The second fundamental theorem of calculus
A definite integral is defined as a limit of sums, which can take a page to compute. If an antiderivative of the integrand is known, can the integral be read off from it in one line?
Where you are 2 in The fundamental theorem of calculus
When you reach for it
Finding \displaystyle \int_0^8 \left(10 + \frac{t^2}{4}\right)dt from the definition (The definite integral as a limit of Riemann sums) took a formula for a sum of squares and a limit. The first fundamental theorem showed that the accumulation function A(x) = \int_a^x f(t)\,dt of a continuous f is an antiderivative of f. The second fundamental theorem turns this around: to find \int_a^b f(x)\,dx, take any antiderivative F of f and subtract, F(b) - F(a). No sums and no limits.
You reach for it whenever a definite integral has an integrand whose antiderivative you can write down: a polynomial or a power (Antiderivatives and indefinite integrals), a trig function, an exponential, \displaystyle \frac{1}{x}, or an inverse trig form (Antiderivatives of trig, exponential and log functions). It is the second half of the fundamental theorem of calculus, and it is how nearly every exact integral in the rest of calculus is found.
The procedure
Day 1: the theorem
Theorem (second fundamental theorem of calculus). Let f be continuous on [a, b], and let F be an antiderivative of f on [a, b]: F is continuous on [a, b] and F'(x) = f(x) for every x in (a, b). Then
The difference is written with an evaluation bar:
When F has several terms they go in parentheses before the bar, so \left(x^3 + x\right)\Big|_0^2 = 10 - 0 = 10.
- Any antiderivative works. Two antiderivatives of f on an interval differ by a constant (a consequence of the mean value theorem), and the constant cancels: \big(F(b) + C\big) - \big(F(a) + C\big) = F(b) - F(a). That is why the +\,C is left out.
- Limits in either order. With the conventions \int_b^a = -\int_a^b and \int_a^a = 0 (Signed area and integrals from geometry), the formula also holds when the lower limit is the larger: both sides change sign when the limits swap, and both are 0 when they are equal.
- The answer is a signed area, because the integral is one. A negative answer means more area lies below the axis than above it.
Day 1: why it is true
Two proofs follow. The first rests on the first fundamental theorem. The second goes straight back to Riemann sums and proves slightly more.
Proof from the first fundamental theorem. Let A(x) = \int_a^x f(t)\,dt for x in [a, b]. By the first fundamental theorem, A is continuous on [a, b] and A'(x) = f(x) on (a, b). So G = F - A is continuous on [a, b] and G'(x) = f(x) - f(x) = 0 on (a, b). For any x in (a, b], the mean value theorem on [a, x] gives a c with G(x) - G(a) = G'(c)(x - a) = 0. So G is a constant K, that is, F(x) = A(x) + K for every x in [a, b], and
Proof from the mean value theorem on each piece. Here f need only be integrable on [a, b], with F continuous on [a, b] and F' = f on (a, b). Take any partition a = x_0 < x_1 < \dots < x_n = b and write \Delta x_i = x_i - x_{i-1}. On each piece [x_{i-1}, x_i], F is continuous and differentiable inside, so the mean value theorem gives a point c_i in (x_{i-1}, x_i) with
Add these n equations. On the left every inside value appears once with each sign and cancels (the sum telescopes):
The right side is a Riemann sum for f with sample points c_i. So every partition, however fine, has a Riemann sum exactly equal to the fixed number F(b) - F(a). Because f is integrable, all Riemann sums whose widest piece is small enough lie as close as we like to \int_a^b f(x)\,dx. A fixed number that is within every distance of the integral is the integral. \blacksquare
The second proof never uses continuity of f, so it proves the stronger statement: if F is continuous on [a, b], F' = f on (a, b), and f is integrable, then \int_a^b f(x)\,dx = F(b) - F(a). Exercise 11 watches the telescoping happen for f(x) = 2x.
Day 2: the antiderivatives in use
Every antiderivative formula becomes a formula for definite integrals. These come from Antiderivatives and indefinite integrals and Antiderivatives of trig, exponential and log functions. The last column matters: the theorem needs an antiderivative on the whole closed interval.
| Integrand | Antiderivative | On any interval where |
|---|---|---|
| x^n, n \ne -1 | \dfrac{x^{n+1}}{n+1} | x^n is defined |
| \dfrac{1}{x} | \ln\lvert x \rvert | x \ne 0 |
| e^{kx}, k \ne 0 | \dfrac{1}{k}e^{kx} | always |
| \cos x | \sin x | always |
| \sin x | -\cos x | always |
| \sec^2 x | \tan x | \cos x \ne 0 |
| \sec x\tan x | \sec x | \cos x \ne 0 |
| \dfrac{1}{1 + x^2} | \arctan x | always |
| \displaystyle \dfrac{1}{\sqrt{1 - x^2}} | \arcsin x | -1 < x < 1 |
Splitting at a break. A piecewise function or an absolute value has different formulas on different pieces, and no one formula is an antiderivative on all of [a, b]. Additivity (Properties of definite integrals) cuts the interval at each break, \int_a^b = \int_a^c + \int_c^b, and the theorem is used on each piece with that piece’s formula. For \lvert g(x) \rvert the breaks are where g changes sign: \lvert g \rvert = g where g \ge 0 and \lvert g \rvert = -g where g \le 0. A jump at a break is allowed too: the stronger form of the theorem asks only that F' = f inside each piece.
Net versus total. The theorem gives signed area, the area above the axis minus the area below. The total area between the graph and the axis is \int_a^b \lvert f(x) \rvert\,dx, found by splitting at the zeros of f and adding the size of each piece. When f is a velocity and F a position, the first is the displacement F(b) - F(a) and the second is the total distance traveled.
Worked examples
Day 1: a polynomial, and the constant that cancels
Evaluate \int_{-1}^{2} \left(x^3 - 6x + 1\right)dx with the antiderivative \displaystyle F(x) = \frac{x^4}{4} - 3x^2 + x, and again with F(x) + 5.
SolutionThe integrand is a polynomial, so it is continuous on [-1, 2]. Reading the power rule backward term by term gives F, and F'(x) = x^3 - 6x + 1 confirms it. By the theorem,
With F(x) + 5 the two values become -1 and \displaystyle \frac{5}{4}, and \displaystyle -1 - \frac{5}{4} = -\frac{9}{4} again: the 5 is added at both ends and cancels. The answer is negative because the cubic is positive only from -1 to about 0.17 and negative from there to 2, so more of the region lies below the axis.
Day 1: Vusal checks three numbers taken on trust
Three earlier lessons quoted values they could not yet produce: Leonor’s distance \displaystyle \int_0^8 \left(10 + \frac{t^2}{4}\right)dt = \frac{368}{3} meters in Approximating area with Riemann sums, the integral \displaystyle \int_1^2 x^3\,dx = \frac{15}{4} found by a limit of sums in The definite integral as a limit of Riemann sums, and \displaystyle \int_1^2 \frac{1}{x}\,dx = \ln 2 in The trapezoidal rule and Simpson's rule. Vusal confirms each.
Solution(a) The derivative of \displaystyle 10t + \frac{t^3}{12} is \displaystyle 10 + \frac{3t^2}{12} = 10 + \frac{t^2}{4}, so it is an antiderivative, and
(b) \dfrac{x^4}{4}\Big|_1^2 = \dfrac{16}{4} - \dfrac{1}{4} = \dfrac{15}{4}: one line in place of four sum formulas and a limit.
(c) On [1, 2] every x is positive, so \ln x is an antiderivative of \displaystyle \frac{1}{x} there, and \ln x\Big|_1^2 = \ln 2 - \ln 1 = \ln 2 \approx 0.693147. The trapezoid and midpoint sums in that lesson bracketed this number; now it is exact.
Day 1: Sabina rewrites first
(a) Evaluate \displaystyle \int_1^4 \frac{x^2 - 2}{\sqrt{x}}\,dx.
(b) Show that \displaystyle \int_1^b \frac{1}{x^2}\,dx < 1 for every b > 1, and find the b that makes it \displaystyle \frac{3}{4}.
Solution(a) There is no quotient rule for antiderivatives, so divide each term by x^{1/2} first: the integrand is x^{3/2} - 2x^{-1/2}, continuous on [1, 4] since x > 0 there. The power rule gives the antiderivative \displaystyle \frac{2}{5}x^{5/2} - 4x^{1/2}, and with 4^{5/2} = 32,
(b) An antiderivative of x^{-2} is -x^{-1}, and x^{-2} is continuous on [1, b], so
Since b > 0, \displaystyle \frac{1}{b} > 0 and the integral is less than 1. It equals \displaystyle \frac{3}{4} when \displaystyle \frac{1}{b} = \frac{1}{4}, at b = 4. However far right the interval reaches, the area under this curve stays below 1.
Day 2: trig, exponential, logarithm and arctangent
Evaluate (a) \int_0^{\pi/3} \left(2\sin x + \sec^2 x\right)dx, (b) \int_0^{\ln 3} e^{2x}\,dx, (c) \displaystyle \int_1^{e^2} \frac{x + 3}{x}\,dx, (d) \displaystyle \int_0^1 \frac{4}{1 + x^2}\,dx.
Solution(a) On \displaystyle \left[0, \frac{\pi}{3}\right], \cos x > 0, so \tan x is an antiderivative of \sec^2 x on the whole interval. With -2\cos x for 2\sin x,
(b) With k = 2, an antiderivative is \displaystyle \frac{1}{2}e^{2x}, and e^{2\ln 3} = \left(e^{\ln 3}\right)^2 = 9:
(c) Split the quotient: \displaystyle \frac{x + 3}{x} = 1 + \frac{3}{x}. Every x in [1, e^2] is positive, so x + 3\ln x is an antiderivative there:
(d) 4\arctan x is an antiderivative on all of \mathbb{R}:
A quotient of two polynomials, with no trig function and no circle in sight, gives the number \pi.
Day 2: an absolute value and a piecewise integrand
(a) Evaluate \int_0^3 \lvert x^2 - 4 \rvert\,dx.
(b) Let f(x) = e^x for x \le 0 and f(x) = 1 + x for x > 0. Evaluate \int_{-1}^{2} f(x)\,dx.
Solution(a) The expression x^2 - 4 is negative on [0, 2) and positive on (2, 3], so \lvert x^2 - 4 \rvert = 4 - x^2 on [0, 2] and x^2 - 4 on [2, 3]. Split at 2 and use each formula on its own piece:
The integral is \displaystyle \frac{16}{3} + \frac{7}{3} = \frac{23}{3}. Without the absolute value, \displaystyle \left(\frac{x^3}{3} - 4x\right)\Big|_0^3 = -3, the signed area, which is a different number.
(b) Both formulas give 1 at x = 0, so f is continuous, but neither formula is an antiderivative across 0. Split there:
The integral is \displaystyle 5 - \frac{1}{e} \approx 4.632.
Day 2: Murad’s scaffold, displacement and distance
Murad runs a window-washing scaffold up and down the side of a building. Its velocity is v(t) = 3t^2 - 12t + 9 meters per minute for 0 \le t \le 4, positive when it rises. Find its net change in height and the total distance it travels.
A height function is P(t) = t^3 - 6t^2 + 9t, since P'(t) = v(t). The net change is
For distance, factor: v(t) = 3(t - 1)(t - 3), positive on (0, 1), negative on (1, 3) and positive on (3, 4). With P(0) = 0, P(1) = 4, P(3) = 0 and P(4) = 4, the three pieces are
The scaffold rises 4 meters, drops 4 and rises 4 again, so the total distance is \int_0^4 \lvert v(t) \rvert\,dt = 4 + 4 + 4 = 12 meters. The displacement counts the drop as negative, which is why it is only 4.
Day 2: when the theorem does not apply
Sabina writes
Explain what is wrong, and then evaluate \displaystyle \int_{-2}^{-1} \frac{1}{x}\,dx.
The integrand is positive wherever it is defined, so a negative answer is impossible. The theorem needs f continuous on [-1, 1], and \displaystyle \frac{1}{x^2} is not even defined at 0; nor is \displaystyle -\frac{1}{x} an antiderivative on [-1, 1], since it is not defined at 0 either. Worse, the integrand is unbounded near 0. On the piece of a partition touching 0, a sample point close enough to 0 makes f(c_i)\,\Delta x_i as large as we like, so the Riemann sums have no limit and \displaystyle \int_{-1}^{1} \frac{1}{x^2}\,dx does not exist. The area itself has no bound: for each integer n \ge 2,
The second integral is fine. The interval [-2, -1] avoids 0, and on (-\infty, 0), \ln\lvert x \rvert is an antiderivative of \displaystyle \frac{1}{x}:
It is negative because \displaystyle \frac{1}{x} < 0 on the whole interval.
Boundaries
- The theorem needs an antiderivative on the whole closed interval. A formula that is undefined or breaks at a point inside, such as \displaystyle -\frac{1}{x} on [-1, 1] or \tan x on [0, \pi], gives a wrong number with no warning (Example 7, Exercises 23 and 24).
- In the form proved first, f must be continuous on [a, b]. The second proof needs only f integrable and F' = f inside the interval, which covers jumps once the interval is split. An unbounded f is not integrable at all.
- The value is a signed area. Area between the graph and the axis, or distance traveled, needs \lvert f \rvert and a split at each zero.
- The theorem is only as useful as the antiderivatives you can find. e^{-x^2} is continuous, so \int_0^1 e^{-x^2}\,dx exists, but it has no antiderivative built from the usual functions; such integrals are estimated numerically (The trapezoidal rule and Simpson's rule).
- Integrands such as (2x + 1)^5 or x\cos(x^2) have antiderivatives, but finding them takes u-substitution, the next tool.
Where this goes next
With this theorem, finding an exact integral comes down to finding an antiderivative, and the rest of integral calculus is largely about that search: u-substitution reverses the chain rule, and later methods reverse the product rule. Read with F as an amount and f = F' as its rate, the theorem says the integral of a rate is the total change, which accumulation and net change puts to work with rates of flow, growth and motion. The same subtraction gives exact average values.
Exercises
Day 1: polynomials and powers
- Evaluate \int_0^2 \left(4x^3 - 6x + 5\right)dx.
- Evaluate \int_{-2}^{1} \left(x^2 + 2x - 3\right)dx.
- Evaluate \displaystyle \int_1^8 \sqrt[3]{x}\,dx.
- Evaluate \displaystyle \int_1^2 \frac{3}{x^4}\,dx.
- Evaluate \displaystyle \int_1^4 \left(\sqrt{x} + 1\right)^2 dx.
- Evaluate \int_3^0 \left(x^2 - 1\right)dx.
Day 1: unknowns in the limits and the exponent
- Find the number b > 1 for which \int_1^b (2x + 3)\,dx = 14.
- Find \int_0^1 x^n\,dx for every n \ge 0, and use it to find \int_0^1 \left(x^4 + x^9\right)dx.
Day 1: reasoning
- For \int_0^3 x^2\,dx, Vusal uses the antiderivative \displaystyle \frac{x^3}{3} + 7 and Sabina uses \displaystyle \frac{x^3}{3} - 2. Show that they get the same value, and explain why that must happen.
- The function f has a continuous derivative on [1, 4], with f(1) = 3 and \int_1^4 f'(x)\,dx = 5. Find f(4), and say what the theorem means when the integrand is a rate of change.
- Divide [0, 1] into n equal pieces. Show that the midpoint sum for \int_0^1 2x\,dx equals 1 for every n, and explain how this is the second proof at work with F(x) = x^2.
Day 2: trig, exponential, logarithm and inverse trig
- Evaluate \int_0^{\pi/2} \left(3\cos x - x\right)dx.
- Evaluate \int_{\pi/4}^{\pi/3} \sec x\tan x\,dx.
- Evaluate \int_0^1 \left(e^x + e^{-x}\right)dx.
- Evaluate \displaystyle \int_1^e \frac{x^2 + 1}{x}\,dx.
- Evaluate \displaystyle \int_{-1/2}^{\sqrt{2}/2} \frac{1}{\sqrt{1 - x^2}}\,dx.
- Evaluate \displaystyle \int_{-1}^{\sqrt{3}} \frac{2}{1 + x^2}\,dx.
Day 2: splitting at a break
- Evaluate \int_{-1}^{2} \lvert x^2 - x \rvert\,dx.
- Evaluate \int_0^{\pi} \lvert \cos x \rvert\,dx.
- Let \displaystyle f(x) = \sqrt{x} for 0 \le x \le 4 and f(x) = 6 - x for 4 < x \le 6. Evaluate \int_0^6 f(x)\,dx.
Day 2: net and total
- A cart moves along a straight track with velocity v(t) = t^2 - 7t + 10 meters per second for 0 \le t \le 6. Find its displacement and the total distance it travels.
- For f(x) = x^3 - x on [0, 2], find \int_0^2 f(x)\,dx and the total area between the graph and the x-axis.
Day 2: when the theorem applies
- Murad computes \int_0^{\pi} \sec^2 x\,dx = \tan x\Big|_0^{\pi} = 0 - 0 = 0. Explain why the answer cannot be right and what went wrong.
- For x \ne 0 let \displaystyle G(x) = -\arctan\left(\frac{1}{x}\right). Show that \displaystyle G'(x) = \frac{1}{1 + x^2}, compute G(1) - G(-1), and compare with \displaystyle \int_{-1}^{1} \frac{1}{1 + x^2}\,dx. Which is right, and why does the other fail?
- Use the second fundamental theorem to show that \int_{-a}^{a} x^n\,dx = 0 for every odd positive integer n and every a.
Answers
- 14. \left(x^4 - 3x^2 + 5x\right)\Big|_0^2 = 16 - 12 + 10 = 14.
- -9. With \displaystyle F(x) = \frac{x^3}{3} + x^2 - 3x, \displaystyle F(1) = -\frac{5}{3} and \displaystyle F(-2) = \frac{22}{3}, so the integral is \displaystyle -\frac{27}{3} = -9. The parabola has zeros -3 and 1, so it lies below the axis on all of [-2, 1].
- \displaystyle \frac{45}{4}. \displaystyle \frac{3}{4}x^{4/3}\Big|_1^8 = \frac{3}{4}(16 - 1).
- \displaystyle \frac{7}{8}. An antiderivative of 3x^{-4} is -x^{-3}, and \displaystyle -\frac{1}{8} - (-1) = \frac{7}{8}.
- \displaystyle \frac{119}{6}. Expand to x + 2x^{1/2} + 1, with antiderivative \displaystyle \frac{x^2}{2} + \frac{4}{3}x^{3/2} + x, whose values are \displaystyle \frac{68}{3} at 4 and \displaystyle \frac{17}{6} at 1.
- -6. \displaystyle \left(\frac{x^3}{3} - x\right)\Big|_3^0 = 0 - (9 - 3), the negative of \int_0^3 \left(x^2 - 1\right)dx = 6.
- b = 3. \left(x^2 + 3x\right)\Big|_1^b = b^2 + 3b - 4 = 14, so (b + 6)(b - 3) = 0, and b > 1 leaves b = 3.
- \displaystyle \frac{1}{n + 1}, and \displaystyle \frac{3}{10}. \displaystyle \frac{x^{n+1}}{n+1}\Big|_0^1 = \frac{1}{n+1} because 0^{n+1} = 0; then \displaystyle \frac{1}{5} + \frac{1}{10} = \frac{3}{10}.
- Both get 9: (9 + 7) - (0 + 7) = 9 and (9 - 2) - (0 - 2) = 9. The constant is added at both ends and subtracts away, so every antiderivative gives the same difference.
- f(4) = 8. Since f is an antiderivative of the continuous f', \int_1^4 f'(x)\,dx = f(4) - f(1), so 5 = f(4) - 3. The integral of a rate of change over an interval is the total change in the quantity.
- The midpoint of [x_{i-1}, x_i] is \displaystyle c_i = \frac{x_{i-1} + x_i}{2}, so 2c_i\,\Delta x_i = (x_i + x_{i-1})(x_i - x_{i-1}) = x_i^2 - x_{i-1}^2. The sum telescopes to 1^2 - 0^2 = 1. These midpoints are exactly the points the mean value theorem gives for F(x) = x^2 on each piece, since \displaystyle F'(c_i) = 2c_i = \frac{x_i^2 - x_{i-1}^2}{\Delta x_i}, so this midpoint sum is the sum in the second proof.
- \displaystyle 3 - \frac{\pi^2}{8} \approx 1.766. \displaystyle \left(3\sin x - \frac{x^2}{2}\right)\Big|_0^{\pi/2} = 3 - \frac{\pi^2}{8} - 0.
- \displaystyle 2 - \sqrt{2}. \cos x > 0 on \displaystyle \left[\frac{\pi}{4}, \frac{\pi}{3}\right], so \displaystyle \sec x\Big|_{\pi/4}^{\pi/3} = 2 - \sqrt{2}.
- \displaystyle e - \frac{1}{e} \approx 2.350. \left(e^x - e^{-x}\right)\Big|_0^1 = \left(e - e^{-1}\right) - (1 - 1).
- \displaystyle \frac{e^2 + 1}{2}. The integrand is \displaystyle x + \frac{1}{x}, with antiderivative \displaystyle \frac{x^2}{2} + \ln x for x > 0: \displaystyle \left(\frac{e^2}{2} + 1\right) - \left(\frac{1}{2} + 0\right).
- \displaystyle \frac{5\pi}{12}. The interval lies inside (-1, 1), and \displaystyle \arcsin\frac{\sqrt{2}}{2} - \arcsin\left(-\frac{1}{2}\right) = \frac{\pi}{4} + \frac{\pi}{6}.
- \displaystyle \frac{7\pi}{6}. \displaystyle 2\arctan x\Big|_{-1}^{\sqrt{3}} = \frac{2\pi}{3} + \frac{\pi}{2}.
- \displaystyle \frac{11}{6}. x^2 - x = x(x - 1) is \ge 0 on [-1, 0] and [1, 2] and \le 0 on [0, 1]. With \displaystyle F(x) = \frac{x^3}{3} - \frac{x^2}{2}: \displaystyle F(-1) = -\frac{5}{6}, F(0) = 0, \displaystyle F(1) = -\frac{1}{6}, \displaystyle F(2) = \frac{2}{3}, so the pieces have sizes \displaystyle \frac{5}{6}, \displaystyle \frac{1}{6} and \displaystyle \frac{5}{6}.
- 2. \cos x \ge 0 on \displaystyle \left[0, \frac{\pi}{2}\right] and \le 0 on \displaystyle \left[\frac{\pi}{2}, \pi\right]: \sin x\Big|_0^{\pi/2} - \sin x\Big|_{\pi/2}^{\pi} = 1 - (0 - 1) = 2.
- \displaystyle \frac{22}{3}. \displaystyle \frac{2}{3}x^{3/2}\Big|_0^4 = \frac{16}{3} and \displaystyle \left(6x - \frac{x^2}{2}\right)\Big|_4^6 = 18 - 16 = 2.
- Displacement 6 meters, distance 15 meters. v(t) = (t - 2)(t - 5), and \displaystyle P(t) = \frac{t^3}{3} - \frac{7t^2}{2} + 10t has P(0) = 0, \displaystyle P(2) = \frac{26}{3}, \displaystyle P(5) = \frac{25}{6}, P(6) = 6. The distance is \displaystyle \frac{26}{3} + \frac{9}{2} + \frac{11}{6} = 15.
- \int_0^2 f(x)\,dx = 2, total area \displaystyle \frac{5}{2}. With \displaystyle F(x) = \frac{x^4}{4} - \frac{x^2}{2}: F(0) = 0, \displaystyle F(1) = -\frac{1}{4}, F(2) = 2. Since f \le 0 on [0, 1] and f \ge 0 on [1, 2], the area is \displaystyle \frac{1}{4} + \frac{9}{4}.
- Since \sec^2 x \ge 1 wherever it is defined, a value of 0 is impossible. The function \tan x is undefined at \displaystyle \frac{\pi}{2}, inside [0, \pi], so it is not an antiderivative on the whole interval; and \sec^2 x is unbounded near \displaystyle \frac{\pi}{2}, so the integral does not exist.
- By the chain rule, \displaystyle G'(x) = -\frac{1}{1 + 1/x^2}\cdot\left(-\frac{1}{x^2}\right) = \frac{1}{x^2 + 1}. Then \displaystyle G(1) - G(-1) = -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2}, while \displaystyle \arctan x\Big|_{-1}^{1} = \frac{\pi}{2}. The second is right (the integrand is positive). G is undefined at 0 and jumps there: \displaystyle G(x) = \arctan x - \frac{\pi}{2} for x > 0 and \displaystyle \arctan x + \frac{\pi}{2} for x < 0, so it is not an antiderivative on [-1, 1].
- An antiderivative of x^n is \displaystyle F(x) = \frac{x^{n+1}}{n + 1}, and n + 1 is even, so F(-a) = F(a). Hence \int_{-a}^{a} x^n\,dx = F(a) - F(-a) = 0, the symmetry of odd integrands from Signed area and integrals from geometry found by algebra.