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MATH1011 Chap.12 The Definite Integral and Area

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Chapter 12 of 12 · MATH1011

The Definite Integral and Area

A definite integral accumulates signed change

The definite integral is the limit of sums of thin signed rectangles. Contributions above the horizontal axis are positive and contributions below it are negative, so the result is signed area rather than automatic geometric area.

A curve can enclose a large region and still have an integral of zero when equal positive and negative contributions cancel.

The Fundamental Theorem of Calculus connects this accumulation to antiderivatives. If F prime equals f, then the integral of f from a to b is F(b) minus F(a). Keep the order of the limits visible: reversing them changes the sign, and using the same limit twice gives zero.

The arbitrary constant cancels between the two endpoint evaluations and therefore does not appear in a definite answer.

Geometric area requires splitting at sign changes

To find area between a curve and the axis, solve for every crossing inside the interval, split the integral at those points and make each contribution positive.

To find area between two curves, locate their intersections and decide which graph is above on each subinterval. Integrate top minus bottom rather than relying on a sketch that may reverse the order.

Symmetry can shorten the calculation after it has been justified. The integral of an odd function over a symmetric interval is zero, while the integral of an even function doubles the right half.

Those statements concern signed integrals. The geometric area under an odd function over a symmetric interval is generally twice the positive half, not zero.

Approximation and improper limits have different jobs

The trapezoidal rule approximates a finite definite integral from tabulated or sampled values.

Equal subintervals are required for the standard formula, and the endpoint ordinates receive half the combined weight of the interior ordinates. State the step width before substituting so an incorrect number of strips cannot hide inside arithmetic.

An improper integral is not evaluated by writing infinity into an antiderivative.

Replace the infinite endpoint by a finite variable and take a limit, or isolate any vertical singularity with one-sided limits. Convergence requires every required piece to have a finite limit. If one side diverges, the whole improper integral diverges even when informal cancellation looks tempting.

In this chapter

What this chapter covers

  • 01

    Signed accumulation and Riemann sums

  • 02

    The Fundamental Theorem of Calculus

  • 03

    Geometric area from a curve and the axis

  • 04

    Area between two curves

  • 05

    The trapezoidal rule

  • 06

    Improper integrals and convergence

Worked example · free

Separate signed integral from geometric area

Q [9 marks]. AskSia-authored practice. For f(x) = x^2 − 1 on [−2, 2], find the definite integral and the total geometric area between the graph and the x-axis. The marks are an AskSia study allocation, not a University marking scheme.
One crossing separates two lobes with opposite signs.One crossing separates two lobes with opposite signs. From 0 to 3 the area is 9 and enters the integral as −9; from 3 to 4 the area is 7/3 and enters as +7/3. Adding them gives −20/3, while adding their sizes gives 34/3.xy12442-2-4below the axis, area 9above the axis, area 7/3crossing at x = 3
One crossing separates two lobes with opposite signs. From 0 to 3 the area is 9 and enters the integral as −9; from 3 to 4 the area is 7/3 and enters as +7/3. Adding them gives −20/3, while adding their sizes gives 34/3.
  • 2Find the axis crossings and identify the sign on each subinterval.
  • 3Evaluate the signed definite integral using an antiderivative.
  • 4Split at both crossings and make each geometric contribution positive.
The equation x^2 − 1 = 0 gives crossings at x = −1 and x = 1. The function is positive on [−2, −1] and [1, 2], and negative between the crossings. An antiderivative is F(x) = x^3/3 − x. Thus the signed integral from −2 to 2 is F(2) − F(−2) = (8/3 − 2) − (−8/3 + 2) = 4/3. For geometric area, use symmetry and separate the negative middle: twice the integral from 1 to 2 of (x^2 − 1) plus twice the integral from 0 to 1 of (1 − x^2). The first contribution is 2 times 4/3 = 8/3, and the second is 2 times 2/3 = 4/3. The total geometric area is 4, larger than the signed integral because no cancellation is allowed.
Sia tip — Mark every axis crossing before integrating area; the sign of the integrand, not the presence of an integral sign, decides where absolute value is needed.
Glossary

Key terms

Definite Integral
A Definite Integral is the limit of signed sums over an interval.
Signed Area
Signed Area counts regions above the axis positively and regions below it negatively.
Fundamental Theorem
The Fundamental Theorem evaluates a definite integral by subtracting antiderivative values at its endpoints.
Trapezoidal Rule
The Trapezoidal Rule approximates a definite integral by replacing curve segments with straight-sided trapezoids.
Improper Integral
An Improper Integral is defined through a limit when an endpoint is infinite or the integrand becomes unbounded.
FAQ

The Definite Integral and Area FAQ

Why can a definite integral be smaller than the enclosed area?

A definite integral is signed: regions below the axis subtract from regions above it. Geometric area keeps every region positive, so it requires splitting at each sign change and reversing the sign of any below-axis contribution before adding the pieces.

Do I include an arbitrary constant in a definite integral?

No. Any antiderivative constant appears in both endpoint values and cancels in F(b) minus F(a). Include the constant for an indefinite integral, but omit it once fixed limits turn the calculation into a number.

How do I decide whether an improper integral converges?

Replace the improper endpoint or singular boundary with a variable and evaluate the relevant one-sided limit. Every required piece must approach a finite number. A divergent piece makes the integral divergent, even if another piece appears to cancel it informally.

Study strategy

Exam move

On each definite-integral problem, draw a small sign line before using an antiderivative. Label the task as signed integral, geometric area, approximation or convergence so that the method cannot drift. Recompute one trapezoidal estimate from a table and one improper integral from limits each study session, then differentiate the antiderivative used in every exact calculation.

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