1The idea in one paragraph

A definite integral has limits, a lower one a and an upper one b, and its value is a number, not a function. The fundamental theorem of calculus says how to find that number: take any antiderivative F of the integrand and work out F(b) − F(a). The constant of integration cancels, which is why it is left out. When the curve is above the x-axis, that number is the area under it. When the curve is below the axis, the integral counts the area as negative, so a region that is half above and half below can give an integral of zero. To find an area, then, you split the interval where the curve crosses the axis and add the sizes of the pieces. Between two curves, you integrate the top curve minus the bottom curve, which works whichever side of the axis they are on.

2Evaluating a definite integral

The guide states the rule in this form:

∫ g′(x) dx from a to b = g(b) − g(a). Integrate, then subtract the value at the lower limit from the value at the upper limit.

The standard layout puts the antiderivative in square brackets with the limits on the closing bracket. Write it; the bracket line is where the method mark sits.

Figure 1 shows what that 11 measures: the curve is above the x-axis from x = 1 to x = 4, so 11 is the shaded area.

Why is there no + C? Because it would cancel: (F(b) + C) − (F(a) + C) = F(b) − F(a). Any antiderivative gives the same answer.

The same pattern works for every integral from 5.10. Paper 1 chooses limits that make the values exact.

Three things go wrong here more than anywhere else: subtracting in the wrong order, losing a minus sign in front of cos, and forgetting that the lower limit's value may not be zero (cos 0 = 1, e⁰ = 1). Put brackets round each evaluation, as above, and the signs look after themselves.

Rules you can use. These follow straight from F(b) − F(a).

• ∫ from a to a of f(x) dx = 0.
• Swapping the limits changes the sign: ∫ from b to a = −∫ from a to b.
• Adjacent intervals add: ∫ from a to b + ∫ from b to c = ∫ from a to c.
• Constants and sums behave: ∫(k f(x) + g(x)) dx from a to b = k ∫ f(x) dx + ∫ g(x) dx, over the same limits.

So if you are told that ∫ from 1 to 5 of f(x) dx = 7, then ∫ from 1 to 5 of (2f(x) + 3) dx = 2 × 7 + [3x]₁⁵ = 14 + 12 = 26. The constant 3 integrates over an interval of length 4; forgetting that and writing 14 + 3 is the usual error.

3Substitution with limits

When the integral needs the reverse chain rule of 5.10, you have two safe routes.

Route 1: integrate by inspection, then use the x-limits.

Route 2: substitute, and change the limits to u-values. With u = x² + 1, du = 2x dx. When x = 0, u = 1; when x = 1, u = 2.

Route 2 never needs you to substitute back, but you must change the limits: keeping x = 0 and x = 1 with u in the integrand is the classic mistake. A second example, using the guide's integral of sin x ÷ cos x from 5.10:

"Show that" questions often end in a logarithm like this. Use the log laws of Topic 1 to reach the given form: −ln(1/2) = ln 2.

4Integrals only technology can do

The guide says plainly that the value of some definite integrals can only be found with technology. ∫ e^−x² dx is the famous one: e^−x² has an antiderivative, but it cannot be written with the functions you know, so no method on Paper 1 will ever find it. On Paper 2 you do not need one. The GDC's numerical integration command works straight from the limits.

Write the integral, with its limits and dx, before the number. The expression is worth a mark and the number another.

5Area between a curve and the x-axis

Where a curve is below the x-axis, y is negative, so each thin strip under it contributes a negative amount to the integral. The integral is then minus the area. Figure 2 shows y = x² − 2x from x = 0 to x = 3.

The integral from 0 to 3 is zero, but the region obviously has an area: 4/3 + 4/3 = 8/3.

Area = ∫ |y| dx. Without technology: find where the curve crosses the x-axis, integrate each piece separately, and add the sizes of the answers.

The method, every time:

1. Factorise and find the x-intercepts inside the interval.
2. Sketch, so you know which pieces are above and which below.
3. Integrate over each piece separately.
4. Add the absolute values.

A larger example. Find the area of the region enclosed by y = x³ − x² − 6x and the x-axis.

Figure 3 shows the two pieces. The area is 253/12 ≈ 21.1. The single integral from −2 to 3 would have given 16/3 − 63/4 = −125/12, which is not an area at all.

A trigonometric check: over 0 ≤ x ≤ 2π, ∫ sin x dx = 0, because the arch below the axis cancels the arch above it. The area is 2 + 2 = 4.

On Paper 2, the GDC can integrate |f(x)| directly, which makes the splitting automatic. Still write the expression, for example "area = ∫ from −2 to 3 of |x³ − x² − 6x| dx = 21.1", and still sketch: the GDC cannot tell you that you have the wrong limits.

6Area between two curves

Take a thin vertical strip across the region between two curves. Its height is the top curve's y minus the bottom curve's y, and adding up those strips gives the area.

Area between two curves = ∫ (top − bottom) dx, from the left intersection to the right intersection.

This works even when part of the region is below the x-axis: the height of a strip is top − bottom wherever it sits, so the signs look after themselves.

Example. The region enclosed by y = x + 2 and y = x², drawn in Figure 4.

The guide expects you to write a correct expression before calculating. On a question that asks you to "write down an expression for the area", the line "area = ∫ from −1 to 2 of ((x + 2) − x²) dx" is the whole answer, and it needs the limits and the dx.

When the curves cross more than twice, the top one changes. y = x³ and y = x meet where x³ = x, at x = −1, 0 and 1. Figure 5 shows that x³ is on top on −1 < x < 0 and x is on top on 0 < x < 1. So split:

Integrating (x − x³) from −1 to 1 in one go gives 0, the same cancelling trap as section 5. To decide which curve is on top in each interval, test one x-value between the intersections.

On Paper 2, when the intersections cannot be found by hand, the GDC finds them and then the integral. Find the area enclosed by y = 4 − x² and y = eˣ (Figure 6).

Store the intersections in the GDC's memory and use the stored values as limits. Typing −1.96 and 1.06 by hand shifts the answer in the third figure; here it still rounds to 6.43, but on another question it may not. Equally, ∫ from a to b of |g(x) − f(x)| dx on the GDC handles any number of crossings at once.