Maths › Integration › Definite integrals and areas

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 vote negative, so honest area work means knowing where the curve is 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, integrating pieces separately.

WHAT YOU PROBABLY THINK

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 subtract the value at the lower limit from the value at the upper. The constant c cancels in the subtraction, so square-bracket working omits it.

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 limit first, lower limit subtracted, is a convention worth engraving; reversed limits change the sign of everything.

When the curve sits above the axis across the whole interval, that number is the area between curve and axis, and area questions in this topic all start by checking that condition.

The region between the curve y equals 6x minus x squared minus 5 and the x-axis, from its root at 1 to its root at 5, shaded: the definite integral evaluates the area as 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.

YOUR TURN

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 + 9 − 9) − (1 − 3 + 1/3) = 9 + 5/3 = 32/3.

One integral of the gap handles both boundaries at once, and finding the crossings is the simultaneous-equations lesson reporting for duty.

Below the axis

Where a curve runs under the axis, its integral there comes out negative, the height being negative all the way across. That is what the opening lie misses. The definite integral is a signed total, and a region below the axis cancels one above rather than adding to it.

y equals x cubed from minus 2 to 2: the region below the axis on the left cancels the region above on the right, so the integral is zero even though the two regions together cover an area of 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.

TRY IT UNSEEN

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 possible reminder to sketch first, split at the roots, and add sizes.

THE EXAM BIT

  • 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, the region is below the axis; quote areas as positive sizes with the sign explained.
  • Exact fraction answers like 32/3 are expected 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 vanish, so the value is 752/15.

Every step was Year 12 machinery: index rewriting, the reversed power rule, and the square-bracket routine, stacked.

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.

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, integrating pieces separately.

No animated video for this topic yet; these notes stand alone.