The shape of the lesson
By the end of the lesson, learners can
- Undoing a derivative: Differentiating takes the power down and reduces it by one.
- Linear brackets: Reversing the chain rule for a linear inside: raise the power by one, divide by the new power AND by the coefficient of \(x\).
- Finding the constant: The gradient function alone gives a family of curves (Fig. 1).
- Rewrite: roots and fractions as powers; split fractions term by term.
- Powers: \(x^{n} \to \dfrac{x^{n+1}}{n+1}\), \(n \ne -1\).
- Linear bracket: \((ax + b)^{n} \to \dfrac{(ax + b)^{n+1}}{a(n+1)}\).
- Constant: always \(+ c\); find it from a point on the curve.
- Check: differentiate your answer and compare.
One picture every idea returns to
\(\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + c\) and \(\displaystyle\int (ax + b)^{n}\,dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + c\), for any rational \(n \ne -1\). A point on the curve fixes the value of \(c\).
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, Running the derivative backwards.. Integration undoes differentiation: given the gradient function, recover the curve. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Undoing a derivative | Differentiating takes the power down and reduces it by one. Model the worked example A polynomial on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Linear brackets | Reversing the chain rule for a linear inside: raise the power by one, divide by the new power AND by the coefficient of \(x\). Model the worked example Positive and negative powers on the board, one line at a time. Your turn: Find \(\displaystyle\int (4x - 3)^{5}\,dx\). | Pairs |
| 21 to 30 min | Teach and check: Finding the constant | The gradient function alone gives a family of curves (Fig. 1). Model the worked example A curve through a point on the board, one line at a time. | Round Table |
| 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, 20 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 Finding the constant, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Dividing by a fraction is multiplying by its reciprocal: \(\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} = \tfrac{2}{3}x^{\frac{3}{2}}\), not \(\tfrac{3}{2}x^{\frac{3}{2}}\). | And \(\displaystyle\int x^{-1}\,dx\) cannot be done this way (it would divide by zero). |
| Divide by \(a\); do not multiply. \(\displaystyle\int (2x + 1)^{3}\,dx\) is \(\dfrac{(2x + 1)^{4}}{8}\), not \(\dfrac{2(2x + 1)^{4}}{4}\). | This shortcut only works when the inside is LINEAR: \(\displaystyle\int (x^{2} + 1)^{3}\,dx\) needs expanding. |
| Find \(c\) straight after integrating, using the equation of the CURVE (\(y\)), not the gradient. | When you integrate twice, each integration needs its own constant. |
Integration as reverse differentiation: worksheet
6 questions, 20 marks. Written in the style of the papers; no past paper question is reproduced.
Find \(\displaystyle\int (4x^{3} - 6x^{2} + 2)\,dx\).
Find \(\displaystyle\int \left(\dfrac{3}{x^{2}} + \sqrt{x}\right) dx\).
Find \(\displaystyle\int (5 - 2x)^{3}\,dx\).
Find \(\displaystyle\int \dfrac{x^{2} - 4}{x^{2}}\,dx\).
A curve has \(\dfrac{dy}{dx} = 6x^{2} - 2x\) and passes through the point \((2, 9)\). Find the equation of the curve.
A curve is such that \(\dfrac{dy}{dx} = 3\sqrt{x} - 6\) for \(x > 0\), and it passes through the point \((4, 0)\).
(a) Find the equation of the curve.
(b) Find the coordinates of the stationary point and determine its nature.
(c) What does your answer to (b) tell you about the curve and the \(x\)-axis?
Answer key and marking notes
- \(\dfrac{4x^{4}}{4} - \dfrac{6x^{3}}{3} + 2x + c\)
- \(= x^{4} - 2x^{3} + 2x + c\)
- \(3x^{-2} + x^{\frac{1}{2}}\) integrates to \(\dfrac{3x^{-1}}{-1} + \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}\).
- \(= -\dfrac{3}{x} + \tfrac{2}{3}x^{\frac{3}{2}} + c\)
- \(\dfrac{(5 - 2x)^{4}}{4 \times (-2)} + c\)
- \(= -\dfrac{(5 - 2x)^{4}}{8} + c\)
- Split: \(\dfrac{x^{2} - 4}{x^{2}} = 1 - 4x^{-2}\).
- Integrate: \(x - \dfrac{4x^{-1}}{-1} + c\).
- \(= x + \dfrac{4}{x} + c\)
- \(y = 2x^{3} - x^{2} + c\).
- At \((2, 9)\): \(9 = 16 - 4 + c\), so \(c = -3\).
- \(y = 2x^{3} - x^{2} - 3\)
- (a) \(y = \dfrac{3x^{\frac{3}{2}}}{\frac{3}{2}} - 6x + c = 2x^{\frac{3}{2}} - 6x + c\).
- At \((4, 0)\): \(0 = 2 \times 8 - 24 + c\), so \(c = 8\): \(y = 2x^{\frac{3}{2}} - 6x + 8\).
- (b) \(3\sqrt{x} = 6\) gives \(x = 4\), \(y = 0\). \(\dfrac{d^{2}y}{dx^{2}} = \tfrac{3}{2}x^{-\frac{1}{2}} = \tfrac{3}{4} > 0\): minimum \((4, 0)\).
- (c) The minimum lies on the \(x\)-axis, so the curve touches the \(x\)-axis at \((4, 0)\) and is otherwise above it.
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.
Find \(\displaystyle\int (4x^{3} - 6x + 5)\,dx\).
- A\(x^{4} - 3x^{2} + 5x + c\)
- B\(12x^{2} - 6 + c\)
- C\(x^{4} - 3x^{2} + c\)
- D\(x^{4} - 6x^{2} + 5x + c\)
Find \(\displaystyle\int \dfrac{8}{x^{3}}\,dx\).
- A\(-\dfrac{4}{x^{2}} + c\)
- B\(\dfrac{4}{x^{2}} + c\)
- C\(-\dfrac{24}{x^{4}} + c\)
- D\(-\dfrac{2}{x^{4}} + c\)
Find \(\displaystyle\int 3\sqrt{x}\,dx\).
- A\(2x^{\frac{3}{2}} + c\)
- B\(\tfrac{9}{2}x^{\frac{3}{2}} + c\)
- C\(\tfrac{3}{2}x^{-\frac{1}{2}} + c\)
- D\(3x^{\frac{3}{2}} + c\)
A curve has \(\dfrac{dy}{dx} = 6x^{2} - 2\) and passes through the point \((2, 5)\). Find the equation of the curve.
- A\(y = 2x^{3} - 2x - 7\)
- B\(y = 2x^{3} - 2x + 17\)
- C\(y = 2x^{3} - 2x - 17\)
- D\(y = 2x^{3} - 2x\)
Find \(\displaystyle\int (3x - 2)^{5}\,dx\).
- A\(\dfrac{(3x - 2)^{6}}{18} + c\)
- B\(\dfrac{(3x - 2)^{6}}{6} + c\)
- C\(\dfrac{(3x - 2)^{6}}{2} + c\)
- D\(15(3x - 2)^{4} + c\)
Find \(\displaystyle\int \dfrac{6}{\sqrt{4x + 1}}\,dx\).
- A\(3\sqrt{4x + 1} + c\)
- B\(12\sqrt{4x + 1} + c\)
- C\(\tfrac{3}{4}\sqrt{4x + 1} + c\)
- D\(-12(4x + 1)^{-\frac{3}{2}} + c\)
A curve has \(\dfrac{dy}{dx} = 2x - \dfrac{4}{x^{2}}\) and passes through the point \((2, 7)\). Find the equation of the curve.
- A\(y = x^{2} + \dfrac{4}{x} + 1\)
- B\(y = x^{2} - \dfrac{4}{x} + 5\)
- C\(y = x^{2} + \dfrac{4}{x} + 7\)
- D\(y = x^{2} + \dfrac{4}{x} + 4\)
A curve has \(\dfrac{d^{2}y}{dx^{2}} = 12x + 2\) and a stationary point at \((1, 3)\). Find the equation of the curve.
- A\(y = 2x^{3} + x^{2} - 8x + 8\)
- B\(y = 2x^{3} + x^{2}\)
- C\(y = 2x^{3} + x^{2} - 5x + 5\)
- D\(y = 2x^{3} + x^{2} + 8x - 8\)
Quiz key
\(4x^{3} \to x^{4}\), \(-6x \to -3x^{2}\), \(5 \to 5x\); add \(c\). \(12x^{2} - 6 + c\) differentiates instead of integrating; \(x^{4} - 3x^{2} + c\) treats the constant 5 as if it vanished (that happens when differentiating, not integrating); \(-6x^{2}\) raises the power but forgets to divide by 2.
\(\displaystyle\int 8x^{-3}\,dx = \dfrac{8x^{-2}}{-2} + c = -\dfrac{4}{x^{2}} + c\). Check by differentiating: \(-4 \times (-2)x^{-3} = 8x^{-3}\). \(\dfrac{4}{x^{2}}\) loses the sign from dividing by \(-2\); \(-\dfrac{24}{x^{4}}\) is the derivative; \(-\dfrac{2}{x^{4}}\) takes the power to \(-4\) (lowering it) instead of raising it to \(-2\).
\(\displaystyle\int 3x^{\frac{1}{2}}\,dx = 3 \times \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} + c = 3 \times \tfrac{2}{3}x^{\frac{3}{2}} + c = 2x^{\frac{3}{2}} + c\). \(\tfrac{9}{2}x^{\frac{3}{2}}\) multiplies by the new power instead of dividing; \(\tfrac{3}{2}x^{-\frac{1}{2}}\) is the derivative; \(3x^{\frac{3}{2}}\) raises the power but never divides.
\(y = 2x^{3} - 2x + c\). At \((2, 5)\): \(5 = 16 - 4 + c\), so \(c = -7\). \(+17\) comes from \(c = 5 + 12\), a sign slip; \(-17\) substitutes \(x = 2\) into \(\dfrac{dy}{dx}\) (giving 22) instead of into \(y\); leaving out \(c\) gives a curve through \((2, 12)\), not \((2, 5)\).
\(\displaystyle\int (3x - 2)^{5}\,dx = \dfrac{(3x - 2)^{6}}{3 \times 6} + c = \dfrac{(3x - 2)^{6}}{18} + c\). Differentiating back: \(\dfrac{6 \times 3(3x - 2)^{5}}{18} = (3x - 2)^{5}\). \(\dfrac{1}{6}\) forgets to divide by the 3; \(\dfrac{1}{2}\) multiplies by 3 instead of dividing (\(\tfrac{3}{6}\)); \(15(3x - 2)^{4}\) is the derivative.
\(\displaystyle\int 6(4x + 1)^{-\frac{1}{2}}\,dx = \dfrac{6(4x + 1)^{\frac{1}{2}}}{4 \times \frac{1}{2}} + c = 3\sqrt{4x + 1} + c\). \(12\sqrt{4x + 1}\) forgets to divide by 4; \(\tfrac{3}{4}\sqrt{4x + 1}\) multiplies by \(\tfrac{1}{2}\) instead of dividing by it; \(-12(4x + 1)^{-\frac{3}{2}}\) is the derivative.
\(y = x^{2} + 4x^{-1} + c = x^{2} + \dfrac{4}{x} + c\). At \((2, 7)\): \(7 = 4 + 2 + c\), so \(c = 1\). \(x^{2} - \dfrac{4}{x} + 5\) does not divide by the new power \(-1\), so the sign is wrong; \(+7\) sets \(c\) equal to the \(y\)-coordinate; \(+4\) substitutes into \(\dfrac{dy}{dx}\) (which is 3 at \(x = 2\)) instead of into \(y\).
\(\dfrac{dy}{dx} = 6x^{2} + 2x + c_{1}\); stationary at \(x = 1\): \(8 + c_{1} = 0\), \(c_{1} = -8\). Then \(y = 2x^{3} + x^{2} - 8x + c_{2}\); at \((1, 3)\): \(3 = 2 + 1 - 8 + c_{2}\), \(c_{2} = 8\). \(2x^{3} + x^{2}\) ignores the first constant; \(-5x + 5\) sets \(\dfrac{dy}{dx} = 3\) (the \(y\)-value) instead of 0; \(+8x - 8\) gets the sign of \(c_{1}\) wrong.
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.