1
The answer to F′ = f is a family
t², t² + 4 and t² − 7 all have derivative 2t, so all are antiderivatives of 2t. Their graphs are the same curve moved up or down. The lab draws five of them and the one member through a chosen point.
Lesson · Accumulation and the Integral · The exact total
The Constant of Integration
Every antiderivative of a rate differs from every other by a constant, so the answer to the question of which function has this rate is a family and not a function. A total change does not see the constant, but a value of the quantity itself does. One given value is a single equation for the single unknown constant, and it is exactly enough. This is the simplest differential equation with its initial value, which the next courses make harder.
F′ = f is solved by a family: F + C
two antiderivatives differ by a constant
y(t₀) = y₀ gives C = y₀ − F(t₀)
one value, one unknown: exactly enough
C is zero only if F(t₀) already equals y₀
A family of antiderivatives and the value that picks one
Core ideas
What this lesson establishes
Three ideas: the family, the reason it contains every antiderivative, and the arithmetic of picking one member.
2
The family has every antiderivative in it
If F and G both have derivative f, then F − G has derivative 0, and a polynomial with derivative 0 is a constant. So G = F + C for some C, and writing F + C leaves nothing out.
3
One given value fixes the constant
A value y(t₀) = y₀ says F(t₀) + C = y₀. That is one equation, with C appearing once and with coefficient 1, so C = y₀ − F(t₀) and there is exactly one solution.
The material
Why +C, and how one value removes it
The tank that gains water at a known rate, the proof that nothing is missing from the family, and the arithmetic of the constant.
A tank gains water at 2t litres per minute, t minutes after a tap opens. The rate tells you how much water arrives; it does not tell you how much was in the tank to begin with. A tank that held 4 litres at the start and one that held 100 gain water identically. The total change of the previous lesson never needed the starting amount. The amount in the tank at a given time does.
Definition · A family and an initial value
The antiderivatives of f are the family F + C, one for each constant C. An initial value is a given pair y(t₀) = y₀.
The pair of conditions y′ = f(t) and y(t₀) = y₀ is the simplest differential equation there is: a rule for the rate, and one value to start from. The rest of this Subject is that sentence with harder rules.
Take y′ = 2t with y(1) = 5. The family is y = t² + C. The given value says 1² + C = 5, so C = 4 and y = t² + 4. The lab draws the family as parallel copies of one curve and marks the member through the point (1, 5).
y = t² + C, at t = 1
C = −2 y(1) = −1
C = 0 y(1) = 1
C = 4 y(1) = 5 (the one given)
C = 10 y(1) = 11
every member has y′ = 2t
Why nothing is missing, and why one value is enough
Theorem · Two antiderivatives of a polynomial differ by a constant
If F′ = G′ for polynomials F and G, then F − G is a constant.
Let D = F − G. By the sum rule of “The Derivative as a Function”, D′ = F′ − G′ = 0.
Write D = c₀ + c₁t + c₂t² + … + c_k·tᵏ. Its derivative is c₁ + 2c₂·t + … + k·c_k·tᵏ⁻¹, and a polynomial is zero only when every coefficient is zero. So c₁ = 0, 2c₂ = 0, and so on to k·c_k = 0, which forces c₁ = c₂ = … = c_k = 0. Only c₀ is left, and D = c₀.
□
That proof is for polynomials, the only rates the lab accepts. For other functions the same statement is true and is not proved here. One value then does the rest. Without a value the family is every curve; with y(1) = 5 it is one. A second value either repeats the first or contradicts it: t² + 4 already gives y(2) = 8, and asking for y(2) = 9 as well would need C = 5 and C = 4 together.
Example · The constant is not zero because the value is zero
Take y′ = 3t² with y(2) = 0. The family is y = t³ + C, and 2³ + C = 0 gives C = −8, so y = t³ − 8.
The value 0 is the height of the curve at t = 2, not the constant. C is the difference between that height and F(2) = 8.
The constant is zero when the value agrees with F, and that can be arranged. The cubic preset is t³ − t with y(0) = 1/2. Here F = t⁴/4 − t²/2, and F(0) = 0, so C = 1/2 − 0 = 1/2 and y = t⁴/4 − t²/2 + 1/2. Starting at t = 0 makes C equal to the starting value itself, which is why a reader who always starts there can come to think C is simply the start.
Interactive
Antiderivatives
F, its constant, and F(b) − F(a)
Method
Fixing the constant from one value
The only new move is the third. Do the check at the end, because it catches both kinds of slip.
1
Find an antiderivative F of the rate
Reverse the power rule on each term and check by differentiating. Leave the constant out for now.
2
Write the family y = F(t) + C
Every solution of y′ = f is in it. Say so in the line, so that the C is not dropped by habit.
3
Put the given value into the family
Replace t by t₀ and y by y₀: F(t₀) + C = y₀. Evaluate F(t₀) as a fraction. The unknown is only C.
4
Solve for C
C = y₀ − F(t₀). Do not copy y₀ into C unless F(t₀) is zero.
5
Check both conditions
Differentiate the final y and compare with the rate; evaluate it at t₀ and compare with y₀. A solution passes both.
A family and the value that picks one
Score 0 / 0Choose the answer you can justify.
A tank with the rate 2t that holds 5 litres at t = 1
Find the amount y(t) in a tank that gains water at 2t litres per minute and holds 5 litres at t = 1.
y′ = 2t, y(1) = 5
F(t) = t² (check: F′ = 2t)
family: y = t² + C
value: 1² + C = 5
C = 5 − 1 = 4
y = t² + 4
check: y′ = 2t, y(1) = 1 + 4 = 5
the tank held y(0) = 4 at the start
One equation, one unknown. The same rate with a different value gives the same curve moved up or down, and the lab shows the move when the second number changes. The amount at t = 0 was never given; it came out of the formula as 4.
For a rehearsal, take the through-zero preset. The supplied first move is that F = t³. Put t = 2 into t³ + C = 0, solve, and check that your y has derivative 3t² and value 0 at t = 2.
Common mistakes
1
Taking the constant to be zero unless told otherwiseFor y′ = 2t with y(1) = 5 the constant is 4, and for y′ = 3t² with y(2) = 0 it is −8. A zero constant means the curve passes through the plain antiderivative, which happens only if F(t₀) is already y₀. The habit comes from the previous lesson, where the constant dropped out of the total and could be left out.
2
Copying the given value into the constantFrom y(1) = 5 it is tempting to write y = t² + 5. At t = 1 that gives 6. The value is the height of the curve, the constant is the height minus F(t₀), and the two agree only when F(t₀) = 0, as for t₀ = 0 with the polynomials here.
3
Dropping the constant because the total did not need itF(b) − F(a) cancelled the constant, so leaving it out cost nothing. A value of the quantity does not cancel it: the tank's amount at t = 3 is 13 with C = 4 and 9 with C = 0. The rate fixes how the quantity changes; only a value fixes where it is.
Check your understanding
✓
Finish when you can write the family of antiderivatives and fix its constant from one value.
You should be able to find F, write y = F(t) + C, solve F(t₀) + C = y₀ for C as an exact fraction, check the result against both the rate and the given value, and say why one value is exactly enough while none is too few and two are redundant or contradictory.
Note: Sums, antiderivatives and constants are now all in play, and they want a compact notation. “The Integral Sign and Its Rules” supplies it and checks two rules for it on exact fractions.