Riemann Sums, Summation Notation, and Definite Integral Notation

3. Visual Connection

5. Concept Development

Finite sums approximate; limiting sums define

A finite Riemann sum uses a fixed partition and normally approximates net accumulation. A definite integral is the single value approached as every subinterval becomes arbitrarily narrow:

Read sigma notation structurally

In \(\sum_{i=1}^{n}a_i\), the index \(i\) begins at 1 and ends at \(n\). Substitute each integer value of \(i\), generate the corresponding term, and add. The index is a placeholder and has no meaning outside the sum.

A Riemann-sum term is height times width

Each product \(f(x_i^*)\Delta x_i\) uses the function value at a sample point in the \(i\)th subinterval and the length of that subinterval. The sum combines signed rectangles, so a negative height creates a negative contribution.

Regular partitions have a common width

For \(n\) equal subintervals of \([a,b]\),

The outside factor in a limiting sum often reveals \(\Delta x\), and \(n\Delta x=b-a\) reveals the interval length.

Left-endpoint sample points

The left endpoint of the \(i\)th interval is

The shift \(i-1\) is essential: using \(i\) changes the sample points to right endpoints.

Right-endpoint sample points

The right endpoint of the \(i\)th interval is

As \(i\) runs from 1 to \(n\), the first sample is \(a+\Delta x\) and the last is \(b\).

Midpoint sample points

The midpoint of the \(i\)th equal subinterval is

The half-step shift places each sample halfway between consecutive partition endpoints.

General sample points

A Riemann sum does not require a named endpoint rule. Any \(x_i^*\in[x_{i-1},x_i]\) may be chosen. For an integrable function, all valid choices approach the same definite integral as the largest width approaches zero.

Anatomy of definite-integral notation

In \(\int_a^b f(x)\,dx\), \(a\) and \(b\) are the bounds, \(f(x)\) is the integrand, and \(x\) is the variable of integration. A definite integral with constant bounds is a number representing net signed accumulation.

The variable of integration is a dummy variable

Changing the internal symbol does not change the value:

The integrand and differential must use the same internal variable.

Translate a limiting sum into an integral

First identify the width from the factor outside the function. Next identify the sample-point expression inside the function. Recover the starting value and total interval length, then replace the sample expression by a dummy variable in the integrand.

Translate an integral into a limiting sum

Choose a sample rule, compute \(\Delta x=(b-a)/n\), write the sample point, substitute it into \(f\), multiply by \(\Delta x\), sum from 1 to \(n\), and place \(\lim\limits_{n\to\infty}\) in front.

Nonuniform partitions need a mesh condition

For unequal widths, merely letting the number of intervals increase is not enough; one interval could remain wide. The correct limiting condition is \(\max_i\Delta x_i\to0\), ensuring that every subinterval becomes narrow.

Integrability and signed area

Continuous functions on closed intervals are integrable, and some functions with limited discontinuities are also integrable. The resulting definite integral counts area above the axis positively and area below the axis negatively.

Common errors

Frequent errors include forgetting the width factor, reading \(c/n\) as an endpoint instead of an interval length, overlooking a starting-value shift, confusing \(i\) with \(i-1/2\), treating a finite sum as exact, and writing an integral whose differential does not match its integrand.

6. Detailed Worked Example and Error Check

Example 1: Expand a finite sigma expression.

The upper index gives four terms; it is not a value to substitute only once.

Example 2: Integral to a right-endpoint limit. For \(\int_2^5(x^2+1)\,dx\), the width is \(3/n\) and the right endpoint is \(x_i=2+3i/n\). Therefore

Example 3: Integral to a midpoint limit. For \(\int_{-1}^{3}e^x\,dx\), \(\Delta x=4/n\) and

Thus a midpoint representation is

Example 4: Decode a shifted right sum.

The width \(5/n\) gives interval length 5, and \(1+5i/n\) shows a right partition beginning at 1. The limit is \(\int_1^6x^3\,dx\).

Example 5: Decode the width before the function.

Rewrite the product mentally as \(\sum \sqrt{2+i/n}(1/n)\). It is the right-endpoint form of \(\int_2^3\sqrt{x}\,dx\); the factor \(1/n\) is the width, not part of the square-root function.

Example 6: General nonuniform form. If \(a=x_0<\cdots<x_n=b\), \(c_i\in[x_{i-1},x_i]\), and every width shrinks, then

The sample points may vary from interval to interval; integrability makes the limiting value independent of those valid choices.

Example 7: Evaluate from the definition. Use right endpoints for \(\int_0^3(2x+1)\,dx\). Then \(\Delta x=3/n\), \(x_i=3i/n\), and

Using \(\sum i=n(n+1)/2\) and \(\sum1=n\), the expression is \(9(n+1)/n+3\), whose limit is \(12\).