Integrals · Lesson 7 of 9

Definite Integrals and the Fundamental Theorem

“Connect accumulated area to antiderivatives and evaluate definite integrals with the correct limits.”

Learning Objectives

• Distinguish a definite integral from a family of antiderivatives. • Interpret an area function and state both Fundamental Theorems of Calculus. • Evaluate a definite integral as an upper-endpoint value minus a lower-endpoint value. • Apply domain and continuity checks before using the theorem. • Use substitution either by returning to the original variable or by changing both limits.

From a family of functions to one accumulated value

An indefinite integral asks for all functions with a specified derivative. A definite integral asks for an accumulated value between two specified endpoints. When a continuous graph lies above the horizontal axis, this value measures the region beneath the graph and between the endpoints. The notation records both the function being accumulated and the interval.

Definition

Definite integral

The notation $\int_{a}^{b} f (x) d x$ denotes the definite integral from lower limit $a$ to upper limit $b$ . For a continuous function and fixed limits it is a number. Geometrically, contributions above the horizontal axis are positive and those below it are negative.

One way to develop this value is to split an interval into many narrow strips, approximate each strip by a rectangle, and add the signed rectangle areas. As the strip widths approach zero, the sum approaches the definite integral when the limiting process applies. The Fundamental Theorem gives a powerful alternative: compute the same value using an antiderivative, without carrying out a long limiting sum.

A definite integral is therefore not always the total geometric area. If the graph crosses the axis, negative and positive contributions can cancel. To measure the total area in such a case, one integrates the absolute value or splits the interval and changes signs where necessary. The next lesson develops the interval-splitting tools.

The area function has a moving endpoint

Keep the starting point $a$ fixed and allow the right endpoint to move. The accumulated amount now changes with the endpoint and so becomes a function. To avoid confusing the moving endpoint with the internal integration variable, use $t$ inside the integral and $x$ for its upper limit.

Area or accumulation functionLaTeX

A (x) = \int_{a}^{x} f (t) d t

When

f

is positive,

A (x)

is the area from

a

x

beneath the curve. The variable

t

is internal to the integral;

x

controls how far the accumulation extends.

Imagine moving the endpoint slightly to the right. The new thin strip has width approximately the change in $x$ and height approximately $f (x)$ . Its area is approximately their product. Dividing by the change in $x$ suggests that the accumulation function changes at rate $f (x)$ . This gives the intuition behind the first theorem; a full proof is beyond the chapter's treatment.

First Fundamental TheoremLaTeX

A' (x) = f (x), A (x) = \int_{a}^{x} f (t) d t

f

is continuous on

[a, b]

, this holds at interior points. At endpoints, use the appropriate one-sided derivative interpretation when needed. The theorem connects accumulation to instantaneous rate.

Example — Differentiating an accumulation function

Problem
If $G (x) = \int_{0}^{x} t sin t d t$ , find $G' (x)$ and $G (0)$ .

1.The integrand $t sin t$ is continuous, so the first theorem applies directly.
2.Replace the internal variable by the endpoint: $G' (x) = x sin x$ . No integration by parts is needed to find this derivative.
3.At the starting point there is no interval to accumulate over, so $G (0) = 0$ .

The area function— Compare the fixed left boundary a with the moving endpoint x. The light region beyond x is not included in A(x).

Evaluating with any antiderivative

The first theorem says that the accumulation function is an antiderivative. Any other antiderivative differs from it by a constant, and that constant disappears when endpoint values are subtracted. This explains why definite integration can use any convenient member of the antiderivative family.

Second Fundamental TheoremLaTeX

\int_{a}^{b} f (x) d x = [F (x)]_{a}^{b} = F (b) - F (a), F' (x) = f (x)

For the standard theorem here,

f

is continuous on the whole interval. The notation

[F (x)]_{a}^{b}

means evaluate at the upper limit and subtract the value at the lower limit.

Indeed, $[F (b) + C] - [F (a) + C] = F (b) - F (a)$ . There is no arbitrary constant in the final numerical value. To use the theorem, first inspect the integrand throughout the interval, then find an antiderivative, and finally evaluate it at both limits. An expression being defined at the endpoints alone is insufficient.

Example — A polynomial definite integral

Problem
Evaluate $\int_{2}^{3} x^{2} d x$ .

1.The function is continuous on $[2, 3]$ . One antiderivative is $F (x) = x^{3} /3$ .
2.Evaluate in the required order: $F (3) - F (2) = 27/3 - 8/3$ .
3.The definite integral is $19/3$ . It is a number, with no added $C$ .

Example — Check the denominator over the whole interval

Problem
Evaluate $\int_{4}^{9} x / (30 - x^{3/2})^{2} d x$ .

1.On $[4, 9]$ , $x^{3/2}$ ranges from $8$ to $27$ , so $30 - x^{3/2}$ stays positive and never vanishes. The integrand is continuous.
2.Set $u = 30 - x^{3/2}$ . Then $d u = - \frac{3}{2} x d x$ , so an antiderivative is $2/ [3 (30 - x^{3/2})]$ .
3.Evaluate at $9$ and $4$ : $\frac{2}{3} (1/3 - 1/22) = \frac{2}{3} \cdot 19/66 = 19/99$ .

Example — Partial fractions inside a definite integral

Problem
Evaluate $\int_{1}^{2} x / [(x + 1) (x + 2)] d x$ .

1.The denominator is nonzero on $[1, 2]$ . Decompose $x / [(x + 1) (x + 2)] = - 1/ (x + 1) + 2/ (x + 2)$ .
2.One antiderivative is $F (x) = - ln (x + 1) + 2 ln (x + 2)$ .
3.Compute $F (2) - F (1) = - ln 3 + 2 ln 4 + ln 2 - 2 ln 3 = ln (32/27)$ .

Example — A trigonometric definite integral

Problem
Evaluate $\int_{0}^{π /4} sin^{3} (2 t) cos (2 t) d t$ .

1.Choose $u = sin 2 t$ so $d u = 2 cos 2 t d t$ . An antiderivative is $sin^{4} (2 t) /8$ .
2.At $t = π /4$ , this is $1/8$ ; at $t = 0$ , it is $0$ .
3.Therefore the definite integral is $1/8$ . The use of $t$ as the original integration variable changes no integration rule.

Common mistake

Do not apply the endpoint formula across a point where the integrand is undefined. For example, $x x^{2} - 1$ is not real for $- 1 < x < 1$ , so it cannot be treated as a real continuous integrand on $[- 2, 3]$ . Similarly, an antiderivative formula for $1/ x$ cannot justify evaluating across $x = 0$ .

Substitution with and without changed limits

There are two consistent ways to use substitution in a definite integral. You can integrate in the new variable, return to the original variable, and then use the original limits. Or you can transform both limits at the moment you change variables and stay in the new variable throughout. Mixing these routes is the common source of mistakes.

Substitution with transformed limitsLaTeX

\int_{a}^{b} f (g (x)) g' (x) d x = \int_{g (a)}^{g (b)} f (u) d u

Set

u = g (x)

and use the endpoint values of

u

. The transformed differential includes any sign, so decreasing substitutions may produce reversed limits. Assume the functions satisfy the continuity and differentiability conditions needed for the substitution.

Example — Two routes to the same value

Problem
Evaluate $\int_{- 1}^{1} 5 x^{4} x^{5} + 1 d x$ .

1.Choose $u = x^{5} + 1$ , giving $d u = 5 x^{4} d x$ . The new lower limit is $0$ and the new upper limit is $2$ .
2.With changed limits, evaluate $\int_{0}^{2} u^{1/2} d u = [2 u^{3/2} /3]_{0}^{2} = 42 /3$ .
3.With the other route, an antiderivative in $x$ is $2 (x^{5} + 1)^{3/2} /3$ . Evaluating it from $- 1$ to $1$ gives the same result.
4.The lower endpoint causes no difficulty here: the original integrand has a finite continuous value there, and the root is real over the entire interval.

Example — An inverse-tangent substitution with new bounds

Problem
Evaluate $\int_{0}^{1} tan^{- 1} x / (1 + x^{2}) d x$ .

1.Let $u = tan^{- 1} x$ , with $d u = d x / (1 + x^{2})$ . At $x = 0$ , $u = 0$ ; at $x = 1$ , $u = π /4$ .
2.The transformed integral is $\int_{0}^{π /4} u d u = [u^{2} /2]_{0}^{π /4}$ .
3.Its value is $π^{2} /32$ . Using limits $0$ and $1$ on the expression $u^{2} /2$ would answer a different question.

Common mistake

A definite integral with fixed endpoints has no arbitrary constant in its final value. During substitution, either return to the original variable before applying the original limits, or change both limits and remain in the new variable. Write a small endpoint table when there is any doubt.

Quick check

What is $\int_{1}^{2} 2 x d x$ ?

Quick check

If $A (x) = \int_{2}^{x} cos t d t$ , what is $A' (x)$ ?

Quick check

Why does $C$ disappear in definite integration?

Quick check

Under $u = x^{2} + 1$ , what are the new limits for an integral from $x = 0$ to $x = 2$ ?

Quick check

What must be checked before directly applying the standard continuous-function endpoint theorem?

Practice Problems

Evaluate $\int_{- 1}^{1} (x + 1) d x$ .
Evaluate $\int_{2}^{3} d x / x$ .
Evaluate $\int_{0}^{π /2} cos^{2} x d x$ .
Evaluate $\int_{0}^{1} x e^{x^{2}} d x$ using changed limits.
Evaluate $\int_{0}^{π /2} sin ϕ cos^{5} ϕ d ϕ$ and show the transformed endpoint values.
Evaluate $\int_{0}^{2} x x + 2 d x$ using $x + 2 = t^{2}$ .
If $H (x) = \int_{1}^{x} (1 + t^{2}) d t$ , find $H' (x)$ and $H (1)$ without first finding a formula for $H$ .
Explain why evaluating $ln ∣ x ∣$ at $- 1$ and $1$ does not establish $\int_{- 1}^{1} d x / x = 0$ .

Key Takeaways

A fixed-limit definite integral is a value, while an indefinite integral is a family of functions. An accumulation function has derivative equal to its continuous integrand. The endpoint rule is $F (b) - F (a)$ , and the arbitrary constant cancels. Check the entire interval before using the theorem. Substitution works with original-variable endpoint evaluation or with fully transformed limits.

Previous · Lesson 6

Integration by Parts and Radical Forms

Next · Lesson 8

Properties and Symmetry of Definite Integrals

From a family of functions to one accumulated value

The area function has a moving endpoint

Evaluating with any antiderivative

Substitution with and without changed limits

Quiz

Practice Problems

Key Takeaways