval for the height.
  • Midpoint Riemann Sum: Uses the exact middle xx-value of each subinterval for the height.

Example: Right Riemann Sum

Problem: Approximate ∫02x2 dx\int_0^2 x^2 \, dx using 44 rectangles with right endpoints.

  1. Find Δx\Delta x: Δx=2−04=0.5\Delta x = \frac{2 - 0}{4} = 0.5
  2. Identify the subintervals: [0,0.5],[0.5,1],[1,1.5],[1.5,2][0, 0.5], [0.5, 1], [1, 1.5], [1.5, 2]
  3. Find the heights: Evaluate f(x)=x2f(x) = x^2 at the right endpoints (0.5,1,1.5,20.5, 1, 1.5, 2).
  4. Calculate the total area: Area≈Δx[f(0.5)+f(1)+f(1.5)+f(2)]\text{Area} \approx \Delta x [f(0.5) + f(1) + f(1.5) + f(2)] Area≈0.5[0.52+12+1.52+22]\text{Area} \approx 0.5 [0.5^2 + 1^2 + 1.5^2 + 2^2] Area≈0.5[0.25+1+2.25+4]=0.5[7.5]=3.75\text{Area} \approx 0.5 [0.25 + 1 + 2.25 + 4] = 0.5 [7.5] = 3.75

The Trapezoidal Rule

Instead of flat-topped rectangles, you can use trapezoids to better follow the natural slant of the curve.

The general formula for the Trapezoidal Rule with nn subintervals is: Tn=Δx2[f(x0)+2f(x1)+2f(x2)+⋯+2f(xn−1)+f(xn)]T_n = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)] Notice that the first and last endpoints are evaluated once, while all the inner endpoints are multiplied by 22.

Example: Trapezoidal Rule

Problem: Estimate ∫131x dx\int_1^3 \frac{1}{x} \, dx using n=4n = 4.

  1. Find Δx\Delta x: Δx=3−14=0.5\Delta x = \frac{3 - 1}{4} = 0.5
  2. Identify the endpoints: x0=1,x1=1.5,x2=2,x3=2.5,x4=3x_0=1, x_1=1.5, x_2=2, x_3=2.5, x_4=3
  3. Apply the formula: T4=0.52[f(1)+2f(1.5)+2f(2)+2f(2.5)+f(3)]T_4 = \frac{0.5}{2} \left[ f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + f(3) \right] T4=0.25[1+2(23)+2(12)+2(25)+13]T_4 = 0.25 \left[ 1 + 2\left(\frac{2}{3}\right) + 2\left(\frac{1}{2}\right) + 2\left(\frac{2}{5}\right) + \frac{1}{3} \right] T4=0.25[1+43+1+45+13]T_4 = 0.25 \left[ 1 + \frac{4}{3} + 1 + \frac{4}{5} + \frac{1}{3} \right] T4=0.25[2+53+45]=0.25[30+25+1215]=0.25[6715]=6760≈1.1167T_4 = 0.25 \left[ 2 + \frac{5}{3} + \frac{4}{5} \right] = 0.25 \left[ \frac{30 + 25 + 12}{15} \right] = 0.25 \left[ \frac{67}{15} \right] = \frac{67}{60} \approx 1.1167

Connection to Definite Integrals

Approximations are just the beginning. As you increase the number of rectangles (n→∞n \to \infty), the width of each rectangle (Δx\Delta x) shrinks to zero. The approximation gets closer and closer to the exact area.

This limit is the very definition of the definite integral: lim⁡n→∞∑i=1nf(xi∗)Δx=∫abf(x) dx\lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x = \int_a^b f(x) \, dx