The shape of the lesson
By the end of the lesson, learners can
- Evaluating definite integrals: Integrate as usual, write the result in square brackets with the limits, then substitute the top limit and subtract the value at the bottom limit.
- Area under a curve: For a curve above the \(x\)-axis, \(\displaystyle\int_{a}^{b} y\,dx\) is the area between the curve, the axis and the lines \(x = a\), \(x = b\).
- Between a curve and a line: Find where the graphs meet: those \(x\)-values are the limits.
- Improper integrals: Some integrals have an infinite limit, or a function that is undefined at a limit.
- Sketch: where the region is, where the graphs meet, and whether it crosses the axis.
- Limits: from the question or from solving the equations simultaneously.
- Integrate: (upper \(-\) lower), or \(y\) for a region on the \(x\)-axis, or \(x\) in terms of \(y\) against the \(y\)-axis.
- Evaluate: \(F(b) - F(a)\), with brackets around \(F(a)\).
- Below the axis: integrate each part separately and add the sizes.
One picture every idea returns to
\(\displaystyle\int_{a}^{b} f(x)\,dx = F(b) - F(a)\), where \(F\) is any integral of \(f\). The area between \(y = f(x)\) and the \(x\)-axis is \(\displaystyle\int_{a}^{b} y\,dx\) when the curve is above the axis; between two graphs, integrate (top \(-\) bottom).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Integrals that measure.. Put limits on an integral and the constant disappears: what is left is a number, and that number is an area. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Evaluating definite integrals | Integrate as usual, write the result in square brackets with the limits, then substitute the top limit and subtract the value at the bottom limit. Model the worked example A polynomial on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Area under a curve | For a curve above the \(x\)-axis, \(\displaystyle\int_{a}^{b} y\,dx\) is the area between the curve, the axis and the lines \(x = a\), \(x = b\). Model the worked example A curve that crosses the axis on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Between a curve and a line | Find where the graphs meet: those \(x\)-values are the limits. Model the worked example A curve and a line on the board, one line at a time. Your turn: Find the area enclosed by \(y = x^{2}\) and \(y = 2x\). | Pairs |
| 23 to 30 min | Teach and check: Improper integrals | Some integrals have an infinite limit, or a function that is undefined at a limit. Model the worked example Two simple improper integrals on the board, one line at a time. | Rally Coach |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 24 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Improper integrals, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A lower limit of 0 does not always give 0: \((2 \times 0 + 1)^{\frac{3}{2}} = 1\). | Always substitute properly, and put the bottom-limit value in brackets before subtracting it. |
| Sketch first. | If the curve crosses the \(x\)-axis between the limits, one integral gives (area above) \(-\) (area below), not the total area. |
| Upper minus lower. | If you subtract the wrong way you get the negative of the area; an area is never negative, so take the size, but check your sketch. |
| An improper integral only has a value if the integrated expression settles to a limit. \(\left[-\tfrac{1}{x}\right]\) at \(\infty\) is 0, but something like \(\left[2\sqrt{x}\right]\) at \(\infty\) grows without limit, so \(\displaystyle\int_{1}^{\infty} \dfrac{1}{\sqrt{x}}\,dx\) has no value. | Return to Improper integrals: ask a pair to state the correct version and show it on the lab before moving on. |
Definite integrals and area: worksheet
6 questions, 24 marks. Written in the style of the papers; no past paper question is reproduced.
Evaluate \(\displaystyle\int_{1}^{4} (2x + 3)\,dx\).
Find the exact value of \(\displaystyle\int_{1}^{4} \dfrac{x - 1}{\sqrt{x}}\,dx\).
Evaluate (a) \(\displaystyle\int_{1}^{\infty} \dfrac{1}{(2x - 1)^{2}}\,dx\) (b) \(\displaystyle\int_{0}^{4} \dfrac{1}{\sqrt{x}}\,dx\).
Find the area of the region enclosed by the curve \(y = 4x - x^{2}\) and the \(x\)-axis.
The curve \(y = x^{2} - 2x + 3\) and the line \(y = x + 3\) meet at two points.
(a) Find the \(x\)-coordinates of these points.
(b) Find the area of the region enclosed by the curve and the line.
The rectangle \(OABC\) has vertices \(O(0, 0)\), \(A(2, 0)\), \(B(2, 4)\) and \(C(0, 4)\). The curve \(y = x^{2}\) passes through \(O\) and \(B\) and divides the rectangle into two regions.
(a) Find the area of the region between the curve and the \(x\)-axis.
(b) By integrating with respect to \(y\), find the area of the region between the curve and the \(y\)-axis.
(c) Show that your answers are consistent.
Answer key and marking notes
- \(\left[x^{2} + 3x\right]_{1}^{4} = (16 + 12) - (1 + 3)\)
- \(= 28 - 4 = 24\)
- \(\dfrac{x - 1}{\sqrt{x}} = x^{\frac{1}{2}} - x^{-\frac{1}{2}}\).
- Integrate: \(\left[\tfrac{2}{3}x^{\frac{3}{2}} - 2x^{\frac{1}{2}}\right]_{1}^{4}\).
- \(= \left(\tfrac{16}{3} - 4\right) - \left(\tfrac{2}{3} - 2\right) = \tfrac{4}{3} + \tfrac{4}{3}\).
- \(= \dfrac{8}{3}\)
- (a) \(\displaystyle\int (2x - 1)^{-2}\,dx = \dfrac{(2x - 1)^{-1}}{-1 \times 2} = -\dfrac{1}{2(2x - 1)}\).
- As \(x \to \infty\) this tends to 0; at \(x = 1\) it is \(-\tfrac{1}{2}\). Value \(= 0 - \left(-\tfrac{1}{2}\right) = \tfrac{1}{2}\).
- (b) \(\left[2\sqrt{x}\right]_{0}^{4} = 4 - 0 = 4\).
- \(4x - x^{2} = x(4 - x) = 0\) at \(x = 0\) and \(x = 4\); the curve is above the axis between them.
- \(\displaystyle\int_{0}^{4} (4x - x^{2})\,dx = \left[2x^{2} - \tfrac{x^{3}}{3}\right]_{0}^{4}\).
- \(= 32 - \tfrac{64}{3} = \dfrac{32}{3}\)
- (a) \(x^{2} - 2x + 3 = x + 3\) gives \(x^{2} - 3x = 0\), so \(x = 0\) or \(x = 3\).
- (b) Line minus curve: \((x + 3) - (x^{2} - 2x + 3) = 3x - x^{2}\).
- \(\displaystyle\int_{0}^{3} (3x - x^{2})\,dx = \left[\tfrac{3x^{2}}{2} - \tfrac{x^{3}}{3}\right]_{0}^{3} = \tfrac{27}{2} - 9\).
- \(= \dfrac{9}{2}\)
- (a) \(\displaystyle\int_{0}^{2} x^{2}\,dx = \left[\tfrac{x^{3}}{3}\right]_{0}^{2} = \dfrac{8}{3}\).
- (b) \(x = \sqrt{y}\): \(\displaystyle\int_{0}^{4} y^{\frac{1}{2}}\,dy = \left[\tfrac{2}{3}y^{\frac{3}{2}}\right]_{0}^{4} = \tfrac{2}{3} \times 8 = \dfrac{16}{3}\).
- (c) \(\tfrac{8}{3} + \tfrac{16}{3} = 8 = 2 \times 4\), the area of the rectangle, as it must be.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Evaluate \(\displaystyle\int_{1}^{2} (4x^{3} - 3)\,dx\).
- A\(12\)
- B\(8\)
- C\(10\)
- D\(28\)
Evaluate \(\displaystyle\int_{1}^{4} \dfrac{1}{\sqrt{x}}\,dx\).
- A\(2\)
- B\(1\)
- C\(\dfrac{7}{16}\)
- D\(\dfrac{14}{3}\)
Find the area of the region enclosed by the curve \(y = 6x - x^{2}\) and the \(x\)-axis.
- A\(36\)
- B\(108\)
- C\(144\)
- D\(-108\)
Find the area of the region enclosed by the curve \(y = x^{2} - 4\) and the \(x\)-axis.
- A\(\dfrac{32}{3}\)
- B\(-\dfrac{32}{3}\)
- C\(\dfrac{16}{3}\)
- D\(\dfrac{64}{3}\)
Find the total area between the curve \(y = x^{2} - 4x + 3\), the \(x\)-axis and the lines \(x = 0\) and \(x = 3\).
- A\(\dfrac{8}{3}\)
- B\(0\)
- C\(\dfrac{4}{3}\)
- D\(-\dfrac{4}{3}\)
The diagram shows the curve \(y = 4x - x^{2}\) and the line \(y = x\), which meet at the origin and at \((3, 3)\). Find the area of the shaded region between them.
- A\(\dfrac{9}{2}\)
- B\(-\dfrac{9}{2}\)
- C\(9\)
- D\(\dfrac{27}{2}\)
Find the area of the region bounded by the curve \(y = x^{3}\) for \(x \ge 0\), the \(y\)-axis and the line \(y = 8\).
- A\(12\)
- B\(4\)
- C\(16\)
- D\(\dfrac{64}{3}\)
Evaluate \(\displaystyle\int_{4}^{\infty} \dfrac{2}{x^{\frac{3}{2}}}\,dx\).
- A\(2\)
- B\(-2\)
- C\(\dfrac{1}{2}\)
- DThe integral has no finite value
Quiz key
\(\left[x^{4} - 3x\right]_{1}^{2} = (16 - 6) - (1 - 3) = 10 - (-2) = 12\). \(8\) subtracts \(+2\) instead of \(-2\); \(10\) is \(F(2)\) alone, forgetting the lower limit; \(28\) substitutes the limits into \(4x^{3} - 3\) itself, without integrating.
\(\left[2x^{\frac{1}{2}}\right]_{1}^{4} = 2 \times 2 - 2 \times 1 = 2\). \(1\) forgets to divide by the new power \(\tfrac{1}{2}\) (which doubles it); \(\dfrac{7}{16}\) uses the derivative \(-\tfrac{1}{2}x^{-\frac{3}{2}}\) instead of the integral; \(\dfrac{14}{3}\) integrates \(\sqrt{x}\) instead of \(\dfrac{1}{\sqrt{x}}\).
\(6x - x^{2} = x(6 - x) = 0\) at \(x = 0\) and \(x = 6\). Area \(= \left[3x^{2} - \tfrac{1}{3}x^{3}\right]_{0}^{6} = 108 - 72 = 36\). \(108\) leaves out the \(-x^{2}\) term; \(144\) integrates \(6x\) as \(6x^{2}\), forgetting to halve; \(-108\) integrates \(x^{2}\) as \(x^{3}\), forgetting to divide by 3.
\(\displaystyle\int_{-2}^{2} (x^{2} - 4)\,dx = \left[\tfrac{1}{3}x^{3} - 4x\right]_{-2}^{2} = \left(\tfrac{8}{3} - 8\right) - \left(-\tfrac{8}{3} + 8\right) = -\dfrac{32}{3}\). The region is below the axis, so the area is \(\dfrac{32}{3}\). \(-\dfrac{32}{3}\) quotes the integral as the area, but an area is never negative; \(\dfrac{16}{3}\) uses only \(0 \le x \le 2\); \(\dfrac{64}{3}\) integrates \(-4\) as \(+4x\).
\(x^{2} - 4x + 3 = (x - 1)(x - 3)\), so the curve is above the axis for \(0 \le x \le 1\) and below for \(1 \le x \le 3\). \(\displaystyle\int_{0}^{1} = \dfrac{4}{3}\) and \(\displaystyle\int_{1}^{3} = -\dfrac{4}{3}\). Total area \(= \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{8}{3}\). \(0\) is \(\displaystyle\int_{0}^{3}\) in one go, where the two parts cancel; \(\dfrac{4}{3}\) and \(-\dfrac{4}{3}\) are each only one of the two pieces.
The curve is on top: \(\displaystyle\int_{0}^{3} \left((4x - x^{2}) - x\right) dx = \left[\tfrac{3}{2}x^{2} - \tfrac{1}{3}x^{3}\right]_{0}^{3} = \dfrac{27}{2} - 9 = \dfrac{9}{2}\). \(-\dfrac{9}{2}\) subtracts the wrong way round (line minus curve); \(9\) is the whole area under the curve, without removing the triangle under the line; \(\dfrac{27}{2}\) adds the triangle (area \(\tfrac{9}{2}\)) instead of subtracting it.
\(\displaystyle\int_{0}^{8} y^{\frac{1}{3}}\,dy = \left[\tfrac{3}{4}y^{\frac{4}{3}}\right]_{0}^{8} = \tfrac{3}{4} \times 16 = 12\). Check: the rectangle \(2 \times 8 = 16\) minus the area under the curve, \(\displaystyle\int_{0}^{2} x^{3}\,dx = 4\), also gives 12. \(4\) is the area between the curve and the \(x\)-axis (the wrong region); \(16\) is the whole rectangle; \(\dfrac{64}{3}\) multiplies by \(\tfrac{4}{3}\) instead of dividing.
\(\displaystyle\int 2x^{-\frac{3}{2}}\,dx = \dfrac{2x^{-\frac{1}{2}}}{-\frac{1}{2}} = -\dfrac{4}{\sqrt{x}}\). As \(x \to \infty\), \(-\dfrac{4}{\sqrt{x}} \to 0\). So the value is \(0 - \left(-\dfrac{4}{2}\right) = 2\). \(-2\) subtracts in the wrong order; \(\dfrac{1}{2}\) multiplies by \(-\tfrac{1}{2}\) instead of dividing; the integral is finite because \(\dfrac{1}{\sqrt{x}}\) tends to zero.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.