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Definite integrals and areas

Put limits on an integral and the constant cancels, the answer becomes a number, and the number means something: the area between curve and axis. Regions below the axis contribute negatively, so finding an area means establishing where the curve crosses the axis before integrating across it.

Year 12-13EDEXCEL 9MA0 8.3

Builds on Integration as antidifferentiation and Simultaneous equations and inequalities.

IN THIS TOPIC

  • Evaluate definite integrals with the square-bracket routine.
  • Find areas under curves and between a curve and a line.
  • Handle regions below the axis by splitting at the roots.

COMMON MISCONCEPTION

A definite integral is an area, always.

Numbers out of integrals

A definite integral carries limits, ∫ab f(x) dx, and the theorem evaluates it in one move. Antidifferentiate, then take the value at the upper limit and subtract the value at the lower. The constant c cancels in that subtraction, so square-bracket working leaves it out.

WORKED EXAMPLE

The square-bracket routine

Evaluate ∫13 x2 dx.

Antidifferentiate inside brackets: [x3/3] from 1 to 3.

Substitute and subtract: 27/3 − 1/3 = 26/3.

Upper first, lower subtracted. Engrave it. Reversed limits change the sign of everything downstream.

When the curve sits above the axis across the whole interval, that number is the area between curve and axis. Every area question in this topic starts by checking that condition, whether or not the question tells you to.

The region between y equals 6x minus x squared minus 5 and the x-axis, from root 1 to root 5: the definite integral gives the area 32 over 315area = 32/3y = 6x − x² − 5
FIG. 1The arch of y = 6x − x² − 5 between its roots at 1 and 5. One definite integral measures the shaded area, 32/3.

GUIDED PRACTICE

Area between a curve and a line

Find the area of the region enclosed by the parabola y = x2 and the line y = 2x + 3, before opening the working.

Show the working

Intersections first: x2 = 2x + 3 gives (x − 3)(x + 1) = 0, so the region runs from −1 to 3.

Integrate the difference, top minus bottom: ∫(2x + 3 − x2) dx from −1 to 3 = [x2 + 3x − x3/3].

Evaluating, 9 − (−5/3) = 32/3.

One integral of the gap handled both boundaries at once, and the crossings came from solving the two equations simultaneously.

Below the axis

Where a curve runs under the axis its integral there comes out negative, the height being negative all the way across. A definite integral is a signed total, and a region below the axis will cancel one above it instead of adding on.

y equals x cubed from minus 2 to 2: the region below the axis cancels the region above, so the integral is zero though the two areas total 8counts negativecounts positiveintegral 0; area 8: integrate the pieces separately
FIG. 2y = x³ from −2 to 2. The coral lobe counts −4, the amber lobe +4, and the integral across both is 0, while the true area is 8.

INDEPENDENT PRACTICE

Signed against true

Evaluate ∫−22 x3 dx, and find the total area enclosed between y = x3 and the x-axis over the same interval.

Show the working

The integral is [x4/4] from −2 to 2 = 4 − 4 = 0.

For area, split at the axis crossing. Each lobe integrates to 4 in size, so the total area is 4 + 4 = 8.

A zero integral over a region that is plainly not empty is the sharpest reminder you will get. Sketch first, split at the roots, add the sizes.

ASSESSMENT FOCUS

  • Sketch before integrating. The sketch decides whether the region needs splitting, and examiners award marks for the split itself.
  • Square brackets, upper substitution minus lower, no c. The routine is short and every line of it is marked.
  • Between a curve and a line, integrate top minus bottom between the intersection points, found by solving simultaneously.
  • A negative integral is information. It tells you the region is below the axis. Quote areas as positive sizes and explain the sign.
  • Exact fraction answers like 32/3 are wanted exact. Decimalising an exact area loses the accuracy mark.

CHECK YOURSELF

Using algebra, find the exact value of ∫04 (x + 2)2/√x dx.

Show a hint

Expand the bracket, divide each term by x to the half, then integrate powers.

Show the answer

The integrand expands to x3/2 + 4x1/2 + 4x−1/2.

Integrating gives [(2/5)x5/2 + (8/3)x3/2 + 8x1/2] from 0 to 4.

At 4 the terms are 64/5, 64/3 and 16, and at 0 all of them vanish, so the value is 752/15.

Every step used Year 12 methods. Index rewriting, the reversed power rule, the square-bracket routine.

Antidifferentiate, bracket, substitute both limits and subtract; the c never survives.

Integrals are signed; areas are not. Sketch, split at the roots, and add sizes.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 30 KB · print edition · white background

Question workbook

PDF · 31 KB · screen edition · navy

Answer book

PDF · 36 KB · fully worked · navy
6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the definite integrals and areas questions page.

CHECK YOUR PROGRESS

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  • Evaluate definite integrals with the square-bracket routine.
  • Find areas under curves and between a curve and a line.
  • Handle regions below the axis by splitting at the roots.

Open the full revision checklist to see every objective in the course in one place.