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) on an interval [a,b] by dividing the interval into n smaller subintervals and building a shape (usually a rectangle) on each one.
If we divide the interval into n subintervals of equal width, the width of each subinterval is: Îx=nbâaâ
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 x-value of each subinterval for the height.
Example: Right Riemann Sum
Problem: Approximate â«02âx2dx using 4 rectangles with right endpoints.
- Find Îx: Îx=42â0â=0.5
- Identify the subintervals: [0,0.5],[0.5,1],[1,1.5],[1.5,2]
- Find the heights: Evaluate f(x)=x2 at the right endpoints (0.5,1,1.5,2).
- Calculate the total area: AreaâÎx[f(0.5)+f(1)+f(1.5)+f(2)] Areaâ0.5[0.52+12+1.52+22] Areaâ0.5[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 n subintervals is: Tnâ=2Îxâ[f(x0â)+2f(x1â)+2f(x2â)+â¯+2f(xnâ1â)+f(xnâ)] Notice that the first and last endpoints are evaluated once, while all the inner endpoints are multiplied by 2.
Example: Trapezoidal Rule
Problem: Estimate â«13âx1âdx using n=4.
- Find Îx: Îx=43â1â=0.5
- Identify the endpoints: x0â=1,x1â=1.5,x2â=2,x3â=2.5,x4â=3
- Apply the formula: T4â=20.5â[f(1)+2f(1.5)+2f(2)+2f(2.5)+f(3)] T4â=0.25[1+2(32â)+2(21â)+2(52â)+31â] T4â=0.25[1+34â+1+54â+31â] T4â=0.25[2+35â+54â]=0.25[1530+25+12â]=0.25[1567â]=6067ââ1.1167
Connection to Definite Integrals
Approximations are just the beginning. As you increase the number of rectangles (nââ), the width of each rectangle (Îx) shrinks to zero. The approximation gets closer and closer to the exact area.
This limit is the very definition of the definite integral: limnââââi=1nâf(xiââ)Îx=â«abâf(x)dx