P2.8 Definite integration, areas and the trapezium rule
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P2 (Pure Mathematics 2), specification sections 8.1–8.3. Indefinite integration (the power rule and \(+\,c\)) is in P1. Every region on this page is found with \(dx\); integrals of \(e^x\), \(\frac1x\) and trigonometric functions are in P3, and volumes of revolution in P4.
What this page covers
7 points
- Evaluating definite integrals of sums of powers of \(x\), including roots, fractions and unknown limits
- Why there is no \(+\,c\) in a definite integral
- The area between a curve and the \(x\)-axis
- What happens when part of the region is below the axis
- Areas of regions bounded by a curve and a straight line, or between two curves
- The trapezium rule and the effect of using more strips
- Whether the trapezium rule over- or underestimates, and percentage error
In one minute
5 points
- A definite integral \(\int_a^b f(x)\,dx = F(b) - F(a)\), where \(F\) is any integral of \(f\), is a number, and it equals the area under the curve between \(x = a\) and \(x = b\) as long as the curve stays above the \(x\)-axis there.
- The constant \(c\) cancels, so leave it out of definite integrals.
- Below the axis the integral is negative. If the curve crosses the axis, find where, integrate each part separately and add the sizes.
- Between two graphs: area \(= \int_a^b (\text{top} - \text{bottom})\,dx\), with \(a\) and \(b\) where they meet. Composite regions: add or subtract simple pieces such as triangles.
- Trapezium rule: \(\int_a^b y\,dx \approx \frac12 h\{y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\}\), \(h = \frac{b-a}{n}\). Curve bending up: overestimate. Curve bending down: underestimate. More strips: smaller error.
Contents
12 sections
- What does a definite integral calculate?
- How do I find the area under a curve?
- What happens when the region is below the x-axis?
- How do I find areas bounded by curves and lines?
- How does the trapezium rule estimate an area?
- Is the trapezium rule too big or too small, and by how much?
- Exam tips for P2.8
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What does a definite integral calculate?
In P1, integrating gave a family of curves: \(\int 2x\,dx = x^2 + c\), one curve for each value of \(c\). Put two numbers on the integral sign and the answer becomes a single number instead, and that number turns out to be an area.
- Integrate as in P1, write the result in square brackets with the limits attached, then substitute: upper value minus lower value.
- The constant of integration cancels: \((F(b) + c) - (F(a) + c) = F(b) - F(a)\). That is why there is no \(+\,c\) in a definite integral.
- For example, \(\int_1^3 (3x^2 - 4x + 1)\,dx = \Big[x^3 - 2x^2 + x\Big]_1^3 = (27 - 18 + 3) - (1 - 2 + 1) = 12 - 0 = 12\).
- Swapping the limits changes the sign: \(\int_3^1 (3x^2 - 4x + 1)\,dx = -12\).
Find the exact value of \(\displaystyle\int_1^4 \left(2\sqrt{x} - \frac{3}{x^2}\right) dx\).
Write each term as a power of \(x\): \(2x^{1/2} - 3x^{-2}\).
Integrate (raise each power by one, divide by the new power): \[\Big[\tfrac43 x^{3/2} + 3x^{-1}\Big]_1^4 = \Big[\tfrac43 x^{3/2} + \tfrac3x\Big]_1^4\]
Substitute the limits, using \(4^{3/2} = 8\): \(\left(\frac{32}{3} + \frac34\right) - \left(\frac43 + 3\right) = \frac{28}{3} - \frac94\).
Combine over a common denominator: \(\frac{112}{12} - \frac{27}{12}\). \(\frac{85}{12}\)
"Exact" means leave it as a fraction; the decimal 7.0833… would not be in the form asked.
Given that \(k > 1\) and \(\int_1^k (4x - 1)\,dx = 14\), find the value of \(k\).
Integrate and substitute the limits; the upper limit is \(k\): \(\Big[2x^2 - x\Big]_1^k = (2k^2 - k) - (2 - 1) = 2k^2 - k - 1\).
Set equal to 14 and solve the quadratic: \(2k^2 - k - 15 = 0 \Rightarrow (k - 3)(2k + 5) = 0\).
Use the condition \(k > 1\): \(k = -\frac52\) is rejected. \(k = 3\)
Losing a sign when subtracting the lower value. Put brackets round both \(F(b)\) and \(F(a)\) before subtracting: \((\ldots) - (\ldots)\). The lower value often contains negatives, and every one of them changes sign.
How do I find the area under a curve?
Why should an integral give an area? Think of the area under the curve from \(a\) up to a moving right-hand edge \(x\), and call it \(A(x)\). Moving the edge a tiny step \(\delta x\) adds a thin strip of height about \(y\), so \(\delta A \approx y\,\delta x\), and \(\frac{dA}{dx} = y\). The area function is therefore an integral of \(y\), and the area between \(a\) and \(b\) is \(A(b) - A(a)\): exactly the definite integral.
If the region is described as "bounded by the curve and the \(x\)-axis", the limits are where the curve meets the axis: solve \(f(x) = 0\) first.
The curve \(C\) has equation \(y = 6x - x^2 - 5\). The region \(R\) is bounded by \(C\) and the \(x\)-axis. Find the exact area of \(R\).
Find the limits, where the curve meets the \(x\)-axis: \(6x - x^2 - 5 = 0 \Rightarrow x^2 - 6x + 5 = 0 \Rightarrow (x - 1)(x - 5) = 0\), so \(x = 1\) and \(x = 5\).
Integrate: \(\int_1^5 (6x - x^2 - 5)\,dx = \Big[3x^2 - \frac{x^3}{3} - 5x\Big]_1^5\).
Substitute the limits: \(\left(75 - \frac{125}{3} - 25\right) - \left(3 - \frac13 - 5\right) = \frac{25}{3} - \left(-\frac73\right)\). \(\frac{32}{3}\)
The answer is positive, as it must be for a region above the axis: a quick check worth making every time.
Sketch the curve, even roughly, before integrating for an area. The sketch tells you the limits, whether the region is above or below the axis, and whether you need to add or subtract another shape.
What happens when the region is below the x-axis?
Where the curve is below the \(x\)-axis, \(y < 0\), so each thin strip \(y\,\delta x\) is negative and the definite integral there is negative. The area is the size (modulus) of that integral.
If the curve crosses the axis between the limits, the positive and negative parts cancel. For example \(\int_{-2}^{2} x^3\,dx = 0\), although the region between \(y = x^3\) and the axis from \(x = -2\) to 2 has area 8.
The method is therefore: find where the curve meets the axis, integrate each part separately, make each answer positive, then add.
The curve \(y = x(x - 1)(x - 3)\) meets the \(x\)-axis at \(x = 0\), 1 and 3. (a) Evaluate \(\int_0^3 x(x - 1)(x - 3)\,dx\). (b) Find the total area of the regions bounded by the curve and the \(x\)-axis for \(0 \le x \le 3\).
Expand, then integrate: \(x(x - 1)(x - 3) = x^3 - 4x^2 + 3x\), so \(F(x) = \frac{x^4}{4} - \frac43 x^3 + \frac32 x^2\).
(a) One integral from 0 to 3: \(F(3) - F(0) = \left(\frac{81}{4} - 36 + \frac{27}{2}\right) - 0\). \(-\frac94\)
(b) Split at \(x = 1\), where the curve crosses the axis: \(\int_0^1 = F(1) - F(0) = \frac14 - \frac43 + \frac32 = \frac{5}{12}\); \(\int_1^3 = F(3) - F(1) = -\frac94 - \frac{5}{12} = -\frac83\). Add the sizes: \(\frac{5}{12} + \frac83\). Area \(= \frac{37}{12}\)
The answer to (a), \(-\frac94\), is the signed total \(\frac{5}{12} - \frac83\); it is not an area.
Giving a negative number as an area, or integrating straight across a root. An area is never negative, and an integral across a point where the curve crosses the axis gives the difference of the two areas, not their sum.
How do I find areas bounded by curves and lines?
Two approaches cover almost every region you will meet, and both integrate with respect to \(x\).
- Split into pieces. A region bounded by a curve, a straight line and the \(x\)-axis is often "area under the curve" plus (or minus) a triangle or trapezium. Use geometry for straight-edged pieces: it is quicker than integrating.
- Top minus bottom. Between two graphs \(y = f(x)\) (upper) and \(y = g(x)\) (lower) meeting at \(x = a\) and \(x = b\), the area is \(\int_a^b \big(f(x) - g(x)\big)\,dx\). This works even if part of the region is below the \(x\)-axis, because \(f - g\) stays positive.
- Find the \(x\)-coordinates where the graphs meet by solving \(f(x) = g(x)\); these are the limits.
The curve \(y = 2x^2\), \(x \ge 0\), and the line \(y = 12 - 2x\) meet at the point \(P\). The region \(R\) is bounded by the curve, the line and the \(x\)-axis. (a) Find the coordinates of \(P\). (b) Find the exact area of \(R\).
(a) Solve simultaneously: \(2x^2 = 12 - 2x \Rightarrow x^2 + x - 6 = 0 \Rightarrow (x + 3)(x - 2) = 0\); \(x \ge 0\), so \(x = 2\). \(P(2, 8)\)
(b) Split at \(x = 2\). Under the curve from 0 to 2: \(\int_0^2 2x^2\,dx = \Big[\frac23 x^3\Big]_0^2 = \frac{16}{3}\). Under the line from 2 to where it meets the axis (\(12 - 2x = 0 \Rightarrow x = 6\)) is a triangle: \(\frac12 \times (6 - 2) \times 8 = 16\). Add: \(\frac{16}{3} + 16\). Area of \(R = \frac{64}{3}\)
The curve \(C\) has equation \(y = x(x - 3)^2\) and the line \(L\) has equation \(y = x\). (a) Show that \(C\) and \(L\) meet where \(x = 0\), \(x = 2\) and \(x = 4\). (b) Show that the two finite regions enclosed by \(C\) and \(L\) have equal area, and find that area.
(a) Set the equations equal: \(x^3 - 6x^2 + 9x = x \Rightarrow x(x^2 - 6x + 8) = 0 \Rightarrow x(x - 2)(x - 4) = 0\), so \(x = 0, 2, 4\) ✓
(b) Decide which graph is on top in each interval, using a test value. At \(x = 1\): \(C\) gives 4, \(L\) gives 1, so \(C\) is above. At \(x = 3\): \(C\) gives 0, \(L\) gives 3, so \(L\) is above.
Top minus bottom on \([0, 2]\), with \(C - L = x^3 - 6x^2 + 8x\): \(\int_0^2 (x^3 - 6x^2 + 8x)\,dx = \Big[\frac{x^4}{4} - 2x^3 + 4x^2\Big]_0^2 = (4 - 16 + 16) - 0 = 4\).
Top minus bottom on \([2, 4]\) is \(L - C\), the negative of the same expression: \(-\Big[\frac{x^4}{4} - 2x^3 + 4x^2\Big]_2^4 = -\big((64 - 128 + 64) - 4\big) = 4\). Both regions have area 4, so they are equal
\(\int_0^4 (C - L)\,dx = 0\): integrating across the crossing point at \(x = 2\) cancels the two regions exactly, the same trap as for regions below the axis.
Write "area \(= \int_a^b (\text{top} - \text{bottom})\,dx\)" with the actual expressions before simplifying. Subtracting the line from the curve first, then integrating once, is quicker and safer than finding two areas under graphs and subtracting.
How does the trapezium rule estimate an area?
Many functions cannot be integrated with the rules you know at this stage: \(2^x\), \(\sqrt{1 + x^3}\), \(\log_{10} x\). Data from an experiment has no formula at all. The trapezium rule replaces the curve by straight chords between equally spaced points and adds the areas of the trapezia underneath.
In words: half the strip width × (first + last + twice everything in between). Each trapezium has area \(\frac12 h(y_{r} + y_{r+1})\); each inner ordinate is a side of two trapezia, which is why it counts twice.
Use the trapezium rule with 4 strips to estimate \(\int_0^2 2^x\,dx\). Give the ordinates to 4 decimal places and your answer to 4 significant figures.
Strip width and the \(x\)-values: \(h = \frac{2 - 0}{4} = 0.5\); \(x = 0, 0.5, 1, 1.5, 2\).
| \(x\) | 0 | 0.5 | 1 | 1.5 | 2 |
|---|---|---|---|---|---|
| \(y = 2^x\) | 1 | 1.4142 | 2 | 2.8284 | 4 |
Apply the rule, end ordinates once and inner ordinates twice: \(\frac12 \times 0.5 \times \{(1 + 4) + 2(1.4142 + 2 + 2.8284)\} = 0.25 \times \{5 + 12.4852\} = 0.25 \times 17.4852\). \(\approx 4.371\)
Keep the ordinates to at least 4 decimal places in the working; rounding them to 2 d.p. can change the final figure.
Mixing up strips and ordinates. Four strips need five \(x\)-values; \(h = \frac{b - a}{n}\) uses the number of strips, not the number of ordinates.
Is the trapezium rule too big or too small, and by how much?
- If the curve bends upwards (convex, like \(y = x^2\) or \(2^x\)) across the interval, every chord lies above the curve, so the trapezium rule gives an overestimate.
- If the curve bends downwards (concave, like \(y = \sqrt{x}\)), every chord lies below the curve: an underestimate.
- If the curve changes from one to the other inside the interval, the errors partly cancel and you cannot say in general.
- More strips (smaller \(h\)) make the chords hug the curve more closely, so the estimate improves. For \(2^x\) on \([0, 2]\), 2, 4 and 8 strips give 4.5, 4.371 and 4.339, closing in on the true value 4.328 (to 4 s.f.) from above.
(a) Use the trapezium rule with 4 strips to estimate \(\int_1^3 \frac{6}{x^2}\,dx\). (b) State, with a reason, whether your answer is an overestimate or an underestimate. (c) Find the exact value of \(\int_1^3 \frac{6}{x^2}\,dx\), and hence the percentage error in your estimate.
(a) \(h = \frac{3 - 1}{4} = 0.5\).
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|
| \(y = 6/x^2\) | 6 | 2.6667 | 1.5 | 0.96 | 0.6667 |
\(0.25 \times \{(6 + 0.6667) + 2(2.6667 + 1.5 + 0.96)\} = 0.25 \times \{6.6667 + 10.2534\}\). \(\approx 4.23\)
(b) \(y = 6x^{-2}\) bends upwards on \([1, 3]\) (it is decreasing ever more slowly), so each chord lies above the curve. Overestimate
(c) \(\int_1^3 6x^{-2}\,dx = \Big[-\frac6x\Big]_1^3 = -2 - (-6) = 4\). Percentage error \(= \frac{4.23 - 4}{4} \times 100\%\). 5.75%
The exact value is smaller than the estimate, consistent with part (b). Use the unrounded estimate (here it happens to be exactly 4.23).
| Strips \(n\) | Width \(h\) | Estimate | Percentage error |
|---|---|---|---|
| 2 | 1 | 4.833 | 20.8% |
| 4 | 0.5 | 4.23 | 5.75% |
| 8 | 0.25 | 4.059 | 1.49% |
The same integral with 2, 4 and 8 strips: each halving of \(h\) cuts the error to about a quarter.
"State, with a reason, whether … overestimate or underestimate" needs both parts: the shape (curve bends up or down, chords above or below the curve) and the conclusion. A sketch with one strip and its chord is a good reason.
Exam tips for P2.8
Write out \(F(b) - F(a)\) with both brackets before simplifying. A calculator can evaluate a definite integral, but the working is where the method is shown, and "use algebraic integration" means a calculator value alone is not enough.
Limits come from the region: where the curve meets the axis, where the graphs meet, or the given lines \(x = a\) and \(x = b\). Solve for them first and mark them on a sketch.
If an integral for an area comes out negative, the region is below the axis (or you used bottom − top). State the area as a positive value, and split at every crossing point.
Give \(h\), the table of ordinates and the bracket \(\{y_0 + y_n + 2(\ldots)\}\) as separate lines. Keep 4 or more decimal places in the table and round only the final answer, to the accuracy asked.
If \(\int_a^b f(x)\,dx \approx T\), then \(\int_a^b \big(f(x) + k\big)\,dx \approx T + k(b - a)\) and \(\int_a^b mf(x)\,dx \approx mT\). Use your earlier answer rather than starting a new table.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A definite integral needs \(+\,c\). | The constant appears in \(F(b)\) and \(F(a)\) and cancels. | Leave \(+\,c\) out of definite integrals. |
| A definite integral always equals an area. | Parts below the \(x\)-axis count as negative. | Split at the roots and add the sizes of the parts. |
| An integral of zero means there is no region. | Equal areas above and below the axis cancel. | Check where the curve crosses the axis between the limits. |
| Area between two graphs = area under the top one. | That includes the region under the lower graph too. | Integrate (top − bottom) between the meeting points. |
| With \(n\) strips there are \(n\) ordinates. | Each strip has two edges, shared with its neighbours. | \(n\) strips need \(n + 1\) ordinates; \(h = \frac{b-a}{n}\). |
| The trapezium rule always overestimates. | That depends on which way the curve bends. | Bends up: overestimate. Bends down: underestimate. |
How this topic connects
Integration (P1) and further integration (P3, P4)
Every definite integral starts with the indefinite integral from P1: rewriting roots and fractions as powers, then raising each power by one. Definite integrals of \(e^x\), \(\frac1x\) and trigonometric functions come in P3; in P4 they give volumes of revolution and are combined with substitution and integration by parts. The trapezium rule returns for functions that still cannot be integrated exactly.
Variable acceleration (M2)
The area under a velocity–time graph is the displacement, \(\int_{t_1}^{t_2} v\,dt\). When the velocity changes sign, total distance needs the same splitting as for regions below the axis.
IAL Physics
Areas under graphs carry physical meaning: displacement from a velocity–time graph, work done from a force–extension graph, impulse from a force–time graph. When the graph comes from data, a trapezium estimate is the natural method.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\int_a^b f(x)\,dx = F(b) - F(a),\ F' = f\) | evaluating a definite integral | No — learn it |
| \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + c,\ n \ne -1\) | integrating each term before substituting limits | No — learn it |
| Area \(= \int_a^b y\,dx\) (curve above the axis) | area between a curve and the \(x\)-axis | No — learn it |
| Area \(= \int_a^b (y_{\text{top}} - y_{\text{bottom}})\,dx\) | area between two graphs meeting at \(x = a, b\) | No — derive it |
| \(\int_a^b y\,dx \approx \frac12 h\{y_0 + y_n + 2(y_1 + y_2 + \cdots + y_{n-1})\},\ h = \frac{b-a}{n}\) | estimating a definite integral with \(n\) strips | Yes |
| percentage error \(= \frac{|\text{estimate} - \text{exact}|}{\text{exact}} \times 100\%\) | judging the accuracy of an estimate | No — learn it |
Check your understanding
1. Evaluate \(\int_0^2 (x^3 + 1)\,dx\).
\(\Big[\frac{x^4}{4} + x\Big]_0^2 = (4 + 2) - 0 = 6\).
2. Find the exact value of \(\int_1^4 \sqrt{x}\,dx\).
\(\Big[\frac23 x^{3/2}\Big]_1^4 = \frac23(8 - 1) = \frac{14}{3}\).
3. \(\int_{-1}^{1} x^3\,dx = 0\). Does that mean the curve \(y = x^3\) encloses no area with the \(x\)-axis between −1 and 1?
No. The part from −1 to 0 is below the axis and gives \(-\frac14\); the part from 0 to 1 gives \(+\frac14\). They cancel in the integral, but the total area is \(\frac12\).
4. How many ordinates are needed to use the trapezium rule with 6 strips on \(0 \le x \le 3\), and what is \(h\)?
7 ordinates (\(x = 0, 0.5, 1, \ldots, 3\)), with \(h = \frac36 = 0.5\).
5. The curve \(y = \sqrt{1 + x}\) bends downwards for \(x \ge 0\). Does the trapezium rule over- or underestimate \(\int_0^3 \sqrt{1 + x}\,dx\)?
Underestimate: the curve bends downwards, so every chord lies below it and each trapezium misses a sliver of area.
6. Given that \(\int_0^2 2^x\,dx \approx 4.371\) (Worked example 7), estimate \(\int_0^2 (2^x + 3)\,dx\) without a new table.
Adding 3 adds a rectangle of height 3 and width 2: \(4.371 + 3 \times 2 = 10.371 \approx 10.37\).
Specification checklist
- (P2 8.1) I can evaluate a definite integral by integrating and substituting the limits, upper minus lower, and give exact answers as fractions or surds.
- (P2 8.1) I can find an unknown limit or constant from the value of a definite integral.
- (P2 8.2) I can find the area between a curve and the \(x\)-axis, finding the limits where the curve meets the axis.
- (P2 8.2) I can deal with regions below the \(x\)-axis, splitting the integral where the curve crosses the axis.
- (P2 8.2) I can find the area of a region bounded by a curve and a straight line, or by two curves, using top minus bottom or by combining simpler shapes.
- (P2 8.3) I can use the trapezium rule with a given number of strips, and explain how increasing the number of strips improves the estimate.
- (P2 8.3) I can decide from the shape of the curve whether the trapezium rule gives an overestimate or an underestimate, and calculate a percentage error.