The Fundamental Theorem of Calculus and Definite Integrals: Study Guide (Calculus AB 6.7)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so integrals are written in a compact form. ∫ (a to b) f(x) dx means the definite integral of f(x) from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top.

[F(x)] (a to b) is shorthand for F(b) − F(a). On paper, write square brackets around F(x) with a at the bottom right and b at the top right. A vertical bar in place of the brackets means the same thing.

Antiderivatives

An antiderivative of a function f is a function F whose derivative is f. In symbols, F′(x) = f(x).

• x³ is an antiderivative of 3x², because d/dx (x³) = 3x².
• x³ + 7 is also an antiderivative of 3x², because the derivative of 7 is 0.
• −cos x is an antiderivative of sin x, because d/dx (−cos x) = sin x.

So a function has many antiderivatives. They differ only by a constant. (If F′ = G′ on an interval, then (F − G)′ = 0 there, and a function with zero derivative on an interval is constant. That is a consequence of the Mean Value Theorem, Topic 5.1.)

You find antiderivatives by running derivative rules backwards. You know these derivative facts, so you already know these antiderivatives:

Function f(x)	One antiderivative F(x)	Check: F′(x)
xⁿ, for n ≠ −1	xⁿ⁺¹/(n + 1)	xⁿ
1/x, for x > 0	ln x	1/x
eˣ	eˣ	eˣ
cos x	sin x	cos x
sin x	−cos x	sin x
sec²x	tan x	sec²x

Constant multiples and sums work term by term: an antiderivative of 5x⁴ − 2 is x⁵ − 2x. Topic 6.8 builds the full table and the notation for families of antiderivatives. For this topic, these rules are enough.

Always check by differentiating. If F′(x) is not exactly f(x), F is not an antiderivative of f. This check takes seconds and catches most sign errors.

Where the theorem comes from

Topic 6.4 gave you one antiderivative of any continuous function. If f is continuous on an interval containing a, the accumulation function

A(x) = ∫ (a to x) f(t) dt

has A′(x) = f(x). So A is an antiderivative of f, even when no formula for it is known.

Now let F be any other antiderivative of f that you can write down. Because two antiderivatives differ by a constant, F(x) = A(x) + C for some number C. Then

F(b) − F(a) = (A(b) + C) − (A(a) + C) = A(b) − A(a).

A(b) is ∫ (a to b) f(t) dt, and A(a) = ∫ (a to a) f(t) dt = 0. So F(b) − F(a) = ∫ (a to b) f(x) dx. The constant C cancelled, which is why any antiderivative works.

The Fundamental Theorem of Calculus (evaluation form). If f is continuous on [a, b] and F is an antiderivative of f, then ∫ (a to b) f(x) dx = F(b) − F(a).

This is remarkable. The left side is a limit of Riemann sums: infinitely many thinner and thinner rectangles. The right side needs two values of one function.

Picturing it: area under f equals rise in F

Figure 1 checks the theorem with geometry. The trapezoid has area 7.5, and F(3) − F(0) = (4.5 + 3) − 0 = 7.5. Read it as: f is the rate at which F changes, so adding up f over [0, 3] gives the total change in F.

How to evaluate a definite integral

1. Check continuity. Is f continuous on the whole closed interval [a, b]? Look for zero denominators, logs of 0 and points where tan or sec is undefined.
2. Rewrite if needed. Roots become powers (√x = x^(1/2)); fractions with a single term on the bottom become negative powers (2/x² = 2x⁻²).
3. Find an antiderivative F, term by term. No + C is needed.
4. Write it in brackets: [F(x)] (a to b).
5. Substitute and subtract, putting each value in its own brackets: F(b) − F(a). Upper limit first.

Write the working as one chain of equal expressions: ∫ (a to b) f(x) dx = [F(x)] (a to b) = F(b) − F(a) = number. Every “=” must join things that really are equal.

Worked example 1: powers and roots

Question. Evaluate ∫ (1 to 4) (3√x − 2/x²) dx exactly.

1. Continuity. √x is continuous for x ≥ 0 and 2/x² is continuous for x ≠ 0. Both are continuous on [1, 4], so the theorem applies.
2. Rewrite as powers: 3x^(1/2) − 2x⁻².
3. Antidifferentiate each term.
   • 3x^(1/2) → 3 · x^(3/2)/(3/2) = 2x^(3/2)
   • −2x⁻² → −2 · x⁻¹/(−1) = 2x⁻¹ = 2/x So F(x) = 2x^(3/2) + 2/x.
4. Check: F′(x) = 3x^(1/2) − 2x⁻², which is the integrand.
5. Evaluate. ∫ (1 to 4) (3√x − 2/x²) dx = [2x^(3/2) + 2/x] (1 to 4) = (2 · 8 + 2/4) − (2 · 1 + 2/1) = 16.5 − 4 = 25/2

Notes. 4^(3/2) = (√4)³ = 8. The bracket around F(1) matters: forgetting it gives 16.5 − 2 + 2 = 16.5, which is wrong.

Worked example 2: trig, exponential and log integrands

Question. Evaluate exactly:

(a) ∫ (0 to π/2) (cos x − 2 sin x) dx (b) ∫ (0 to ln 3) eˣ dx (c) ∫ (1 to e³) (2/x) dx

(a) The integrand is continuous everywhere. An antiderivative of cos x is sin x. An antiderivative of −2 sin x is 2 cos x, since d/dx (2 cos x) = −2 sin x.

∫ (0 to π/2) (cos x − 2 sin x) dx = [sin x + 2 cos x] (0 to π/2) = (1 + 0) − (0 + 2) = −1

A negative answer is allowed. It means the integrand is negative over more of the interval than it is positive, in the signed-area sense. Here cos x − 2 sin x is positive only until tan x = 1/2 (about x = 0.464). The region above the axis has area about 0.236 and the region below has area about 1.236. Their signed sum is −1.

(b) eˣ is its own antiderivative and is continuous everywhere.

∫ (0 to ln 3) eˣ dx = [eˣ] (0 to ln 3) = e^(ln 3) − e⁰ = 3 − 1 = 2

(c) 2/x is continuous on [1, e³] because 0 is not in the interval. An antiderivative for x > 0 is 2 ln x.

∫ (1 to e³) (2/x) dx = [2 ln x] (1 to e³) = 2 · 3 − 2 · 0 = 6

When the theorem does not apply

The condition “f is continuous on [a, b]” is not decoration. Look at ∫ (−1 to 2) (3/x⁴) dx. An antiderivative of 3x⁻⁴ is −x⁻³ = −1/x³, so a careless calculation gives

[−1/x³] (−1 to 2) = (−1/8) − (1) = −9/8.

That cannot be right. The integrand 3/x⁴ is positive wherever it is defined, so a signed area under it cannot be negative. The mistake is that 3/x⁴ is not continuous on [−1, 2]: it is undefined at x = 0 and grows without bound there. The theorem’s condition fails, so F(b) − F(a) means nothing here.

Rule: before you subtract, scan the interval for points where the integrand is undefined. If there is one, stop. (BC students meet integrals like this again as improper integrals in Topic 6.13.)

The theorem in context: net change

Apply the theorem to a function F and its derivative F′. F is an antiderivative of F′, so if F′ is continuous on [a, b],

∫ (a to b) F′(x) dx = F(b) − F(a)

Integrating a rate of change gives the net change in the amount. Rearranged, F(b) = F(a) + ∫ (a to b) F′(x) dx: the final amount is the starting amount plus the net change.

Worked example 3: net change from a rate

Question. A drone’s height above the ground is H(t) metres, t seconds after a timer starts. Its vertical velocity is v(t) = 3t² − 12t + 9 metres per second for 0 ≤ t ≤ 4. At t = 0 the drone is 20 metres up. (The data are invented.)

(a) Find ∫ (0 to 4) v(t) dt and say what it means. (b) Find H(4).

(a) v is a polynomial, so it is continuous. An antiderivative is t³ − 6t² + 9t.

∫ (0 to 4) v(t) dt = [t³ − 6t² + 9t] (0 to 4) = (64 − 96 + 36) − 0 = 4

Since v = H′, this is H(4) − H(0). Between t = 0 and t = 4 seconds, the drone’s height increases by a net 4 metres.

(b) H(4) = H(0) + ∫ (0 to 4) v(t) dt = 20 + 4 = 24 metres.

Interpretation. v(t) = 3(t − 1)(t − 3), so the drone rises for 0 < t < 1, falls for 1 < t < 3 and rises again for 3 < t < 4. Over those pieces the height changes by +4, −4 and +4 metres. The integral gives the net change, +4, not the total distance moved, which is 12 metres. Total distance is a Unit 8 idea.

Reading a limit of Riemann sums as an integral

Topic 6.3 showed that a limit of Riemann sums is a definite integral. Now you can evaluate it. Consider

lim (n → ∞) Σ (i = 1 to n) (3/n)(1 + 3i/n)²

Match the parts: Δx = 3/n, so the interval has length 3. The sample point is xᵢ = 1 + 3i/n, which starts near 1 and ends at 1 + 3 = 4. The function is f(x) = x². So the limit is

∫ (1 to 4) x² dx = [x³/3] (1 to 4) = 64/3 − 1/3 = 21

Without the theorem, you would need a formula for Σ i² and a long limit calculation.

Common misconceptions

• Subtracting in the wrong order. It is F(upper) − F(lower). Reversing gives the negative of the answer.
• Dropping brackets. F(b) − F(a) must subtract every term of F(a). Put F(a) in brackets.
• Differentiating instead. Writing f(b) − f(a), or f′, instead of F(b) − F(a).
• “I must add + C.” You can, but it cancels. It is not wrong, just unnecessary.
• Ignoring continuity. Using the theorem across a point where the integrand is undefined, as in ∫ (−1 to 2) (3/x⁴) dx, gives nonsense.
• “A definite integral is always positive.” It is a signed quantity. A negative answer is fine when more of the region is below the axis.
• Confusing net change with total change. ∫ (a to b) v(t) dt is net change in position, not distance travelled.
• Mixing up the two forms of the theorem. d/dx ∫ (a to x) f(t) dt = f(x) gives a function. ∫ (a to b) f(x) dx = F(b) − F(a) gives a number.
• Antiderivative sign slips. The antiderivative of sin x is −cos x, not cos x. Differentiate your F to check.

Where this leads

You can now evaluate definite integrals exactly whenever you can find an antiderivative. In Topic 6.8, Finding Antiderivatives and Indefinite Integrals, you will build the full table of basic antiderivatives and the notation ∫ f(x) dx = F(x) + C. Topics 6.9 onwards add techniques, starting with substitution, for integrands the basic rules cannot handle. Look back at Topic 6.4 for the accumulation form of the theorem, and at Topic 6.6 for the properties used alongside it. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.