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Cambridge International AS & A Level · Topic 1.8
A-Level Maths 9709 / Topic 1.8 / AS & A Level

Running the derivative backwards.

Integration undoes differentiation: given the gradient function, recover the curve. Add one to the power, divide by the new power, and never forget the constant, because a whole family of curves shares the same gradient.

Topic 1.8All integration lessons →
The key idea

\(\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + c\) 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 \(c\).

x y −11 c = −2c = 0c = 2 c = −4 (1, −2)
Fig. 1: every curve y = x² + x + c has gradient 2x + 1; only c = −4 passes through (1, −2)
Section 01

Undoing a derivative.

Differentiating takes the power down and reduces it by one. Integrating does the reverse: increase the power by one, then divide by the new power. Check any answer by differentiating it.

Worked example A polynomial
Find \(\displaystyle\int (2x^{3} - 5x + 1)\,dx\).
  1. \(2x^{3} \to \dfrac{2x^{4}}{4} = \dfrac{x^{4}}{2}\).
  2. \(-5x \to -\dfrac{5x^{2}}{2}\), and \(1 \to x\).
  3. Add the constant of integration.
Answer: \(\dfrac{x^{4}}{2} - \dfrac{5x^{2}}{2} + x + c\)
Worked example Roots and reciprocals
Find \(\displaystyle\int \left(6\sqrt{x} - \dfrac{4}{x^{3}}\right) dx\).
  1. Rewrite: \(6x^{\frac{1}{2}} - 4x^{-3}\).
  2. \(6x^{\frac{1}{2}} \to 6 \times \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} = 4x^{\frac{3}{2}}\).
  3. \(-4x^{-3} \to -4 \times \dfrac{x^{-2}}{-2} = 2x^{-2}\).
  4. Check: differentiating \(4x^{\frac{3}{2}} + 2x^{-2}\) gives \(6x^{\frac{1}{2}} - 4x^{-3}\).
Answer: \(4x^{\frac{3}{2}} + \dfrac{2}{x^{2}} + c\)
Examiner trap

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).

Section 02

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 \(x\). Differentiating the answer brings the factor \(a\) back, cancelling the division.

Worked example Positive and negative powers
Find (a) \(\displaystyle\int (2x + 1)^{3}\,dx\) (b) \(\displaystyle\int \dfrac{1}{(3x - 2)^{2}}\,dx\).
  1. (a) \(\dfrac{(2x + 1)^{4}}{4 \times 2} = \dfrac{(2x + 1)^{4}}{8}\).
  2. (b) Rewrite as \((3x - 2)^{-2}\). New power \(-1\); divide by \(-1\) and by \(3\).
  3. \(\dfrac{(3x - 2)^{-1}}{-1 \times 3} = -\dfrac{1}{3(3x - 2)}\).
Answer: (a) \(\dfrac{(2x + 1)^{4}}{8} + c\)   (b) \(-\dfrac{1}{3(3x - 2)} + c\)
Your turn
Find \(\displaystyle\int (4x - 3)^{5}\,dx\).
Examiner trap

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.

Stage 1 · Learn

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Find \(\displaystyle\int x^{4}\,dx\).

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Find \(\displaystyle\int 6x^{2}\,dx\).

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Find \(\displaystyle\int x^{-2}\,dx\).

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Which curve has gradient \(2x\) and passes through \((0, 5)\)?

Section 03

Finding the constant.

The gradient function alone gives a family of curves (Fig. 1). One known point on the curve picks out the right member: substitute it and solve for \(c\).

Worked example A curve through a point
A curve has \(\dfrac{dy}{dx} = 2x + 1\) and passes through \((1, -2)\). Find its equation.
  1. \(y = x^{2} + x + c\).
  2. Substitute \((1, -2)\): \(-2 = 1 + 1 + c\), so \(c = -4\).
Answer: \(y = x^{2} + x - 4\)
Worked example With a bracket
A curve has \(\dfrac{dy}{dx} = \dfrac{3}{\sqrt{2x + 1}}\) and passes through \((4, 5)\). Find its equation.
  1. \(\dfrac{dy}{dx} = 3(2x + 1)^{-\frac{1}{2}}\).
  2. Integrate: \(y = \dfrac{3(2x + 1)^{\frac{1}{2}}}{\frac{1}{2} \times 2} + c = 3\sqrt{2x + 1} + c\).
  3. Substitute \((4, 5)\): \(5 = 3\sqrt{9} + c = 9 + c\), so \(c = -4\).
Answer: \(y = 3\sqrt{2x + 1} - 4\)
Worked example From the second derivative
A curve has \(\dfrac{d^{2}y}{dx^{2}} = 6x - 4\) and a stationary point at \((1, 2)\). Find its equation.
  1. Integrate once: \(\dfrac{dy}{dx} = 3x^{2} - 4x + c_{1}\).
  2. Stationary at \(x = 1\) means \(\dfrac{dy}{dx} = 0\): \(3 - 4 + c_{1} = 0\), so \(c_{1} = 1\).
  3. Integrate again: \(y = x^{3} - 2x^{2} + x + c_{2}\).
  4. Through \((1, 2)\): \(1 - 2 + 1 + c_{2} = 2\), so \(c_{2} = 2\).
Answer: \(y = x^{3} - 2x^{2} + x + 2\)
Examiner trap

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.

Stage 2 · Exam

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Find \(\displaystyle\int (2x + 1)^{4}\,dx\).

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A curve has \(\dfrac{dy}{dx} = \dfrac{3}{\sqrt{2x + 1}}\) and passes through \((4, 5)\). Find the equation of the curve.

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Find \(\displaystyle\int \left(6\sqrt{x} - \dfrac{4}{x^{3}}\right) dx\).

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A curve has \(\dfrac{d^{2}y}{dx^{2}} = 6x - 4\) and a stationary point at \((1, 2)\). Find the equation of the curve.

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Six original Cambridge-style questions
Integrating powers and linear brackets, splitting a fraction, and finding the equation of a curve from its gradient.