Application of Integrals · Lesson 2 of 3

Areas When a Curve Crosses an Axis

“Separate signed accumulation from geometric area and learn when to split an integral.”

Learning Objectives

• Distinguish the signed value of a definite integral from nonnegative geometric area. • Locate zeros and determine the sign of a curve on each part of an interval. • Calculate areas below an axis and across several axis crossings. • Use piecewise formulas for absolute-value graphs. • Check polynomial and trigonometric areas through symmetry and simple geometry.

Why an integral can be negative

A region cannot have negative geometric area. A definite integral, however, adds signed contributions: a positive function contributes positively and a negative function contributes negatively when the bounds run from left to right. The distinction matters as soon as a graph lies below the horizontal axis.

At a point below the axis, the coordinate $f (x)$ is negative, but the physical distance from the graph up to the axis is $0 - f (x) = - f (x)$ . That positive distance is the height of the geometric strip. Integration has not failed; we simply have to integrate a length rather than a signed coordinate.

Definition

Signed integral and geometric area

The definite integral records positive and negative contributions according to the sign of the function. The geometric area between a graph and an axis adds nonnegative strip areas. They agree directly on an interval where the graph lies on or above the axis, but need not agree elsewhere.

A region entirely below the axisLaTeX

f (x) \leq 0 on [a, b] ⟹ A = - \int_{a}^{b} f (x) d x = \int_{a}^{b} f (x) d x

Here

a < b

. The single absolute value is sufficient because all contributions to this integral have the same sign.

Example — Making a below-axis height positive

Problem
Find the area between $y = - x^{2} - 1$ , the horizontal axis, and the lines $x = 0$ and $x = 1$ .

1.The graph is below the axis on the whole interval because $- x^{2} - 1 < 0$ . The strip height is $x^{2} + 1$ , not $- x^{2} - 1$ .
2.Reverse the sign of the integrand. This gives $A = \int_{0}^{1} (x^{2} + 1) d x$ , equal to the negative of the original signed integral.
3.Evaluating gives $A = [x^{3} /3 + x]_{0}^{1} = 1/3 + 1 = 4/3$ square units. The signed integral of the original function is $- 4/3$ , which describes signed accumulation rather than geometric area.

Splitting a region where its sign changes

When a graph crosses the horizontal axis, one integral over the entire interval mixes positive and negative contributions. Taking an absolute value after those contributions have cancelled cannot restore the lost area. Separate the pieces first, then add their positive sizes.

Suppose a continuous function is negative from $a$ to a zero $c$ , then positive from $c$ to $b$ . The total region consists of a below-axis piece and an above-axis piece. Their areas are added even though their signed integrals have opposite signs.

Two pieces on opposite sides of the axisLaTeX

A = - \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x if f \leq 0 on [a, c], f \geq 0 on [c, b]

The minus sign turns the negative first integral into a positive area. If the order of the signs is reversed, the sign corrections must also be reversed.

Below the axis and across the axis— Use the distance to the axis for every strip, and add the pieces rather than cancelling them.

More generally, list the zeros inside the required interval and check the sign on each intervening piece. You can test one interior point when the function is continuous and has no zeros within that piece. A function cannot change from positive to negative there without passing through zero. A zero is a candidate dividing point; if the curve only touches the axis and keeps its sign, splitting there is optional.

Total area written with an absolute valueLaTeX

A = \int_{a}^{b} ∣ f (x) ∣ d x = i \sum \int_{c_{i}}^{c_{i + 1}} f (x) d x

The partition points

c_{i}

include the interval endpoints and enough intervening zeros that

f

keeps one sign on each piece. This is a compact statement of adding positive strip areas, not a rule to ignore the sign analysis.

The position of the absolute value matters

In general, $\int_{a}^{b} f (x) d x$ is not equal to $\int_{a}^{b} ∣ f (x) ∣ d x$ . The first takes the magnitude after cancellation. The second changes each negative strip into a positive length before adding. They agree when the function keeps one sign, but can differ greatly when it changes sign.

Example — A line that cuts the axis

Problem
Find the area between $y = 3 x + 2$ , the horizontal axis, and the lines $x = - 1$ and $x = 1$ .

1.Find the intercept: $3 x + 2 = 0$ gives $x = - 2/3$ . The line is negative on $[- 1, - 2/3)$ and positive on $(- 2/3, 1]$ .
2.Let $F (x) = 3 x^{2} /2 + 2 x$ . Its endpoint values are $F (- 1) = - 1/2$ , $F (- 2/3) = - 2/3$ , and $F (1) = 7/2$ .
3.The first signed contribution is $F (- 2/3) - F (- 1) = - 2/3 + 1/2 = - 1/6$ . Its geometric area is $1/6$ .
4.The second contribution is $F (1) - F (- 2/3) = 7/2 + 2/3 = 25/6$ . It is already positive.
5.Add the areas: $A = 1/6 + 25/6 = 13/3$ square units. By contrast, the unsplit signed integral is $- 1/6 + 25/6 = 4$ .
6.The two pieces are triangles. Their base-height calculations give $(1/3) \times 1/2 = 1/6$ and $(5/3) \times 5/2 = 25/6$ , confirming the result geometrically.

A line crossing between the bounds— The smaller left triangle lies below the axis, but its area is still added.

Trigonometric curves with several crossings

An interval can contain more than one sign change. A trigonometric graph makes the pattern easy to see, but its exact zeros must still be used as integration bounds. Work in radians so that the usual sine and cosine antiderivatives apply.

Example — Three pieces of a cosine curve

Problem
Find the area between $y = cos x$ and the horizontal axis from $x = 0$ to $x = 2 π$ .

1.The zeros inside the interval are $π /2$ and $3 π /2$ . Cosine is positive on the first piece, negative on the middle piece, and positive on the last piece.
2.Add $\int_{0}^{π /2} cos x d x$ , subtract $\int_{π /2}^{3 π /2} cos x d x$ , and then add $\int_{3 π /2}^{2 π} cos x d x$ . Subtraction turns the middle negative contribution into a positive area.
3.Using the antiderivative $sin x$ , the signed integrals are $1 - 0 = 1$ , $- 1 - 1 = - 2$ , and $0 - (- 1) = 1$ .
4.Their positive sizes add to $A = 1 + 2 + 1 = 4$ square units. The signed integral over the whole period is $1 - 2 + 1 = 0$ , so its magnitude would give the wrong area.

Area across a full cosine period— The orange middle region contributes negatively to the integral and positively to the area.

Example — A sine curve over one period

Problem
Find the area between $y = sin x$ and the horizontal axis on $[0, 2 π]$ .

1.The interior zero is $x = π$ . Sine is positive on $(0, π)$ and negative on $(π, 2 π)$ .
2.Start with $\int_{0}^{π} sin x d x$ and subtract $\int_{π}^{2 π} sin x d x$ . An antiderivative is $- cos x$ .
3.The first signed integral is $- cos π + cos 0 = 2$ . The second is $- cos 2 π + cos π = - 2$ .
4.Therefore $A = 2 + 2 = 4$ square units. The equal lobe areas offer a symmetry check, but their equality does not mean the total geometric area is zero.

Absolute values and piecewise formulas

An absolute value can change an algebraic formula without making the resulting graph negative. The safe approach is to locate where the expression inside the absolute value changes sign, then write the correct formula on each side. Do not remove the absolute-value bars until that sign decision has been made.

Example — A graph with a corner above the axis

Problem
Sketch $y = ∣ x + 3∣$ and evaluate $\int_{- 6}^{0} ∣ x + 3∣ d x$ .

1.The expression inside the bars is zero at $x = - 3$ . For $- 6 \leq x \leq - 3$ , it is nonpositive, so $∣ x + 3∣ = - x - 3$ . For $- 3 \leq x \leq 0$ , use $∣ x + 3∣ = x + 3$ .
2.Thus the graph is a V with vertex $(- 3, 0)$ and endpoint heights $3$ . Both arms lie on or above the horizontal axis.
3.Split at the formula change. Calculate $\int_{- 6}^{- 3} (- x - 3) d x$ and $\int_{- 3}^{0} (x + 3) d x$ , then add their values.
4.The first integral is $[- x^{2} /2 - 3 x]_{- 6}^{- 3} = 9/2$ . The second is $[x^{2} /2 + 3 x]_{- 3}^{0} = 9/2$ .
5.The total is $9$ . Geometrically, these are two triangles of base $3$ and height $3$ , so the same answer is $2 (3 \times 3/2) = 9$ square units.

An absolute-value corner— The formula changes at the vertex even though both pieces have nonnegative height.

This split differs slightly from the cosine split. For the V-shaped graph, the function itself stays nonnegative, but the formula inside its absolute value changes sign. For the cosine graph, it is the function value that becomes negative. Both require careful partitions, for different reasons.

Polynomial regions and cancellation

A polynomial can cross an axis without having a corner. Sign checking still matters, and symmetry can be used only after deciding whether the question asks for a signed integral or an area. Two examples show how very different answers can arise from the same bounds.

Example — Unequal regions of a cubic

Problem
Find the area between $y = x^{3}$ and the horizontal axis from $x = - 2$ to $x = 1$ .

1.The graph crosses the axis at zero, is negative on $[- 2, 0)$ , and is positive on $(0, 1]$ .
2.The antiderivative is $x^{4} /4$ . The left signed integral is $[x^{4} /4]_{- 2}^{0} = 0 - 4 = - 4$ , so the left area is $4$ .
3.The right area is $[x^{4} /4]_{0}^{1} = 1/4$ . Adding gives $A = 4 + 1/4 = 17/4$ square units.
4.The whole signed integral is $1/4 - 4 = - 15/4$ . Taking its magnitude gives $15/4$ , still smaller than the actual area because the smaller positive contribution has cancelled some of the negative contribution.

Example — A symmetric graph with zero signed integral

Problem
Find the area between $y = x ∣ x ∣$ and the horizontal axis from $x = - 1$ to $x = 1$ .

1.For $x < 0$ , $∣ x ∣ = - x$ , so $x ∣ x ∣ = - x^{2}$ . For $x \geq 0$ , $x ∣ x ∣ = x^{2}$ . The function is therefore negative on the left and positive on the right.
2.For the geometric area, take the negative of $\int_{- 1}^{0} (- x^{2}) d x$ and add $\int_{0}^{1} x^{2} d x$ .
3.The left signed integral is $- 1/3$ and the right signed integral is $1/3$ . Their areas add to $1/3 + 1/3 = 2/3$ square units.
4.The graph is symmetric under a half-turn about the origin, so the signed contributions cancel. Reflecting the below-axis piece upward, or integrating $∣ x ∣^{2} = x^{2}$ from $- 1$ to $1$ , shows why the area does not cancel.

Equal lobes with opposite signs— Symmetry explains both the cancellation of the integral and the doubling of one lobe’s area.

Example — Touching an axis does not force a sign correction

Problem
Must the area under $y = x^{2}$ on $[- 1, 1]$ be split at zero?

1.The curve reaches zero at the origin but is nonnegative on both sides. There is no below-axis piece.
2.A single integral gives $A = \int_{- 1}^{1} x^{2} d x = [x^{3} /3]_{- 1}^{1} = 2/3$ square units.
3.Splitting into two integrals is allowed, and symmetry gives $2 \int_{0}^{1} x^{2} d x$ . It is not necessary for a sign correction. A zero tells us to inspect the sign, not automatically to attach a minus sign.

Check your understanding

These questions ask you to distinguish the region, its signed integral, and the formula used on each interval. A correct numerical antiderivative cannot repair an incorrect sign decision.

Quick check

If a continuous function is negative on $[a, b]$ with $a < b$ , the geometric area is:

Quick check

Where must the line $y = 3 x + 2$ on $[- 1, 1]$ be split to correct its sign?

Quick check

The area between $y = cos x$ and the axis on $[0, 2 π]$ is:

Quick check

The value of $\int_{- 6}^{0} ∣ x + 3∣ d x$ is:

Quick check

For $y = x^{3}$ on $[- 2, 1]$ , the geometric area is:

Quick check

For $y = x ∣ x ∣$ on $[- 1, 1]$ , which pair gives the signed integral followed by the geometric area?

Quick check

A continuous curve touches the axis at an interior point but remains nonnegative. Which statement is correct?

Quick check

Which expression always gives the geometric area for a continuous graph on $[a, b]$ , with $a < b$ ?

Practice Problems

Find the signed integral and geometric area for $y = - 2 x$ on $[0, 3]$ .
Sketch $y = x - 1$ on $[0, 3]$ . Find its intercept, both signed pieces, and the total area.
Find the area between $y = sin x$ and the axis on $[0, π]$ . Compare it with the area on $[0, 2 π]$ .
Find the signed integral and geometric area for $y = cos x$ on $[0, π]$ .
Write $∣ x - 2∣$ as two formulas and calculate $\int_{0}^{4} ∣ x - 2∣ d x$ . Check using triangles.
Find the area between $y = x^{3}$ and the axis on $[- 1, 2]$ . Explain why it equals the area on $[- 2, 1]$ .
For $y = x ∣ x ∣$ on $[- 2, 2]$ , calculate the signed integral and the geometric area separately.
A student claims the area below $y = cos x$ on $[π /2, 3 π /2]$ is $- 2$ . Explain the sign error and correct it.
Does a graph crossing zero always require changing its algebraic formula? Use $x^{3}$ to explain the distinction between a formula change and an area sign correction.
Find the area below $y = x^{2}$ on $[- 2, 2]$ and explain why the zero at the origin does not create cancellation.

The signed integral and area in the first question are $- 9$ and $9$ . For the line, the signed pieces are $- 1/2$ and $2$ , giving area $5/2$ . The sine areas are $2$ and $4$ . The cosine question gives integral $0$ and area $2$ . The absolute-value integral is $4$ . The cubic area is $17/4$ . The symmetric $x ∣ x ∣$ question gives integral $0$ and area $16/3$ . The below-axis cosine area is $2$ . The final parabola area is $16/3$ square units.

Key Takeaways

• An integral adds signed contributions; geometric area adds nonnegative strip sizes. • On a below-axis piece, reverse the sign of the integral before adding its area. • For a continuous graph, find its zeros and check the signs between them. Split at sign changes. • The universal area expression is $A = \int_{a}^{b} ∣ f (x) ∣ d x$ , with $a < b$ . • The magnitude of one unsplit integral can miss area after cancellation. • Absolute-value formulas and symmetry are useful only when their signs and regions have been understood.

Previous · Lesson 1

Finding Area with Strips and Symmetry

Next · Lesson 3

Chapter Summary and Practice

Why an integral can be negative

Splitting a region where its sign changes

Trigonometric curves with several crossings

Absolute values and piecewise formulas

Polynomial regions and cancellation

Check your understanding

Quiz

Practice Problems

Key Takeaways