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Wize AP Calculus (BC) Textbook > Integration & Accumulation of Change

Approximating Areas & Riemann Sums

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Approximating Areas & Riemann Sums

Areas of planar regions can be approximated (estimated) using finite sums and this process is easier when the region is bounded by the graph of a function.

To approximate the area under the curve f(x)f\left(x\right) on the interval [a, b]\left[a,\ b\right], we divide the area into rectangles with equal widths Δx\Delta x. Each rectangle will have height f(x)f\left(x\right). Then, the approximate area is the sum of the areas of all of these rectangles.


Wize Tip
Depending on how we draw these rectangles, we will get slightly different area approximations.

Common Approximation Rules/Methods

  • Right Endpoint Rule: The area is estimated using rectangles whose right endpoint touches the graph of the function.
  • Left Endpoint Rule: The area is estimated using rectangles whose left endpoint touches the graph of the function.
  • Midpoint Rule: The area is estimated using rectangles whose midpoint touches the graph of the function.
  • Lower Sum: The area is estimated using rectangles lying inside the region, with one point touching the graph.

The sum of the areas of these rectangles is called a Riemann Sum.
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Riemann Sum

Let f(x)f(x) be a function and xix_i a set of points in its domain. We can estimate the area under the graph of f(x)f(x) on the interval [a,b][a,b] using the Riemann Sum

∑i=1nf(xi)Δx\boxed{\sum_{i=1}^{n}f(x_i)\Delta x }
Where Δx\Delta x is a constant of the form

Δx=b−an\displaystyle \boxed{\Delta x=\frac{b-a}{n}}

  • xi=a+iΔxx_i=a+i\Delta x is a right hand rule
  • xi=a+(i−1)Δxx_i=a+(i-1)\Delta x is a left hand rule
  • xi=a+(i−1/2)Δxx_i=a+(i-1/2)\Delta x is a midpoint rule
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Example: Riemann Sums

Estimate the area under the curve and over [0,1][0,1] of the function y=1−x2y=1-x^2 using 4 rectangles.
First divide [0,1][0,1] into 4 equal sub-intervals:
Δx=1−04=14 \Delta x = \dfrac{1-0}{4} = \dfrac{1}{4}
So, the sub-intervals are
[0,14],[14,12],[12,34],[34,1]\left[0,\frac{1}{4}\right],\left[\frac{1}{4},\frac{1}{2}\right],\left[\frac{1}{2},\frac{3}{4}\right],\left[\frac{3}{4},1\right]

If we like to use the left endpoint method, we should take x0=0x_0=0, x1=14x_1=\frac{1}{4}, x2=12x_2=\frac{1}{2}, x3=34x_3=\frac{3}{4}
Consider the table


Now the approximate area is
A≈∑i=14f(xi−1)Δx=[f(x0)+f(x1)+f(x2)+f(x3)]Δx=[1+1516+34+716](14)=0.78125 \begin{aligned} A \approx \sum_{i=1}^4f(x_{i-1})\Delta x=& [f(x_0)+f(x_1)+f(x_2)+f(x_3)]\Delta x\\ =& \left[1+\frac{15}{16}+\frac{3}{4}+\frac{7}{16}\right]\left(\frac{1}{4}\right) = 0.78125 \end{aligned}

*If you use another rule/method such as the midpoint rule or right endpoint rule to approximate the area, you'll get a slightly different approximation
Use the left-endpoint Riemann sum, with n=4n=4, to estimate the area under the curve f(x)=ex−xf\left(x\right)=e^x-x on the interval [1, 13]\left[1,\ 13\right].