Unit 6 · Topic 6.7

6.7 The Fundamental Theorem of Calculus and Definite Integrals

The second part of the Fundamental Theorem of Calculus lets you compute a definite integral exactly: find an antiderivative F and calculate F(b) − F(a). It also says the integral of a rate of change gives the total change in the quantity.

Key terms

Fundamental Theorem of Calculus
antiderivative
F(b) − F(a)
net change
definite integral

Antiderivatives

An antiderivative of f is any function F with F′(x) = f(x). For example, x³ is an antiderivative of 3x², and so are x³ + 7 and x³ − 2. Any two antiderivatives of the same function on an interval differ by a constant.

From 6.4, if f is continuous on an interval containing a, then F(x) = ∫ₐˣ f(t) dt is one antiderivative of f. Every continuous function has one.

The Fundamental Theorem of Calculus (evaluation form)

If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then

∫ₐᵇ f(x) dx = F(b) − F(a).

The usual shorthand is F(x)|ₐᵇ or [F(x)]ₐᵇ, meaning “evaluate at b, then subtract the value at a.” Any antiderivative works, because the constant cancels: (F(b) + C) − (F(a) + C) = F(b) − F(a). So you don't need + C here.

Both conditions matter. If f isn't continuous on [a, b], such as 1/x² on [−1, 1], plugging into F(b) − F(a) can give a confident, wrong answer. (That integral is improper. BC students learn to handle these in 6.13.)

Net change: the same theorem in words

Read the theorem with F as a quantity and f = F′ as its rate: ∫ₐᵇ F′(x) dx = F(b) − F(a). The integral of a rate of change over an interval is the net change in the quantity. Rearranged, this is the formula you'll use constantly in applications:

F(b) = F(a) + ∫ₐᵇ F′(x) dx

In words: final amount = starting amount + accumulated change. This works whether or not you can find a formula for F, because a calculator can evaluate the integral numerically.

Antiderivatives you should know for this topic

You'll build the full list in 6.8. These are enough to start, along with one more: an antiderivative of 1/x is ln|x|. Before you integrate, rewrite roots and fractions as powers, so √x becomes x^(1/2) and 2/x² becomes 2x⁻².

When you evaluate, substitute the upper limit, substitute the lower limit, then subtract. Keep each value in its own brackets so a negative sign reaches every term.

f(x)	an antiderivative F(x)
xⁿ (n ≠ −1)	xⁿ⁺¹/(n + 1)
eˣ	eˣ
sin x	−cos x
cos x	sin x
sec² x	tan x
1/(1 + x²)	arctan x

Worked examples

Try each one yourself first, then open the solution.

Example 1
Evaluating with F(b) − F(a)
Evaluate ∫₁⁴ (3√x − 2/x²) dx.
Show the solutionHide the solution
1. Step 1: Rewrite with exponents: 3x^(1/2) − 2x^(−2).
2. Step 2: Antiderivative: 3 · x^(3/2)/(3/2) − 2 · x^(−1)/(−1) = 2x^(3/2) + 2/x.
3. Step 3: At x = 4: 2(8) + 2/4 = 16.5.
4. Step 4: At x = 1: 2(1) + 2 = 4.
5. Step 5: Subtract: 16.5 − 4 = 12.5.
Answer: 25/2 (that is, 12.5)
Example 2
Net change from a known value
F is differentiable, F(2) = 7 and ∫₂⁵ F′(x) dx = 4. Find F(5).
Show the solutionHide the solution
1. Step 1: Fundamental Theorem: ∫₂⁵ F′(x) dx = F(5) − F(2).
2. Step 2: So 4 = F(5) − 7, and F(5) = 11.
Answer: F(5) = 11
Example 3Calculator allowed
Calculator: amount at a later time
A tank holds 20 gallons at t = 0. Water flows in at a rate of A′(t) = 5 sin(t²/4) gallons per minute. How much water is in the tank at t = 3 minutes?
Show the solutionHide the solution
1. Step 1: There's no elementary antiderivative of sin(t²/4), so use the calculator's numerical integration.
2. Step 2: Set up: A(3) = A(0) + ∫₀³ A′(t) dt = 20 + ∫₀³ 5 sin(t²/4) dt.
3. Step 3: With the calculator in radian mode, ∫₀³ 5 sin(t²/4) dt ≈ 7.782.
4. Step 4: A(3) ≈ 20 + 7.782 = 27.782.
Answer: About 27.782 gallons

Common mistakes

Subtracting in the wrong order. It's F(top) − F(bottom).
Dropping parentheses when F(a) has several terms, so only the first term gets subtracted. Write [F(b)] − [F(a)] with brackets.
Applying F(b) − F(a) when f has a vertical asymptote inside [a, b].
Forgetting the starting amount in an applied question. The integral gives only the change.

On the exam

On calculator free-response questions, write the integral setup first, such as 20 + ∫₀³ 5 sin(t²/4) dt, then the decimal. Round to three decimal places, and keep full precision in intermediate steps.
Without a calculator, show the antiderivative and the substitution of both limits.

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