Antiderivatives and indefinite integrals
A rate is known, such as how many centimeters a kelp frond grows each day, and the amount is wanted, how long the frond is. How is a derivative run backward, and how many answers are there?
Where you are 16 in Integrals
Why this exists
Every derivative rule so far answers one question: given an amount, find its rate. Science usually hands over the reverse. Gwion, a marine biologist, records how fast a kelp frond grows each day, not how long it is. Macsen reads a train’s speed from its instruments and wants to know how far it travels. Each time the derivative is known and the function is wanted.
Undoing a derivative is called antidifferentiation, and it raises two questions that differentiation never did. Is there more than one answer? Yes: adding a constant to a function changes nothing about its rate. Then how is the right answer picked out of all of them? By one known value, such as the frond’s length on the first day. This lesson answers both questions for powers of x and the functions built from them.
Definition
Day 1: antiderivatives and the indefinite integral
A function F is an antiderivative of f on an interval I if F'(x) = f(x) for every x in I.
For example, x^4 is an antiderivative of 4x^3 on (-\infty, \infty), and so are x^4 + 10 and x^4 - \pi, because a constant has derivative 0.
The indefinite integral of f is the whole collection of antiderivatives of f on an interval. It is written
where F is any one antiderivative and C stands for an arbitrary constant, the constant of integration. The sign \int is the integral sign, f(x) is the integrand, and dx names the variable. So \int 4x^3\,dx = x^4 + C. The same sign with limits attached, \int_a^b f(x)\,dx, is the definite integral of Riemann sums, a single number rather than a family of functions. Why the two share a symbol is the content of the fundamental theorem of calculus.
Day 2: initial conditions
An initial condition is one known value of the unknown function, y(x_0) = y_0. A problem that gives a derivative together with an initial condition and asks for the function is an initial value problem. Its answer is a particular antiderivative, one member of the family, as opposed to the general antiderivative F(x) + C. When f'' is given, two antidifferentiations are needed, and two conditions to fix the two constants they bring.
Intuition
All antiderivatives of the same f have the same slope above every x, so their graphs are copies of one curve moved straight up or down. Figure 1 shows five antiderivatives of x^2 - 1. Above any x the five curves are equally steep: level at x = -1 and x = 1, rising with slope 3^2 - 1 = 8 at x = 3. A known point, here (3, 5), lies on exactly one of them.
For a power, the power rule takes x^{n+1} to (n + 1)x^n: it lowers the exponent by one and multiplies by the old exponent. Run backward, it raises the exponent by one and divides by the new exponent. Derivatives of sums and constant multiples are taken term by term, so antiderivatives are found term by term too.
Results
Day 1: the family and the basic rules
Every antiderivative on an interval. If F is an antiderivative of f on an interval I, then the antiderivatives of f on I are exactly the functions F(x) + C, one for each constant C.
The power rule for antiderivatives. For every real number n \ne -1, on any interval where x^n and x^{n+1} are defined,
The case n = 0 gives \int k\,dx = kx + C for a constant k.
Constant multiple and sum rules. For a constant k \ne 0,
Checking. F(x) + C is the indefinite integral of f on I exactly when F'(x) = f(x) on I. Every antiderivative can be checked by differentiating it.
Day 2: picking one antiderivative
One condition, one function. If F is an antiderivative of f on an interval I and x_0 is in I, exactly one antiderivative of f on I takes the value y_0 at x_0, namely F(x) + y_0 - F(x_0).
Two conditions for a second derivative. If G'' = f'' on I, then f(x) = G(x) + C_1 x + C_2. Two conditions, either f(x_0) and f'(x_0) or the values of f at two different points, fix C_1 and C_2.
Why it is true
The family. The lesson on the mean value theorem proved that two functions with equal derivatives on an interval differ by a constant there. If G is any antiderivative of f on I, then G' = f = F', so G = F + C. Conversely (F + C)' = F' + 0 = f, so every F + C is an antiderivative.
The power rule. Since n \ne -1, the number n + 1 is not 0 and the division is allowed. The power rule for derivatives, valid for every real exponent where the power is defined, gives
and the family result supplies every other antiderivative.
Sum and constant multiple. If F' = f and G' = g, the derivative rules give (kF)' = kf and (F \pm G)' = f \pm g. Constants combine: k times an arbitrary constant is still arbitrary when k \ne 0, and C_1 \pm C_2 is again one arbitrary constant. That is why a single +\,C is written at the end.
One condition. Every antiderivative is F + C, and F(x_0) + C = y_0 holds for the one value C = y_0 - F(x_0).
Two conditions. Antidifferentiating f'' = G'' once gives f' = G' + C_1, and again gives f = G + C_1 x + C_2. Given f'(x_0), the first equation fixes C_1, and then f(x_0) fixes C_2. Given f(x_1) and f(x_2) with x_1 \ne x_2, subtracting the two equations leaves C_1(x_2 - x_1) equal to a known number, which fixes C_1, and either equation then fixes C_2.
Worked examples
Day 1: reading the power rule backward
Find (a) \int x^7\,dx, (b) \int \dfrac{1}{x^3}\,dx and (c) \int 5\,dt, and check each answer by differentiating.
Solution(a) With n = 7, raise the exponent to 8 and divide by 8:
Check: \dfrac{d}{dx}\left[\dfrac{x^8}{8} + C\right] = \dfrac{8x^7}{8} = x^7.
(b) The rule is about powers, so write \dfrac{1}{x^3} = x^{-3}. Here n = -3 \ne -1, so the new exponent is -2 and the divisor is -2:
Check: \dfrac{d}{dx}\left[-\dfrac{1}{2}x^{-2}\right] = -\dfrac{1}{2}(-2)x^{-3} = x^{-3}. The integrand is undefined at 0, so this holds on (0, \infty) and on (-\infty, 0), each interval on its own, and the C may be a different constant on each (Boundaries).
(c) The dt says the variable is t. The constant 5 is 5t^0, so the case n = 0 applies:
Check: \dfrac{d}{dt}\big[5t + C\big] = 5.
Day 1: Efa and Macsen get different answers
Efa and Macsen each find \int \left(x^2 + 2x + 1\right)dx. Efa works term by term and writes \dfrac{x^3}{3} + x^2 + x + C. Macsen notices that the integrand is (x + 1)^2 and writes \dfrac{(x + 1)^3}{3} + C. Who is right?
SolutionBoth are. Differentiate each. Efa’s gives x^2 + 2x + 1 term by term. Macsen’s gives, by the chain rule, \dfrac{3(x + 1)^2 \cdot 1}{3} = (x + 1)^2 = x^2 + 2x + 1.
Two antiderivatives of the same function on the interval (-\infty, \infty) must differ by a constant. Expanding Macsen’s function shows the constant:
Macsen’s antiderivative is Efa’s plus \dfrac{1}{3}. Since C is arbitrary, the two families are the same set of functions: Macsen’s with constant C is Efa’s with constant \displaystyle C + \frac{1}{3}. Two correct indefinite integrals can look different, but they never differ by anything other than a constant.
Day 1: term by term, with fractional and negative powers
Find (a) \int \left(8x^3 - 6x^2 + 4x - 5\right)dx and (b) \int \left(4x^{1/3} - 6x^{-4}\right)dx.
Solution(a) The sum rule splits the integral into four, and the constant multiple rule keeps each coefficient outside:
Each of the four integrals brings its own constant, but their sum is one arbitrary constant, so one C is written. Check: the derivative is 8x^3 - 6x^2 + 4x - 5.
(b) For x^{1/3} the new exponent is \displaystyle \frac{4}{3}, and dividing by \displaystyle \frac{4}{3} is multiplying by \displaystyle \frac{3}{4}, so \displaystyle 4 \cdot \frac{3}{4}x^{4/3} = 3x^{4/3}. For x^{-4} the new exponent is -3, so -6 \cdot \dfrac{x^{-3}}{-3} = 2x^{-3}. Together,
on (0, \infty) and on (-\infty, 0) separately. Check: \displaystyle 3 \cdot \frac{4}{3}x^{1/3} + 2(-3)x^{-4} = 4x^{1/3} - 6x^{-4}.
Day 2: rewrite, then antidifferentiate
Find (a) \displaystyle \int \left(3\sqrt{x} - \dfrac{2}{\sqrt{x}}\right)dx, (b) \int \dfrac{2x^4 - 3x + 5}{x^3}\,dx and (c) \int (2t + 1)(t - 3)\,dt.
SolutionThe rules apply to sums of constant multiples of powers, so each integrand is first put in that form.
(a) Roots become exponents: \displaystyle \sqrt{x} = x^{1/2} and \displaystyle \dfrac{1}{\sqrt{x}} = x^{-1/2}. Then
for x > 0. Check: \displaystyle 2 \cdot \frac{3}{2}x^{1/2} - 4 \cdot \frac{1}{2}x^{-1/2} = 3\sqrt{x} - \dfrac{2}{\sqrt{x}}.
(b) The denominator is a single term, so each term of the numerator can be divided by it: for x \ne 0 the integrand is 2x - 3x^{-2} + 5x^{-3}. Then
on (0, \infty) and on (-\infty, 0) separately, each with its own C.
There is no quotient rule run backward: the antiderivative of the top over the antiderivative of the bottom is not an antiderivative of the quotient. Splitting works here only because the bottom is one term; a bottom with several terms calls for dividing first.
(c) Expand by the distributive property: (2t + 1)(t - 3) = 2t^2 - 5t - 3. Then
Antidifferentiating each factor and multiplying would give \displaystyle (t^2 + t)\left(\frac{1}{2}t^2 - 3t\right), a polynomial of degree 4 whose derivative has degree 3, not the degree 2 of the integrand. Products need methods of their own, substitution and integration by parts.
Day 2: the one curve through a point
Find the curve whose slope at every x is x^2 - 1 and which passes through (3, 5). In symbols: find f with f'(x) = x^2 - 1 for all x and f(3) = 5.
SolutionThe general antiderivative, by the power rule term by term, is
These are the curves of Figure 1. The condition f(3) = 5 picks one:
So f(x) = \dfrac{x^3}{3} - x - 1. Check: f'(x) = x^2 - 1 and f(3) = 9 - 3 - 1 = 5. No other answer exists: any function with this derivative is \displaystyle \frac{x^3}{3} - x + C for some C, and only C = -1 meets the condition.
Day 2: two antidifferentiations, two constants
Find f given f''(x) = 12x^2 - 6x, f(0) = 1 and f(2) = 13.
SolutionAntidifferentiate once to get f', then again to get f, keeping each constant:
The antiderivative of the constant C_1 is C_1 x, not another constant. Both conditions are on f, so both go into the second equation. At x = 0 every term but C_2 vanishes, so C_2 = 1. At x = 2,
So f(x) = x^4 - x^3 + 2x + 1. Check: f'(x) = 4x^3 - 3x^2 + 2, f''(x) = 12x^2 - 6x, f(0) = 1, f(2) = 16 - 8 + 4 + 1 = 13.
The general answer x^4 - x^3 + C_1 x + C_2 is one curve plus any line, a family with two free numbers, which is why one condition could not have settled it.
Day 2: Gwion’s kelp frond
Gwion grows a kelp frond in a tank. t days after his first measurement it grows at \displaystyle L'(t) = 12 + 3\sqrt{t} centimeters per day, for 0 \le t \le 16, and at t = 0 it is 50 centimeters long. (a) Find its length L(t). (b) How long is it on day 16? (c) How much does it grow between day 4 and day 9?
Solution(a) Write \displaystyle 3\sqrt{t} = 3t^{1/2} and antidifferentiate term by term:
The initial condition gives L(0) = 0 + 0 + C = 50, so C = 50 and L(t) = 12t + 2t^{3/2} + 50 centimeters. The constant is the starting length, measured in centimeters like L itself.
(b) L(16) = 192 + 2 \cdot 64 + 50 = 370 centimeters, that is 3.7 meters.
(c) The growth is a difference of lengths:
The 50 cancels. Any antiderivative of L' gives the same change, whatever its constant, so the change could have been found without the initial condition. Computing a change as a difference of antiderivative values is the idea behind evaluating a definite integral.
Boundaries
- The power rule excludes n = -1. The formula would divide by n + 1 = 0. The function \dfrac{1}{x} does have antiderivatives, \ln\lvert x \rvert + C, but they come from the derivative of the logarithm, in Antiderivatives of trig, exponential and log functions.
- “Differ by a constant” needs an interval. On a domain made of two intervals, such as x \ne 0 for x^{-2}, each piece may carry its own constant (Exercise 24). An answer like -\dfrac{1}{x} + C is read on each interval separately.
- The power must be of the variable itself. \int (3x + 1)^5\,dx is not \dfrac{(3x + 1)^6}{6} + C, whose derivative is 3(3x + 1)^5 by the chain rule. Expanding works; u-substitution is quicker.
- No product or quotient rule in reverse. The antiderivative of a product is not the product of antiderivatives (Example 4 and Exercise 23).
- Roots need their domain. \displaystyle \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} + C holds for x > 0, where both sides make sense.
- The initial point must lie in the interval on which the antiderivative is taken; a condition at x = 1 says nothing about x < 0 for an integrand undefined at 0.
- Existence is not a formula. Every function continuous on an interval has an antiderivative there, a fact of the fundamental theorem stated here without proof. That antiderivative need not have a formula built from familiar functions; e^{-x^2} has none, a theorem also stated here without proof.
Where this goes next
Antiderivatives of trig, exponential and log functions reads the derivatives of sine, cosine, the exponential and the logarithm backward, filling the gap at n = -1. u-substitution reverses the chain rule, which handles powers of expressions such as (3x + 1)^5. Gwion’s subtraction in Example 7 becomes The second fundamental theorem of calculus: the area under a rate equals the change in any antiderivative. The first fundamental theorem of calculus goes the other way, building an antiderivative out of area. And an equation such as \dfrac{dy}{dx} = x^2 - 1 is the simplest of the equations studied in Differential equations and their solutions, where initial value problems return.
Exercises
Day 1: the power rule and notation
- Find \int x^9\,dx.
- Find \int t^{-5}\,dt.
- Find \int x^{3/5}\,dx.
- Find \displaystyle \int \sqrt{2}\,dx.
- Decide whether F is an antiderivative of f: (a) F(x) = \dfrac{(x^2 + 3)^2}{4} and f(x) = x^3 + 3x; (b) F(x) = \dfrac{x^4}{4} - \dfrac{1}{x} and f(x) = x^3 - \dfrac{1}{x^2} for x > 0.
- Find the general antiderivative of g(w) = w^{-2/3} for w > 0.
Day 1: sums and constant multiples
- Find \int \left(6x^5 - 4x^3 + 2\right)dx.
- Find \int \left(9t^2 - 10t + 3\right)dt.
- Find \int \left(10x^{-6} - 3x^{1/2}\right)dx for x > 0.
- Find \int \left(\dfrac{3}{4}x^2 - \dfrac{1}{2}x\right)dx.
Day 2: rewriting first
- Find \displaystyle \int 5\sqrt[4]{x}\,dx.
- Find \displaystyle \int \dfrac{1}{x^2\sqrt{x}}\,dx.
- Find \int \dfrac{x^3 - 2x^2 + 4}{x^2}\,dx for x > 0.
- Find \int \left(x^2 - 2\right)^2 dx.
- Find \displaystyle \int \sqrt{x}\,(x - 6)\,dx.
Day 2: initial conditions
- Find f if f'(x) = 4x^3 - 2x and f(1) = 6.
- Find f if \displaystyle f'(x) = 6\sqrt{x} for x > 0 and f(4) = 20.
- Find the curve through (2, 1) whose slope at every x is 3x^2 - 4.
Day 2: second derivatives and rates
- Find f if f''(x) = 6x + 2, f'(0) = -1 and f(0) = 3.
- Find f if f''(x) = 12x^2, f(1) = 0 and f(-1) = 4.
- Macsen’s train is moving at 24 meters per second when it starts braking, and from then on its velocity changes at -1.5 meters per second per second. Find its velocity v(t) and its distance s(t) from the point where braking began, t seconds later. When does it stop, and how far has it gone?
- Efa’s frond grows at L'(t) = 20 - 0.8t centimeters per day for 0 \le t \le 25, and L(0) = 60 centimeters. Find L(t) and L(25), and say what happens to the growth rate at t = 25.
Reasoning
- Show with f(x) = x and g(x) = x^2 that \int f(x)g(x)\,dx is not the product of an antiderivative of f and an antiderivative of g.
- Let F(x) = -\dfrac{1}{x} + 2 for x > 0 and F(x) = -\dfrac{1}{x} - 5 for x < 0. Show that F'(x) = \dfrac{1}{x^2} at every point of its domain, and explain why F is not -\dfrac{1}{x} + C for a single constant C. Why does this not contradict the Results?
- Suppose F''(x) = G''(x) for every x in an interval I. Prove that F(x) - G(x) = mx + b on I for some constants m and b.
Answers
- \dfrac{x^{10}}{10} + C. Raise the exponent to 10 and divide by 10.
- -\dfrac{1}{4t^4} + C. The new exponent is -4: \dfrac{t^{-4}}{-4}, valid on t > 0 and on t < 0.
- \dfrac{5}{8}x^{8/5} + C. The new exponent is \displaystyle \frac{8}{5}, and dividing by it is multiplying by \displaystyle \frac{5}{8}.
- \displaystyle \sqrt{2}\,x + C. The integrand is a constant, the case n = 0.
- (a) Yes. By the chain rule F'(x) = \dfrac{2(x^2 + 3) \cdot 2x}{4} = x(x^2 + 3) = x^3 + 3x. (b) No. F'(x) = x^3 + \dfrac{1}{x^2}, since \dfrac{d}{dx}\left[-x^{-1}\right] = x^{-2}; an antiderivative of f is \dfrac{x^4}{4} + \dfrac{1}{x}.
- 3w^{1/3} + C. The new exponent is \displaystyle \frac{1}{3}: \dfrac{w^{1/3}}{1/3} = 3w^{1/3}.
- x^6 - x^4 + 2x + C. Term by term: \displaystyle 6 \cdot \frac{x^6}{6}, \displaystyle -4 \cdot \frac{x^4}{4}, 2x.
- 3t^3 - 5t^2 + 3t + C. Term by term: \displaystyle 9 \cdot \frac{t^3}{3}, \displaystyle -10 \cdot \frac{t^2}{2}, 3t.
- -2x^{-5} - 2x^{3/2} + C. 10 \cdot \dfrac{x^{-5}}{-5} = -2x^{-5} and -3 \cdot \dfrac{x^{3/2}}{3/2} = -2x^{3/2}.
- \dfrac{1}{4}x^3 - \dfrac{1}{4}x^2 + C. \displaystyle \frac{3}{4} \cdot \frac{x^3}{3} and \displaystyle -\frac{1}{2} \cdot \frac{x^2}{2}.
- 4x^{5/4} + C for x > 0. \displaystyle \sqrt[4]{x} = x^{1/4}, and 5 \cdot \dfrac{x^{5/4}}{5/4} = 4x^{5/4}.
- \displaystyle -\dfrac{2}{3x\sqrt{x}} + C for x > 0. The integrand is x^{-5/2}, and \displaystyle \dfrac{x^{-3/2}}{-3/2} = -\frac{2}{3}x^{-3/2}.
- \dfrac{x^2}{2} - 2x - \dfrac{4}{x} + C. Dividing each term by x^2 gives x - 2 + 4x^{-2}, and 4 \cdot \dfrac{x^{-1}}{-1} = -\dfrac{4}{x}.
- \dfrac{x^5}{5} - \dfrac{4x^3}{3} + 4x + C. Expand: (x^2 - 2)^2 = x^4 - 4x^2 + 4.
- \dfrac{2}{5}x^{5/2} - 4x^{3/2} + C for x > 0. Distribute: x^{3/2} - 6x^{1/2}, then \displaystyle \frac{2}{5}x^{5/2} and \displaystyle -6 \cdot \frac{2}{3}x^{3/2}.
- f(x) = x^4 - x^2 + 6. The family is x^4 - x^2 + C, and 1 - 1 + C = 6.
- f(x) = 4x^{3/2} - 12. The family is \displaystyle 6 \cdot \frac{2}{3}x^{3/2} + C = 4x^{3/2} + C; 4 \cdot 8 + C = 20 gives C = -12.
- y = x^3 - 4x + 1. The family is x^3 - 4x + C, and 8 - 8 + C = 1.
- f(x) = x^3 + x^2 - x + 3. f'(x) = 3x^2 + 2x + C_1 with C_1 = f'(0) = -1; then f(x) = x^3 + x^2 - x + C_2 with C_2 = f(0) = 3.
- f(x) = x^4 - 2x + 1. f(x) = x^4 + C_1 x + C_2, so 1 + C_1 + C_2 = 0 and 1 - C_1 + C_2 = 4. Adding gives C_2 = 1, and then C_1 = -2.
- v(t) = 24 - 1.5t and s(t) = 24t - 0.75t^2. The constants come from v(0) = 24 and s(0) = 0. It stops when 24 - 1.5t = 0, at t = 16 seconds, after s(16) = 384 - 192 = 192 meters.
- L(t) = 20t - 0.4t^2 + 60 and L(25) = 500 - 250 + 60 = 310 centimeters. The rate L'(25) = 20 - 20 = 0: the frond stops growing at t = 25, where the model ends.
- The product f(x)g(x) = x^3 has antiderivative \dfrac{x^4}{4} + C. The product of antiderivatives \dfrac{x^2}{2} \cdot \dfrac{x^3}{3} = \dfrac{x^5}{6} has derivative \dfrac{5x^4}{6}, not x^3, and adding constants to either factor cannot fix the degree: the derivative of the product of the two antiderivatives always has degree 4.
- On each piece F is -x^{-1} plus a constant, so F'(x) = x^{-2} = \dfrac{1}{x^2} at every x \ne 0. A single C would need C = 2 (from x > 0) and C = -5 (from x < 0). The Results speak of one interval, and the domain x \ne 0 is two intervals; the mean value theorem argument works on each one separately, giving a separate constant on each.
- Let h = F - G. Then h'' = F'' - G'' = 0 on I, so h' has derivative 0 and is a constant m on I. Then h and mx have the same derivative m on I, so they differ by a constant b: h(x) = mx + b. Both steps use the fact that equal derivatives on an interval force a constant difference.