The definite integral as a limit of Riemann sums
Every Riemann sum is only an estimate, and a finer one is better. Is there one exact number that all the estimates close in on, and can it be found without adding infinitely many rectangles by hand?
Where you are 3 in The definite integral
Why this exists
Approximating area with Riemann sums trapped areas between left and right sums and watched the error fall as n grew: for Leonor’s velocity the sums crept toward \displaystyle \frac{368}{3}, a value taken on trust. A sum with a fixed n is still an estimate. What is wanted is the exact amount, and a way to compute it.
Limits supply both. A limit is the value an expression approaches as its input moves; here the input is the number of pieces n, and it grows without bound. If the Riemann sums approach one number however the rectangles are chosen, that number is the exact amount, and it gets a name and a symbol.
Definition
Day 1: the definite integral
Let f be defined on [a, b] with a < b. Cut [a, b] into n pieces of equal width and pick a sample point x_i^* in piece i:
The definite integral of f from a to b is
provided the limit exists and is the same number for every choice of sample points. When it is, f is integrable on [a, b].
The general version allows unequal pieces. For a partition P: a = x_0 < x_1 < \dots < x_n = b with widths \Delta x_i = x_i - x_{i-1}, the mesh \|P\| is the largest width. Then f is integrable with integral I when every Riemann sum \sum f(x_i^*)\,\Delta x_i is as close to I as we like once \|P\| is small enough, for every partition and every choice of sample points:
Equal widths are the case \displaystyle \|P\| = \Delta x = \frac{b-a}{n}, which shrinks to 0 as n \to \infty.
The parts. In \int_a^b f(x)\,dx the sign \int is a stretched S, for sum. The function f(x) is the integrand; a and b are the lower and upper limits of integration; dx marks x as the variable of integration and recalls the width \Delta x. The integral is a number. It depends on f, a and b, not on the letter x, which is a dummy variable: \int_0^3 x^2\,dx, \int_0^3 t^2\,dt and \int_0^3 u^2\,du are the same number, just as \sum_{i=1}^{n} i and \sum_{k=1}^{n} k are the same sum (Sigma notation).
Day 2: reading a limit of sums
The definition runs both ways. In a limit of the form
the factor that multiplies every term and shrinks like \displaystyle \frac{1}{n} is \displaystyle \Delta x = \frac{b-a}{n}, which gives the length b - a. The expression in the term that grows with i in steps of \Delta x is x_i; its value at i = 0 is a. What is done to x_i is f. Shifting the interval and the function together gives another correct reading of the same limit.
Intuition
Take f(x) = x^3 on [1, 2] and right-endpoint rectangles. With four pieces the corners poking above the curve are large.
With sixteen pieces each corner is a thin sliver. There are more of them, but each shrinks faster than their number grows, so their total shrinks.
The integral is the number left when the slivers are gone. Changing the sample points moves every rectangle top up or down within its piece, but for a continuous f the whole range of heights in a narrow piece is small, so every choice closes in on the same number. Where the graph is below the axis, f(x_i^*) is negative and so is its term, which is why the integral measures signed area (Signed area and integrals from geometry).
Results
Integrability (stated without proof). If f is continuous on [a, b], then f is integrable on [a, b]. More generally, a function that is bounded on [a, b] and has only finitely many discontinuities there, such as finitely many jumps, is integrable.
Computing with right endpoints. If f is integrable on [a, b], then with \displaystyle \Delta x = \frac{b-a}{n},
The same holds with left endpoints \sum_{i=0}^{n-1} f(a + i\,\Delta x)\,\Delta x or midpoints.
Sum formulas. For every positive integer n,
Sums split. For constants c and d, \sum (c\,p_i + d\,q_i) = c\sum p_i + d\sum q_i, because a finite sum may be regrouped and reordered (the associative and commutative properties) and a common factor pulled out (the distributive property).
Why it is true
The computing rule. Integrability says the sums approach \int_a^b f(x)\,dx for every choice of sample points as the mesh goes to 0. Right endpoints with equal widths are one such choice, with mesh \displaystyle \frac{b-a}{n} \to 0, so their limit is that same number. The rule needs integrability first; Example 7 shows a function whose right sums converge although it is not integrable.
The sum of squares. Expanding gives (i+1)^3 - i^3 = 3i^2 + 3i + 1. Add this over i = 1, \dots, n. On the left every middle cube appears once with each sign and cancels, leaving (n+1)^3 - 1:
Solving,
so \displaystyle \sum i^2 = \frac{n(n+1)(2n+1)}{6}. The formula for \sum i is Exercise 8, and the one for \sum i^3, by induction, is Exercise 10.
Worked examples
Day 1: a trapezoid from both ends
Compute \int_1^3 (2x + 1)\,dx from the definition with right endpoints and with left endpoints, and check the answer with geometry.
SolutionThe integrand is continuous, so it is integrable and either kind of endpoint may be used. Here \displaystyle \Delta x = \frac{3 - 1}{n} = \frac{2}{n} and \displaystyle x_i = 1 + \frac{2i}{n}, so \displaystyle f(x_i) = 2\left(1 + \frac{2i}{n}\right) + 1 = 3 + \frac{4i}{n}.
Right endpoints run i = 1 to n. Splitting the sum and using \sum 1 = n and \displaystyle \sum i = \frac{n(n+1)}{2}:
Left endpoints are x_0, \dots, x_{n-1}, so i runs 0 to n-1 and \displaystyle \sum_{i=0}^{n-1} i = \frac{(n-1)n}{2}:
As n \to \infty, \displaystyle \frac{4}{n} \to 0, so both limits are 10 and \int_1^3 (2x + 1)\,dx = 10. Check: the region is a trapezoid with parallel sides f(1) = 3 and f(3) = 7 and width 2, with area \displaystyle 2 \cdot \frac{3 + 7}{2} = 10. The right sums sit \displaystyle \frac{4}{n} above and the left sums \displaystyle \frac{4}{n} below, as the rising line predicts.
Day 1: Ximo confirms Leonor’s distance
Approximating area with Riemann sums found Leonor’s on-ramp distance \displaystyle \int_0^8 \left(10 + \frac{t^2}{4}\right)dt to be about 122.67 meters and quoted the exact value \displaystyle \frac{368}{3} without proof. Ximo wants to see it come out of the definition.
SolutionHere \displaystyle \Delta t = \frac{8}{n} and \displaystyle t_i = \frac{8i}{n}, so \displaystyle v(t_i) = 10 + \frac{1}{4}\cdot\frac{64i^2}{n^2} = 10 + \frac{16i^2}{n^2}. The velocity is continuous, so right endpoints give the integral:
Two checks against the earlier table: n = 4 gives \displaystyle 80 + \frac{64}{3}\cdot\frac{45}{16} = 140, and n = 100 gives 123.3088, both values of R_n found there by adding terms. For the limit, divide top and bottom by n^2: the fraction \displaystyle \frac{(n+1)(2n+1)}{n^2} equals \displaystyle (1 + \frac{1}{n})(2 + \frac{1}{n}), which approaches 2. So
Day 1: Severino needs all three formulas
Compute \int_1^2 x^3\,dx from the definition.
SolutionWith \displaystyle \Delta x = \frac{1}{n} and \displaystyle x_i = 1 + \frac{i}{n}, the cube of a sum (Special products) gives
Multiply by \displaystyle \frac{1}{n}, sum, and use each formula:
As a check, n = 1 gives 1 + 3 + 3 + 1 = 8 = f(2) \cdot 1, the single right rectangle. As n \to \infty the four terms approach 1, \displaystyle \frac{3}{2}, 1 and \displaystyle \frac{1}{4}, so
Day 2: reading the integral out of a limit
Write each limit as a definite integral, in two ways for the second.
(a) \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{\pi}{2n}\cos\left(\frac{\pi i}{2n}\right)
(b) \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}\cdot\frac{1}{1 + \frac{4i}{n}}
Solution(a) The factor in front is \displaystyle \Delta x = \frac{\pi}{2n}, so \displaystyle b - a = \frac{\pi}{2}. Inside the cosine, \displaystyle \frac{\pi i}{2n} = i\,\Delta x, which is 0 at i = 0, so a = 0, \displaystyle b = \frac{\pi}{2} and f = \cos:
The cosine is continuous, so the limit exists and these right sums give the integral.
(b) Here \displaystyle \Delta x = \frac{4}{n}, so the interval has length 4. If \displaystyle x_i = \frac{4i}{n}, the interval is [0, 4] and the term is \displaystyle \frac{1}{1 + x_i}. If instead \displaystyle x_i = 1 + \frac{4i}{n}, the interval is [1, 5] and the term is \displaystyle \frac{1}{x_i}. Both are right sums of the same numbers:
The second region is the first slid one unit right, so the areas agree. The limit fixes the numbers, not the picture.
Day 2: one integral, three sums
Write \int_{-1}^{2} e^{-x^2}\,dx as a limit of right, left and midpoint sums. Compare the three sums at n = 100.
SolutionThe integrand is continuous, so it is integrable and every choice of sample points has the same limit. With \displaystyle \Delta x = \frac{3}{n} and \displaystyle x_i = -1 + \frac{3i}{n}, the midpoint of piece i is \displaystyle m_i = -1 + \frac{3(2i - 1)}{2n}:
Adding the terms by computer at n = 100 gives L_{100} \approx 1.6341, R_{100} \approx 1.6236 and M_{100} \approx 1.6289; the integral is about 1.6289. No formula from algebra gives this area, and none will: the function e^{-x^2} has no antiderivative written with the usual functions. The limit is the only definition the number has, and sums like these are how it is computed.
Day 2: Ofelia’s integral that is not an area
Ofelia computes \int_0^3 (x^2 - 2x)\,dx from the definition. Explain her answer using the graph.
SolutionWith \displaystyle \Delta x = \frac{3}{n} and \displaystyle x_i = \frac{3i}{n}, \displaystyle f(x_i) = \frac{9i^2}{n^2} - \frac{6i}{n}, so
As n \to \infty, \displaystyle \frac{9(n+1)}{2n^2} \to 0, so \int_0^3 (x^2 - 2x)\,dx = 0. For n = 6 the formula gives \displaystyle \frac{63}{72} = 0.875; the figure shows those six rectangles.
The region between the graph and the axis is not empty. Since x^2 - 2x = x(x - 2), the graph is below the axis on (0, 2) and above it on (2, 3), and the terms from the first stretch are negative. The integral is the area above minus the area below (Signed area and integrals from geometry), and here the two are equal, each \displaystyle \frac{4}{3}, as The fundamental theorem of calculus will let us check. An integral equals the area of a region only when f \ge 0 on the interval.
Day 2: a function that is not integrable
Let D(x) = 1 when x is rational and D(x) = 0 when x is irrational, on [0, 1]. Severino computes its right sums with equal widths and gets 1 every time. Show that D is not integrable all the same.
SolutionSeverino’s sample points \displaystyle x_i = \frac{i}{n} are rational, so every term is \displaystyle 1 \cdot \frac{1}{n} and R_n = 1 for every n.
Every interval of positive width contains both rational and irrational numbers (stated here; it follows from the density of each set). So in any partition, however fine, the sample points may all be chosen irrational instead, and then every term is 0 and the sum is 0. Sums of 1 and sums of 0 occur at every mesh, so the sums do not approach one number as \|P\| \to 0. The definition fails: D is not integrable on [0, 1].
D is bounded, but it is discontinuous at every point, so it lies outside the integrability theorem. The right-endpoint rule gave a wrong “answer” because it was used without first knowing that D is integrable.
Boundaries
- The computing rule with right, left or midpoint sums needs f known to be integrable. A single family of sums can converge for a function that is not integrable (Example 7).
- Integrability is guaranteed for continuous functions, and for bounded functions with finitely many discontinuities. A function discontinuous everywhere may fail.
- An integrable function must be bounded on [a, b]. Let \displaystyle f(x) = \frac{1}{x^2} on (0, 1], with any value at 0: the piece next to 0 can be given a sample point so close to 0 that its single term is as large as we like, so no limit exists. Such integrals are handled later as limits of integrals.
- The definition takes a < b. The conventions for \int_b^a and \int_a^a are in Signed area and integrals from geometry.
- The integral equals an area only when f \ge 0 on [a, b]; in general it is a signed area.
Where this goes next
Computing integrals from the limit works for polynomials and little else. The fundamental theorem of calculus replaces the limit with an antiderivative, and The second fundamental theorem of calculus puts that to work: \displaystyle \frac{15}{4} and \displaystyle \frac{368}{3} come out in a line. Properties of definite integrals collects the rules that follow from the sums, such as splitting an interval. The definition itself is what makes integrals the tool for building up an amount from small pieces: totals from a rate, volumes by slices, work and arc length all start as a Riemann sum and pass to the limit.
Exercises
Day 1: integrals from the definition
- Compute \int_0^4 (3x + 2)\,dx from the definition with right endpoints, and check with geometry.
- Compute \int_2^5 (10 - 2x)\,dx from the definition with right endpoints, and check with geometry.
- Compute \int_0^4 x^2\,dx from the definition.
- Compute \int_1^3 x^2\,dx from the definition.
- Compute \int_0^2 (x^3 + 1)\,dx from the definition.
- Compute \int_0^3 2x^2\,dx from the definition with left endpoints, using \displaystyle \sum_{i=0}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6}.
- For the right sums R_n of Exercise 3, find the smallest n with \displaystyle R_n - \frac{64}{3} < 0.1.
Day 1: reasoning
- Prove \displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2} by writing the sum forward and backward and adding.
- Prove from the definition that \int_a^b c\,dx = c(b - a) for any constant c.
- Prove \displaystyle \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 by induction on n.
Day 2: from a limit to an integral
Write each limit as a definite integral. Give two readings where asked.
- \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n}\left(1 + \frac{2i}{n}\right)^4 (two readings)
- \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{3}{n}\,e^{3i/n}
- \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{2}{n}\sqrt{1 + \left(\frac{2i}{n}\right)^3}
- \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=0}^{n-1} \frac{5}{n}\ln\left(2 + \frac{5i}{n}\right)
- \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{3}{n}\sqrt{9 - \left(\frac{3i}{n}\right)^2}, and find its value with geometry.
Day 2: from an integral to a limit
- Write \displaystyle \int_2^6 \frac{x}{x^2 + 1}\,dx as a limit of right sums and as a limit of midpoint sums.
- Write \int_0^{\pi} x\sin x\,dx as a limit of left sums.
Day 2: signed area
- Compute \int_0^4 (x - 3)\,dx from the definition and explain the sign with the graph.
- Compute \int_0^3 (x^2 - 4)\,dx from the definition. Given that the area below the axis is \displaystyle \frac{16}{3}, find the area above it.
Day 2: reasoning
- Let g(x) = 1 on [0, 1] except \displaystyle g\left(\frac{1}{2}\right) = 5. Show from the definition that \int_0^1 g(x)\,dx = 1.
- Let h(x) = x when x is rational and h(x) = 0 when x is irrational. Show that h is not integrable on [0, 1].
- For a function f on [0, 1], show that the right sum \displaystyle \sum_{i=1}^{n} f(\frac{i}{n})\,\frac{1}{n} and the left sum \displaystyle \sum_{i=0}^{n-1} f(\frac{i}{n})\,\frac{1}{n} differ by \displaystyle \frac{f(1) - f(0)}{n}, and conclude that they have the same limit.
- Evaluate \displaystyle \displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} \frac{i}{n^2} two ways: with a sum formula, and by recognizing an integral and using geometry.
Answers
- 32. \displaystyle \Delta x = \frac{4}{n}, \displaystyle f(x_i) = \frac{12i}{n} + 2, so \displaystyle R_n = \frac{4}{n}\left(2n + \frac{12}{n}\cdot\frac{n(n+1)}{2}\right), which is \displaystyle 32 + \frac{24}{n} \to 32. A trapezoid with sides 2 and 14 and width 4 has area 32.
- 9. \displaystyle \Delta x = \frac{3}{n}, \displaystyle f(x_i) = 6 - \frac{6i}{n}, so \displaystyle R_n = 18 - \frac{9(n+1)}{n} = 9 - \frac{9}{n} \to 9. A triangle with base 3 and height 6 has area 9.
- \displaystyle \frac{64}{3}. \displaystyle R_n = \frac{4}{n}\sum \frac{16i^2}{n^2} \displaystyle = \frac{64}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} \displaystyle = \frac{32(n+1)(2n+1)}{3n^2} \to \frac{64}{3}.
- \displaystyle \frac{26}{3}. \displaystyle x_i^2 = 1 + \frac{4i}{n} + \frac{4i^2}{n^2}, so \displaystyle R_n = 2 + \frac{4(n+1)}{n} + \frac{4(n+1)(2n+1)}{3n^2}, which approaches \displaystyle 2 + 4 + \frac{8}{3}.
- 6. \displaystyle R_n = \frac{2}{n}\sum\left(\frac{8i^3}{n^3} + 1\right) \displaystyle = \frac{4(n+1)^2}{n^2} + 2 \to 6.
- 18. \displaystyle L_n = \frac{3}{n}\sum_{i=0}^{n-1} \frac{18i^2}{n^2} \displaystyle = \frac{54}{n^3}\cdot\frac{(n-1)n(2n-1)}{6} \displaystyle = \frac{9(n-1)(2n-1)}{n^2} \to 18.
- n = 321. Expanding, \displaystyle R_n = \frac{64}{3} + \frac{32}{n} + \frac{32}{3n^2}. At n = 320 the excess is \displaystyle 0.1 + \frac{32}{307{,}200} > 0.1; at n = 321 it is about 0.09979.
- Let S = 1 + 2 + \dots + n and also S = n + (n-1) + \dots + 1. Adding term by term, each of the n columns sums to n + 1, so 2S = n(n+1).
- Every Riemann sum is \sum c\,\Delta x_i = c\sum \Delta x_i = c(b - a), since the widths add to the length of [a, b], whatever the partition and sample points. A constant sequence has that constant as its limit.
- For n = 1 both sides are 1. If the formula holds for n, then \displaystyle \sum_{i=1}^{n+1} i^3 = \frac{n^2(n+1)^2}{4} + (n+1)^3 \displaystyle = \frac{(n+1)^2(n^2 + 4n + 4)}{4} \displaystyle = \left(\frac{(n+1)(n+2)}{2}\right)^2, the formula for n + 1.
- \int_1^3 x^4\,dx (with \displaystyle x_i = 1 + \frac{2i}{n}), or \int_0^2 (1 + x)^4\,dx (with \displaystyle x_i = \frac{2i}{n}).
- \int_0^3 e^x\,dx: \displaystyle \Delta x = \frac{3}{n} and \displaystyle x_i = \frac{3i}{n}.
- \displaystyle \int_0^2 \sqrt{1 + x^3}\,dx: \displaystyle \Delta x = \frac{2}{n} and \displaystyle x_i = \frac{2i}{n}.
- \int_2^7 \ln x\,dx: \displaystyle \Delta x = \frac{5}{n}, and \displaystyle x_i = 2 + \frac{5i}{n} for i = 0 to n - 1 are left endpoints of [2, 7].
- \displaystyle \int_0^3 \sqrt{9 - x^2}\,dx = \frac{9\pi}{4}: the graph is the quarter of the circle of radius 3 in the first quadrant, area \displaystyle \frac{1}{4}\pi(3^2).
- \displaystyle \Delta x = \frac{4}{n}: \displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \frac{x_i}{x_i^2 + 1}\cdot\frac{4}{n} with \displaystyle x_i = 2 + \frac{4i}{n}, and the same with \displaystyle m_i = 2 + \frac{2(2i - 1)}{n} in place of x_i.
- \displaystyle \lim_{n \to \infty} \sum_{i=0}^{n-1} x_i \sin(x_i)\,\frac{\pi}{n} with \displaystyle x_i = \frac{\pi i}{n}, the left endpoints of [0, \pi].
- -4. \displaystyle R_n = \frac{4}{n}\sum\left(\frac{4i}{n} - 3\right) \displaystyle = \frac{8(n+1)}{n} - 12 \to -4. The line is below the axis on [0, 3), a triangle of area \displaystyle \frac{9}{2}, and above on (3, 4], area \displaystyle \frac{1}{2}: \displaystyle \frac{1}{2} - \frac{9}{2} = -4.
- -3. \displaystyle R_n = \frac{3}{n}\sum\left(\frac{9i^2}{n^2} - 4\right) \displaystyle = \frac{9(n+1)(2n+1)}{2n^2} - 12, which approaches 9 - 12. Area above \displaystyle = -3 + \frac{16}{3} = \frac{7}{3}.
- A Riemann sum for g differs from the same sum for the constant 1 only in terms whose sample point is \displaystyle \frac{1}{2}. At most two pieces contain \displaystyle \frac{1}{2}, and each such term is larger by 4\,\Delta x_i \le 4\|P\|. So every sum lies within 8\|P\| of 1, which goes to 0 with the mesh.
- Every piece contains rationals and irrationals. Rational right endpoints with equal widths give \displaystyle \sum \frac{i}{n}\cdot\frac{1}{n} = \frac{n+1}{2n} \to \frac{1}{2}; irrational sample points give 0 at every mesh. No single limit, so h is not integrable.
- The two sums share the terms for i = 1 to n - 1. The right sum adds \displaystyle f(1)\frac{1}{n} and the left sum \displaystyle f(0)\frac{1}{n}, so the difference is \displaystyle \frac{f(1) - f(0)}{n}, which goes to 0. If either sum has a limit, the other has the same one.
- \displaystyle \frac{1}{2}. First, \displaystyle \frac{1}{n^2}\cdot\frac{n(n+1)}{2} = \frac{n+1}{2n} \to \frac{1}{2}. Second, \displaystyle \frac{i}{n^2} = \frac{i}{n}\cdot\frac{1}{n} is a right sum for \int_0^1 x\,dx, a triangle with base and height 1.