Reading a Sigma Limit as an Integral
A limit of sums can look like an algebra problem involving \(n\), but it often describes a familiar geometric process: add the areas of many thin rectangles and let their widths approach zero. The key is to identify the interval being divided, the width of each subinterval, and the function value sampled on each one.
For example, in a sum containing \(f(2+3i/n)\), the expression \(2+3i/n\) suggests sample points on an interval beginning at \(2\) and ending at \(5\). The accompanying factor determines whether the sum uses the interval width \(3/n\), or a different multiple of that width. Both pieces matter: a correct integrand with incorrect bounds, or correct bounds with an incorrect factor, gives the wrong integral.
The Riemann-Sum Pattern
For a continuous function \(f\) on \([a,b]\), divide the interval into \(n\) equal subintervals. Each has width \(\Delta x=(b-a)/n\). Using right endpoints, the sample point in subinterval \(i\) is \(x_i=a+i\Delta x\). The definite integral is the limit of the resulting Riemann sums:
The same integral can be represented with left endpoints or midpoints, provided the points belong to the same partition and the width factor remains correct. The choice of sample point does not change the limit for a continuous function. For converting a given sigma limit, the practical task is to match its pieces to this pattern.
When the index runs from \(1\) to \(n\), a sample point of the form \(a+(b-a)i/n\) is the right endpoint of subinterval \(i\). The expression \(a+(b-a)(i-1)/n\) gives left endpoints. The expression \(a+(b-a)(i-\tfrac12)/n\) gives midpoints.
A Reliable Conversion Process
For each sigma limit, work through these questions before writing an integral:
Read the input of the function as \(x_i\), often in the form \(a+ci/n\) or a shifted-index version.
Use the starting value and total change in the sample-point expression to identify the interval and its width \((b-a)/n\).
The factor outside the function value must match the subinterval width. If it does not, account for the difference in the integrand.
Verify that its bounds, sample points, and width reconstruct the original summand.
Worked Examples
Worked Example: Directly Matching the Width
Convert the following limit to a definite integral:
The function value is sampled at \(x_i=2+3i/n\). At \(i=1\), the sample point is \(2+3/n\), and at \(i=n\), it is \(2+3=5\). Thus the interval is \([2,5]\), divided into \(n\) equal pieces.
The interval length is \(5-2=3\), so the width is \(\Delta x=3/n\). This matches the factor multiplying the squared function value. The function is \(f(x)=x^2\).
Check: With \(x_i=2+3i/n\), the term \(f(x_i)\Delta x\) is \((2+3i/n)^2(3/n)\), exactly the original summand. The bounds match the first and last partition endpoints.
Worked Example: A Coefficient Different from the Width
Convert:
The sample point is \(x_i=1+2i/n\). It starts from \(1\) and increases by \(2\) across the \(n\) subintervals, so the interval is \([1,3]\). Its width is \(\Delta x=(3-1)/n=2/n\).
The coefficient \(1/n\) is only half the width \(2/n\). Write it as \(\tfrac12\Delta x\). The function values are \(x_i^3\), so each term is \(\tfrac12 x_i^3\Delta x\). Therefore the corresponding integrand is \(\tfrac12 x^3\).
Check: At \(x_i=1+2i/n\), the integral’s Riemann-sum term is \(\tfrac12 x_i^3(2/n)=x_i^3/n\). This equals the original term, and the interval endpoints are \(1\) and \(3\).
Worked Example: Midpoint Samples
Convert:
The sample point is \(x_i=1+4(i-\tfrac12)/n\). The half-step shift indicates midpoint samples. The full interval change is \(4\), so the endpoints are \(1\) and \(1+4=5\), and each subinterval has width \(\Delta x=4/n\).
The coefficient \(4/n\) equals the width. The expression being squared is \(x_i\), so the integrand is \(x^2\). The midpoint samples and width give a Riemann sum for the integral on \([1,5]\).
Check: The midpoint of subinterval \(i\) is \(1+(i-\tfrac12)(4/n)\), which is the given sample point. Multiplying its squared value by the width \(4/n\) reproduces each summand.
Worked Example: An Index Beginning at Zero
Convert:
Here the index begins at \(0\), but the shift \(i+\tfrac12\) still places each sample at a subinterval midpoint. The sample points start halfway across the first subinterval and end halfway across the last. The underlying interval begins at \(-1\) and has length \(5\), so its right endpoint is \(-1+5=4\).
There are \(n\) equal subintervals, each with width \(5/n\), which matches the coefficient. The sampled function is \(f(x)=\cos x\). Thus the limit is the integral on \([-1,4]\).
Check: For \(i=0\), the sample is \(-1+5/(2n)\), the first midpoint. For \(i=n-1\), it is \(-1+5(n-\tfrac12)/n=4-5/(2n)\), the last midpoint. Each term is the cosine of its midpoint multiplied by the width \(5/n\).
Why the Conversion Works
The conversion follows directly from the Riemann-sum definition. Suppose the sample points have the form \(x_i=a+i(b-a)/n\), and the summand is \(f(x_i)(b-a)/n\). For every \(n\), the sum is exactly the right-endpoint Riemann sum for \(f\) on \([a,b]\). As \(n\) increases, the partition width \((b-a)/n\) approaches zero. Since a continuous function on a closed interval is integrable, these sums approach its definite integral.
If the coefficient is \(c/n\) rather than \((b-a)/n\), write \(c/n=(c/(b-a))\Delta x\). Each term then has the form \(\bigl((c/(b-a))f(x_i)\bigr)\Delta x\), so the integrand must include the factor \(c/(b-a)\). This is why the coefficient cannot be ignored even when the sample-point expression makes the interval clear.
Common Mistake: Treating Every \(1/n\) as the Width
A frequent error is to assume that a coefficient \(1/n\) always means the interval width is \(1/n\). Consider the sum in the second example. Its sample points \(1+2i/n\) run across an interval of length \(2\), so the width is \(2/n\), not \(1/n\). Writing \(\int_1^3 x^3\,dx\) would double the contribution represented by each Riemann-sum term.
The correction is to use the actual width \(2/n\) and adjust the integrand by a factor of \(1/2\). Always infer the width from the interval endpoints, then compare it with the coefficient written in the sum.
What to Verify Before You Finish
A correct conversion should pass three checks. First, the sample-point expression must place the function values on the interval named by the bounds. Second, the index range and any half-step shift must agree with the chosen endpoint or midpoint convention. Third, multiplying the proposed integrand value by the interval width must recover the original summand. These checks are often quicker and more reliable than trying to evaluate a limit by algebra.
The conversion identifies what a limit represents; it does not require evaluating the integral. Once the bounds and integrand are correct, the expression is in definite-integral form. Keeping that distinction clear prevents an unnecessary calculation from obscuring the main task.
Check Your Understanding
For each limit, identify its interval, sample-point type, and corresponding definite integral. Do not evaluate the integrals.
- \(\displaystyle \lim_{n\to\infty}\sum_{i=1}^{n}\left(3+\frac{2i}{n}\right)\frac{2}{n}\)
- \(\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\left(4-\frac{3i}{n}\right)^2\)
- \(\displaystyle \lim_{n\to\infty}\frac{6}{n}\sum_{i=0}^{n-1}\sqrt{2+\frac{6(i+\tfrac12)}{n}}\)
- For a sum with sample points \(x_i=-2+7(i-1)/n\), what are the interval endpoints, the width, and the sample-point type?
- In \(\displaystyle \sum_{i=1}^{n} f(5+4i/n)(2/n)\), what adjustment to the integrand is needed when converting the limit to an integral?
