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How do you write a series of added rectangular areas in summation notation?

The series of added rectangular areas is written in summation notation by identifying the common width factor and the general term for the height. Since each rectangle has a width of 1n\frac{1}{n} and the height of the kk-th rectangle is f(kn)f\left(\frac{k}{n}\right), the total area approximation is the sum of these products from k=1k=1 to nn. This is compressed into the expression 1n∑k=1nf(kn)\frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right).

Conditions

  • The interval [0,1][0,1] is divided into nn equal parts.
  • The height of the kk-th rectangle is evaluated at the right endpoint kn\frac{k}{n}.
  • The index kk ranges from 11 to nn.

Reasoning, step by step

  1. Identify the width of each rectangle as 1n\frac{1}{n}.
  2. Identify the height of the kk-th rectangle as f(kn)f\left(\frac{k}{n}\right).
  3. Write the area of the kk-th rectangle as the product 1nf(kn)\frac{1}{n} f\left(\frac{k}{n}\right).
  4. Sum these areas for all kk from 11 to nn.
  5. Factor out the constant width 1n\frac{1}{n} to obtain the compact summation notation: 1n∑k=1nf(kn)\frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right).

Example

The video rewrites the expanded form 1nf(1n)+1nf(2n)+⋯+1nf(nn)\frac{1}{n}f\left(\frac{1}{n}\right)+\frac{1}{n}f\left(\frac{2}{n}\right)+\cdots+\frac{1}{n}f\left(\frac{n}{n}\right) as 1n∑k=1nf(kn)\frac{1}{n}\sum_{k=1}^{n} f\left(\frac{k}{n}\right), explaining that 1n\frac{1}{n} is the common width and f(kn)f\left(\frac{k}{n}\right) is the height of the kk-th rectangle.

Common misconceptions

  • Forgetting to include the width 1n\frac{1}{n} inside or outside the summation correctly.
  • Confusing the index kk with the limit nn; kk is the running index, while nn is the total number of partitions.
  • Assuming the summation starts at k=0k=0; for right-endpoint evaluation on [0,1][0,1], it typically starts at k=1k=1.

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