Definite integrals: the evaluation formula and signed area

2026-10-22 huynhvietkhanh Comments: 0

Apply the Fundamental Theorem of Calculus with its hypotheses, evaluate upper minus lower, and distinguish a signed integral from geometric area.

High schoolAntiderivatives & Integrals

Content updated: October 22, 2026

An indefinite integral represents a family of antiderivatives, whereas a definite integral is a number. The Newton–Leibniz evaluation formula connects them through the difference of endpoint values; its sign also depends on the orientation of the limits and the sign of the function.

Objectives: apply the formula with its hypotheses; evaluate the upper endpoint minus the lower; calculate area when a graph changes sign.

Prerequisites: basic antiderivatives, continuity, and function signs.

On this page: Idea · Method · Examples · Practice · Further learning

1. The definite integral and the evaluation formula

Let $a<b$ and let $f$ be continuous on $[a;b]$. For a partition $a=x_0<x_1<\cdots<x_n=b$, set $\Delta x_i=x_i-x_{i-1}$ and choose $\xi_i\in[x_{i-1};x_i]$. The definite integral is the limit of the Riemann sums $\sum_{i=1}^n f(\xi_i)\Delta x_i$ as the largest subinterval length tends to $0$. Each term records the signed contribution on a small interval; continuity ensures that the limit is independent of the sample points and partitions.

If $F$ is an antiderivative of $f$ on an open interval containing $[a;b]$, the Fundamental Theorem of Calculus, Part 2 (Newton–Leibniz evaluation formula), states

$$\int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a).$$

This lesson uses continuity as a convenient sufficient hypothesis for direct application; it does not assert that every discontinuous function is nonintegrable. The bracketed expression is $F(x)$, into which $b$ and $a$ are then substituted. Replacing the antiderivative by $F+C$ leaves the difference unchanged because the constants cancel.

By convention, $\int_a^a f(x)\,dx=0$ and $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$. An upper limit below the lower limit is therefore meaningful. Do not reverse their order without changing the sign.

2. From calculation to area

Find an antiderivative on a suitable interval, check it by differentiation, and evaluate at the limits. Integration is linear and additive over intervals: if $a<c<b$, the integral on $[a;b]$ is the sum of those on $[a;c]$ and $[c;b]$. These properties allow splitting expressions and intervals when needed.

For $a<b$, the geometric area bounded by the graph, the horizontal axis, and the lines $x=a,x=b$ is

$$S=\int_a^b|f(x)|\,dx.$$

If $f\ge0$ on the interval, $S$ is the integral of $f$; if $f\le0$, $S$ is the negative of that integral. When the function keeps one sign, write $S=|\int_a^b f(x)\,dx|$. With a sign change, signed contributions can cancel while geometric areas add; the absolute value of the entire integral is generally no longer the area.

If finitely many sign-change points are known, split there and add the areas of the pieces. Also check continuity on the integration interval; do not pass through a rational-function pole as though using an ordinary definite integral.

3. Examples and sign checks

Example 1. Cosine on a quarter-circle interval

$$\int_0^{\pi/2}\cos x\,dx=[\sin x]_0^{\pi/2}=1-0=1.$$

Cosine is nonnegative on the interval, so the result is also the area under the graph. Angles use radians, in agreement with the derivative formula.

Example 2. Integrating the square function

$$\int_0^1x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13.$$

The expression $x^2/2$ has derivative $x$, rather than $x^2$. Check that the chosen antiderivative differentiates to $x^2$ before evaluating the limits.

Additional Example 3. A zero integral with positive area

For $f(x)=x$ on $[-1;1]$, the integral is $[x^2/2]_{-1}^1=0$. Split the area at $0$: the left part is $1/2$ and the right is also $1/2$, so $S=1$. The absolute integral value $0$ therefore cannot replace area in this sign-changing example.

4. Practice exercises

Evaluate $\int_1^2(2x-1)\,dx$.
Evaluate $\int_1^0x^2\,dx$.
Calculate the integral of $f=-2$ on $[0;3]$ and the area between the graph and horizontal axis.

Show hints and answers

$[x^2-x]_1^2=(4-2)-(1-1)=2$.
Reversing the limits changes the sign, so the answer is $-1/3$.
The integral is $[-2x]_0^3=-6$, whereas the area is $6$, since the function is nonpositive on the interval.

5. Common mistakes

Do not append C to the numerical result of a definite integral. Do not subtract the upper endpoint value from the lower. Check an antiderivative by differentiation. Do not identify area with the absolute value of the whole integral when the function changes sign; retain continuity or examine the appropriate integral type separately.