Antiderivatives and substitution

A rate does not tell us where the motion started

During the first two seconds of a motion, suppose velocity is v(t)=(2 m/s²)t. A position s(t)=(1 m/s²)t²+5 m has that rate, but so do the positions with starting constants 0 m and −2 m. Each increases by 4 m between t=0 and t=2; their starting and final positions are different.

Here t has units seconds, s has units metres, and its derivative has units metres per second. The condition s(0)=5 m selects the first position and gives s(2)=9 m. Finding a function from its derivative is called finding an antiderivative. The extra starting value supplies information that the rate alone cannot contain.

The three curves beside this explanation differ by vertical shifts. Their equal endpoint differences explain the distinction between position and displacement. The formula establishes their rates; parallel-looking drawings alone do not prove that two derivatives agree.

Each curve has rate (2 m/s²)t and displacement 4 m over these two seconds. The initial position chooses the vertical shift.

Prepare the notation and check one proposed answer

Use the published preparation links for function inputs and domains, substitution into formulas, and powers and positive roots. We define the calculus facts needed here. No earlier calculus page is required to read this lesson.

An interval contains every real number between any two of its members. An open interval excludes its finite endpoints; the whole real line is also an open interval. A function assigns one output to each input in its stated domain. For h≠0, the quotient [F(x+h)−F(x)]/h is the average rate over an allowed change h.

The derivative F′(x) is a finite number L approached by that quotient for all sufficiently small positive and negative h. Precisely, for every ε>0 some δ>0 ensures that 0<|h|<δ implies the quotient differs from L by less than ε. The symbol |r| means the nonnegative size of r. On an open interval both directions are permitted near every input. Continuity at x means that F(x+h) tends to F(x) as h tends to zero. A finite derivative implies continuity: the quotient is bounded near zero, and multiplying it by h gives an output change tending to zero.

For F(x)=x²+5 the quotient is 2x+h, hence F′(x)=2x. The constant disappears from the difference. Start by checking this candidate and the first substitution; return to the constant proof, integral bridge and domain gaps after the notation becomes familiar.

An antiderivative is a function; the indefinite integral is a family

Let f be a real function on an open interval I. An antiderivative of f is a function F differentiable at every x in I with F′(x)=f(x). We write ∫f(x) dx=F(x)+C for all such functions when one F is known. The integrand is the expression f(x) inside the integral; dx identifies the integration variable x. The indefinite-integral symbol denotes this family, not one accumulated numerical value.

Why are these all the possibilities? If G′=F′ on I, set H=G−F. For any x₁<x₂ in I, the closed segment [x₁,x₂] stays in I. The mean value theorem says a function continuous on that segment and differentiable inside it has H(x₂)−H(x₁)=H′(c)(x₂−x₁) for some interior c. Differentiability implies continuity, and H′=0; therefore the difference is zero.

Thus G−F has one constant value throughout I. Conversely, differentiating F+C gives F′, since C does not depend on x. This proof needs the interval condition: it cannot bridge an input missing from the domain.

An initial value F(a)=b, with a in that interval, selects C=b−F₀(a) from a known primitive F₀. For f(x)=2x and F(0)=5, F₀=x² gives C=5. Check both requirements: the derivative must equal f throughout the interval, and the chosen value must satisfy the initial condition.

Build simple primitives and differentiate them back

For a nonnegative integer m, ∫xᵐ dx=x^(m+1)/(m+1)+C on the real line. The integer-power derivative follows by expanding (x+h)ⁿ−xⁿ: after division by h, the constant term is nx^(n−1) and every remaining term contains h. This proves the reverse check for n=m+1. For m=0, x⁰ here means the constant function 1, including its value at zero.

The sum and constant-multiple derivative rules follow by adding and scaling difference quotients. Consequently, if F′=f and G′=g on the same interval, (αF+βG)′=αf+βg. For example, a primitive of 3x²−4x+2 is x³−2x²+2x; one final C describes the family.

Negative integer powers obey the same rule away from zero, except m=−1. Differentiating a reciprocal follows by subtracting 1/F(x+h)−1/F(x) with a common denominator; its derivative is −F′(x)/F(x)² wherever F≠0. Applying this to integer powers proves the negative-power rule. At m=−1 the proposed denominator m+1 is zero, so that formula is unavailable.

For t>0, define ln t=∫_1^t (1/u) du, a continuous signed integral. The local integral argument below gives (ln t)′=1/t. Thus ∫(1/x) dx=ln|x|+C on either interval x>0 or x<0. On the negative interval, |x|=−x and differentiation gives (1/(−x))(−1)=1/x. Zero is excluded.

Always verify an expression on its entire claimed interval. A few matching sample values do not establish equality of derivatives. The explorer's polynomial cases compare every coefficient exactly; its small table illustrates that identity and does not replace it.

The two axes have separate stated scales. Use the exact table and proof for values; sampled curves are illustrations.

A missing input separates the constants

On the disconnected domain ℝ without zero, a primitive of 1/x can equal ln(−x)+C₋ for x<0 and ln x+C₊ for x>0. The two constants may be different. There is no interval inside the domain joining a negative input to a positive input, so the preceding mean value argument imposes no relation between them.

The expression ln|x|+C with one constant gives some primitives on that domain but not the entire family. If an initial value is supplied at x=1, it fixes C₊ and leaves C₋ free. State the domain before declaring that one condition determines the whole answer.

This gap also prevents a proper continuous-integral calculation across zero. The function 1/x is undefined there and unbounded near it. The proper Riemann integral used in this lesson does not apply on [−1,1]; subtracting two logarithm endpoint values cannot repair the missing hypothesis. No improper-integral or principal-value convention is being introduced.

A value on one half-line fixes only that side's constant. The mean value theorem cannot cross the missing input.

Connect primitives with continuous signed accumulation

For continuous f on a finite closed interval [a,b] with a<b, the definite integral is the common limit of sums Σf(ξᵢ)(xᵢ−xᵢ₋₁). A partition has a=x₀<⋯<xₙ=b, each tag ξᵢ lies in its closed cell, and its mesh is the largest cell width. The limit must be the same for all tags and all partitions as mesh tends to zero. Precisely, the common number I satisfies: for every ε>0 some δ>0 makes |S−I|<ε for every tagged sum S of mesh less than δ. This is an all-tag, all-partition requirement. It measures signed accumulation: negative values contribute negatively.

Continuity on a closed bounded interval gives uniform continuity. Here is the compactness step behind that claim: if no single distance works for a requested accuracy, there are pairs with distance tending to zero but output separation at least that accuracy. A convergent subsequence of the first inputs exists in [a,b]; the paired inputs approach the same limit, contradicting continuity there. Thus the oscillation within every sufficiently short cell is uniformly small. An upper sum uses the largest value in each closed cell times its width; a lower sum uses the smallest. They differ by at most that oscillation times b−a. All tagged sums lie between them. Common refinements compare any two partitions, so their shrinking gaps establish the same limit for all tags and all sufficiently small meshes.

Splitting a sum at an intermediate input and taking limits gives additivity. Also, if |f|≤M on a segment, the integral's absolute size is at most M times its length. Define ∫_b^a f=−∫_a^b f and ∫_a^a f=0. These conventions preserve additivity and signed direction.

Define A(x)=∫_c^x f(t) dt on an open interval where f is continuous. Additivity gives [A(x+h)−A(x)]/h=(1/h)∫_x^{x+h} f(t) dt. Subtract f(x). The integral bound gives a difference of size at most the largest |f(t)−f(x)| between x and x+h, which tends to zero by continuity. The argument works for negative h because both the integral orientation and h change sign. Therefore A′(x)=f(x).

If F is continuous on [a,b] and F′=f inside it, the mean value theorem supplies a tag in each cell for which F(xᵢ)−F(xᵢ₋₁)=f(ξᵢ)(xᵢ−xᵢ₋₁). Summing telescopes to F(b)−F(a). The all-tag integral limit then proves ∫_a^b f=F(b)−F(a). Endpoint continuity of F is required. We use continuous integrands here, so all the stated existence and limit conditions hold.

Substitution is a reverse chain rule

Suppose g is differentiable on an open interval I, its values lie in an open interval J, and F is differentiable on J with F′=f. The chain rule gives [F(g(x))]′=f(g(x))g′(x). Hence ∫f(g(x))g′(x) dx=F(g(x))+C on I. This conclusion does not require g to be one-to-one or g′ to be nonzero.

To see the chain rule locally, put k=g(x+h)−g(x). Differentiability of F gives F(g(x)+k)−F(g(x))=f(g(x))k+r(k), where r(0)=0 and r(k)/|k|→0 for k≠0. Differentiability of g makes k/h→g′(x), so this quotient is bounded near zero and k→0. Dividing the identity by h makes its remainder tend to zero; if k=0 it is already zero. The derivative is therefore the stated product.

The notation u=g(x), du=g′(x) dx records the linear response of u to an input differential dx. For a finite input change h, the actual change is generally g(x+h)−g(x), not exactly g′(x)h. In a substitution calculation, du is a compact record of the chain-rule factor, not an ordinary number with which to cancel a zero derivative.

For ∫2x(x²+1)² dx, choose u=x²+1. The complete factor 2x dx becomes du, leaving ∫u² du=u³/3+C. Return to x to get (x²+1)³/3+C. Its derivative is (1/3)·3(x²+1)²·2x, exactly the original integrand, including at x=0.

Keep every factor and finish in one variable

For ∫(3x−2)² dx, u=3x−2 has du=3 dx. Because 3 is a nonzero constant, the integrand becomes (1/3)u² du. Integrating gives u³/9, and back-substitution gives (3x−2)³/9+C. Using u³/3 instead would produce a derivative three times too large.

For ∫x(x²+1)² dx, the needed 2x dx is twice the given factor. Write x dx=(1/2)du, so the answer is (x²+1)³/6+C. This identity scales the whole differential factor; it never divides by x and remains valid when x=0.

For ∫(x²+1)² dx, choosing u=x²+1 does not supply the missing 2x. The proposed (x²+1)³/3 fails: at x=0 its derivative is 0 while the integrand is 1. The integral is nevertheless straightforward after expansion: x⁵/5+2x³/3+x+C. A failed substitution does not mean no antiderivative exists.

If x remains after a change of variable, express it consistently or choose another method. For example, with u=x−1 on x>1, x=u+1 and ∫x/√(x−1) dx becomes ∫(u+1)/√u du. The answer is (2/3)(x−1)^(3/2)+2√(x−1)+C on that interval, where u^(3/2) means u√u for positive u. Its derivative is (3/2)√u by the product and root derivative rules. Do not leave an x factor in an integral marked du.

Roots and logarithms keep their domain conditions

For z>0, rationalising [√(z+h)−√z]/h gives 1/[√(z+h)+√z]→1/(2√z). Therefore the derivative of √z is 1/(2√z), and a primitive of 1/√z is 2√z on z>0. The endpoint z=0 is excluded from this derivative formula.

For ∫x/√(x²+4) dx, use u=x²+4 and x dx=(1/2)du. Since u≥4 for every real x, the root denominator never vanishes. The result is √(x²+4)+C, whose derivative is x/√(x²+4). In particular ∫_0^1 x/√(x²+4) dx=√5−2.

For ∫2x/(x²+4) dx, the same positive u gives ∫(1/u) du=ln u+C, hence ln(x²+4)+C on the whole real line. For ∫1/(3x−2) dx, the constant factor instead gives (1/3)ln|3x−2|+C on an interval excluding x=2/3. Constants on the two sides may differ.

These are mathematical formulas in real-number variables. In a physical logarithm model, its argument must be dimensionless, such as a ratio to a stated reference unit. Integration constants carry the units of the recovered output, rather than the units of its rate.

The proposed primitive √(x²+4) differentiates to x/√(x²+4) on all real inputs. The domain check is part of the answer.

Change the bounds along with the variable

A sufficient definite-integral hypothesis is: g has a continuous derivative on an open neighbourhood of the closed interval between a and b, and f is continuous on an open interval J containing the entire image of that closed interval. Then ∫_a^b f(g(x))g′(x) dx=∫_{g(a)}^{g(b)} f(u) du. Both integrands are continuous on their relevant intervals. The formula also covers reversed and equal bounds with the preceding orientation convention.

Why does it hold? The continuous-integral bridge gives an antiderivative F of f on J. The chain rule gives derivative f(g)g′ for F∘g. Evaluating the original integral gives F(g(b))−F(g(a)); evaluating the transformed integral gives exactly the same difference. This proves the formula even when g turns around or has zero derivative.

For ∫_0^1 2x(x²+1)² dx, u=x²+1 changes the lower bound to 1 and upper bound to 2. Thus ∫_1^2 u² du=(8−1)/3=7/3. After changing variables, evaluate the u-primitive at u-bounds. Alternatively, return to the x-primitive and evaluate it at x-bounds. Mixing u³/3 with the original bounds 0 and 1 gives the wrong answer.

A definite integral produces a number, so no extra +C remains in its answer: the same constant cancels in an endpoint difference. Preserve the differential's units as well. A rate measured in metres per second multiplied by a time width in seconds contributes displacement in metres.

Above zero · Below zero

The two axes have separate stated scales. Use the exact table and proof for values; sampled curves are illustrations.

Decreasing bounds and a returning path keep their signs

For ∫_0^1 −2(1−2x)² dx, use u=1−2x and du=−2 dx. The bounds become 1 and −1, in that order. The value is ∫_1^{−1} u² du=[u³/3]_1^{−1}=−2/3. Sorting the transformed bounds without a compensating minus sign changes the integral.

For ∫_{−1}^1 2x(x²+1)² dx, u=x²+1 begins at 2, falls to 1 and returns to 2. The endpoint formula gives ∫_2^2 u² du=0. On the original interval the negative half contributes −7/3 and the positive half contributes 7/3. This cancellation is justified by the signed chain-rule formula; it does not say that the integrand or the geometric region is zero.

The unsigned area of that region is ∫_{−1}^1 |2x(x²+1)²| dx=14/3. It is a different question. Replacing g′ with |g′| would accumulate both directions positively, so it is not the signed substitution formula proved here. An endpoint-only picture is especially misleading for that unsigned question.

Use the explorer to predict derivative agreement and oriented bounds before revealing them. Its exact polynomial identities and fractions certify the retained finite cases. The general substitution proof supplies the all-input conclusion under its hypotheses; sampled curves alone cannot do so.

Above zero · Below zero

The two axes have separate stated scales. Use the exact table and proof for values; sampled curves are illustrations.

Worked examples

Try it yourself

1. Which of x²−2, x²+5 and x³ are antiderivatives of 2x on ℝ?

   Show the answer and reasoning

   x²−2 and x²+5; x³ is not.

   Their derivatives are 2x, 2x and 3x² respectively. The required derivative identity must hold at every real input.

2. For F′(x)=2x on ℝ and F(1)=4, find F.

   Show the answer and reasoning

   F(x)=x²+3.

   The family is x²+C. Substituting x=1 gives 1+C=4, hence C=3.

3. Find ∫(3x²−4x+2) dx and verify it.

   Show the answer and reasoning

   x³−2x²+2x+C.

   The derivatives of the three terms are 3x², −4x and 2. Their sum reproduces the integrand.

4. Find ∫x(x²+1)² dx. Is (x²+1)³/3+C correct?

   Show the answer and reasoning

   The answer is (x²+1)³/6+C; the proposed result is twice too large in its derivative.

   For u=x²+1, x dx=(1/2)du. Differentiating the answer gives (3/6)(x²+1)²·2x=x(x²+1)².

5. Does u=x²+1 turn ∫(x²+1)² dx directly into ∫u² du?

   Show the answer and reasoning

   No. The differential factor 2x is missing.

   The naive candidate differentiates to 2x(x²+1)². Expansion instead gives the valid primitive x⁵/5+2x³/3+x+C.

6. For ∫x/√(x−1) dx on x>1, after u=x−1 what replaces the remaining x?

   Show the answer and reasoning

   x=u+1, giving ∫(u+1)/√u du.

   The transformed integral must contain only the new variable. Here u>0 and du=dx; no branch ambiguity arises.

7. Find ∫2x/(x²+4) dx and state its real domain.

   Show the answer and reasoning

   ln(x²+4)+C on ℝ.

   The inner derivative is 2x and the inner value is always positive. Differentiating the logarithm gives the original fraction.

8. Does F(1)=0 uniquely determine a primitive of 1/x on all of ℝ without zero?

   Show the answer and reasoning

   No. It fixes the positive-side constant but not the negative-side constant.

   On each connected interval the family has one constant. There is no interval in the domain that joins the two sides of zero.

9. Evaluate ∫_0^1 2x(x²+1)² dx and name the transformed bounds.

   Show the answer and reasoning

   The u-bounds are 1 and 2, and the value is 7/3.

   Use u=x²+1 and evaluate u³/3 at 2 and 1. Evaluating at the old bounds would mix variables.

10. What is ∫_1^0 2x(x²+1)² dx?

    Show the answer and reasoning

    −7/3.

    Reversing the original bounds negates the integral; the transformed bounds become 2 and 1 in that order.

11. If g(a)=g(b), does the signed substitution formula always give zero under its stated continuity hypotheses?

    Show the answer and reasoning

    Yes, for the integrand f(g(x))g′(x). It does not force unsigned area to be zero.

    The transformed integral has equal bounds. The returning quadratic example has signed value 0 and unsigned area 14/3.

12. Does g′(0)=0 invalidate the substitution u=x²+1 for 2x(x²+1)²?

    Show the answer and reasoning

    No. The chain-rule identity remains valid at zero.

    The proof multiplies by g′ and never divides by it. Both the original integrand and the composite derivative are zero there.

13. Can ln|x| endpoint subtraction evaluate a proper integral of 1/x on [−1,1]?

    Show the answer and reasoning

    No. The integrand is undefined and unbounded near zero.

    The continuous proper-integral hypotheses fail. Matching logarithm endpoint values do not establish an integral across the pole.

14. For u=x²+1 at x=1 and finite change h=0.1, are the actual change Δu and differential du equal?

    Show the answer and reasoning

    No. Δu=0.21 while du=2x h=0.2.

    The exact change is 2xh+h². Differential notation records the linear term; it does not turn every finite change into an exact product.

15. A rate is measured in litres per minute. What units does its time integral have, and what units does an integration constant have?

    Show the answer and reasoning

    Both the accumulated quantity and the integration constant have units litres.

    Each rate times a time width contributes litres. Adding a constant selects a recovered amount, so it must carry the same output units.

What to carry forward

• An antiderivative differentiates to the original function throughout its stated interval; verify the answer backwards.
• A primitive family has one additive constant per interval. Domain gaps can leave independent constants.
• Substitution reverses the chain rule and retains the entire inner-derivative factor, including its sign and any constant multiplier.
• Write the transformed integrand in one variable. Differential notation records a rule; it does not justify cancelling a zero derivative.
• For continuous signed integrals, transform both bounds or substitute back before evaluating the original bounds.
• A decreasing or returning substitution preserves oriented accumulation. Unsigned area is a separate integral.
• State root, logarithm and pole exclusions and preserve input, rate and recovered-output units.