Translate a sum limit into an integral
Sums that approach an area
The curve is \(y = \frac1x\) on the interval from 1 to 3, which is cut into \(n\) pieces. Each teal rectangle is as tall as the curve at the right end of its piece.
Each piece is \(\frac2n\) wide and the right end of piece \(i\) is \(1 + \frac{2i}{n}\), so the sum of the rectangle areas is \(S_n = \sum_{i=1}^{n}\frac2n\cdot\frac{1}{1 + 2i/n}\). For \(n = 2\), \(S_n \approx 0.833\), and for \(n = 4\), \(S_n = 0.95\).
Increase \(n\). The rectangles become thinner and fit the curve better. For \(n = 10\), \(S_n \approx 1.035\), and for \(n = 100\), \(S_n \approx 1.092\).
The sums approach the area of the faint teal region under the curve, about 1.099. This limit of the sums is written as the definite integral \(\int_1^3 \frac1x\,dx\).
The parts of a sum become the parts of an integral
A sum of rectangle areas is a Riemann sum. Its limit has the form \(\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\,\Delta x\), where \(f(x_i)\) is the orange height of rectangle \(i\) and \(\Delta x\) is its green width. The rectangles lie on the interval from \(a\) to \(b\).
When \(n\) grows, the width \(\Delta x\) becomes very small, and in the integral this small width is written \(dx\).
Press Write as an integral. The sum sign becomes the integral sign \(\int\), a stretched letter S for sum, and the sample point \(x_i\) becomes the variable \(x\).
The interval does not change, so its ends \(a\) and \(b\) become the bounds of the integral.
Notation
Replace each part of the sum by the matching part of the integral.
\[\lim_{n\to\infty}\sum_{i=1}^{n} \cy{f(x_i)}\,\cg{\Delta x} = \int_{a}^{b} \cy{f(x)}\,\cg{dx}\]The function comes from the sample point \(x_i\) inside the sum: change \(x_i\) to \(x\) everywhere, and what is left is \(f(x)\).
The bounds are the first edge \(a\) and the last edge \(b\), not the 1 and \(n\) on the sum sign. To find them, put \(i = 0\) and \(i = n\) into the formula for the sample point.
As a check, the width is \(\frac{b-a}{n}\), so its numerator must equal the length \(b - a\) of the interval.
When it fails
The translation needs a function that is continuous on the interval from \(a\) to \(b\). Move the left end to 0, where \(\frac1x\) is not defined and the curve rises without bound.
With the left end at 0, each term is \(\frac3n\cdot\frac{1}{3i/n} = \frac1i\), so each sum is \(1 + \frac12 + \dots + \frac1n\). This is about 2.93 for \(n = 10\), 5.19 for \(n = 100\) and 7.49 for \(n = 1000\). The sums grow without limit, so there is no definite integral.
Example 1: A cube
Write \(\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac2n\left(\frac{2i}{n}\right)^3\) as a definite integral.
The width is \(\frac2n\), and the sample point in the other factor is \(x_i = \frac{2i}{n}\). Change it to \(x\), and the function is \(f(x) = x^3\).
\[\left(\dfrac{2i}{n}\right)^3 \ \to\ \cy{x^3}\]The sample point formula gives 0 for \(i = 0\) and 2 for \(i = n\), so the bounds are 0 and 2.
\[\dfrac{2\cdot0}{n} = 0, \qquad \dfrac{2n}{n} = 2\]As a check, \(b - a = 2\) is the numerator of the width \(\frac2n\).
\[\lim_{n\to\infty}\sum_{i=1}^{n}\cg{\frac2n}\cy{\left(\frac{2i}{n}\right)^3} = \int_0^2 \cy{x^3}\,\cg{dx}\]
Example 2: A logarithm
Write \(\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac3n\ln\!\left(2 + \frac{3i}{n}\right)\) as a definite integral.
The width is \(\frac3n\), and the sample point is \(x_i = 2 + \frac{3i}{n}\). Change it to \(x\), and the function is \(f(x) = \ln x\).
A tempting wrong move: The index runs from 1 and \(n\) grows without bound, so the bounds are 1 and infinity.
\[\int_1^{\infty} \ln x\,dx\]There are \(n\) pieces, each \(\frac3n\) wide, so together they cover a length of 3 for every \(n\). The rectangles stay between 2 and 5 and do not spread along the red arrow, which marks the wrong interval.
The bounds are the first and last edges: the sample point formula gives 2 for \(i = 0\) and 5 for \(i = n\). As a check, \(5 - 2 = 3\) is the numerator of the width.
\[\begin{aligned} &\lim_{n\to\infty}\sum_{i=1}^{n}\frac3n\ln\!\left(2 + \frac{3i}{n}\right) \\ &\quad = \int_2^5 \ln x\,dx \end{aligned}\]
Practice
Write \(\displaystyle\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac2n\,e^{1 + 2i/n}\) as a definite integral.
Each rectangle is read at the left end of its piece, but the last piece still ends at the upper bound.
Solution
The width is \(\frac2n\), and the sample point is \(x_i = 1 + \frac{2i}{n}\). Change it to \(x\), and the function is \(f(x) = e^x\).
\[e^{1 + 2i/n} \ \to\ \cy{e^x}\]The sample point formula gives 1 for \(i = 0\) and 3 for \(i = n\), so the bounds are 1 and 3. The sum stops at \(i = n - 1\) because each rectangle is read at the left end of its piece, but the last piece still ends at 3. As a check, \(3 - 1 = 2\) is the numerator of the width.
\[\lim_{n\to\infty}\sum_{i=0}^{n-1}\frac2n\,e^{1 + 2i/n} = \int_1^3 e^x\,dx\]
Questions
Questions 1 and 2 use the figure, which shows a sum with \(n = 4\) pieces. Each piece is 1 wide, as the green bracket shows, and is read at its right end, where an orange dot marks the sample point.
As the number of pieces grows, the sums approach an integral. What is its lower bound?
- 1
- 2
- 0
Answer
1
What do the sums of the areas of the rectangles approach as the number of pieces grows?
- The area under the curve from 1 to 5
- The area under the curve from 2 to 5
- The height of the curve at 5
Answer
The area under the curve from 1 to 5
Which integral equals \(\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac5n\cdot\frac{1}{(4 + 5i/n)^2}\)?
- \(\displaystyle\int_4^9\frac{dx}{x^2}\)
- \(\displaystyle\int_0^5\frac{dx}{x^2}\)
- \(\displaystyle\int_4^9\frac{dx}{(4 + x)^2}\)
Answer
\(\displaystyle\int_4^9\frac{dx}{x^2}\)
Summary
You read the function from the sample point, the bounds from the first and last edges, and the width as \(dx\).
This works because the sums approach the area under the curve, and the integral sign stands for this limit.
You use it when a limit of Riemann sums has to be written as a definite integral.
This is one of four skills in topic 6.3 of Which & Why: AP Calculus.