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AS & A Level · Practice questions · Integration as reverse differentiation

Up one power, plus c.

Six original Cambridge-style questions on integration as the reverse of differentiation, including \((ax + b)^{n}\) and finding the constant of integration.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Rewrite, integrate, add c, then fix c.

01
Recall
[2 marks]

Find \(\displaystyle\int (4x^{3} - 6x^{2} + 2)\,dx\).

  • \(\dfrac{4x^{4}}{4} - \dfrac{6x^{3}}{3} + 2x + c\) ✓
  • \(= x^{4} - 2x^{3} + 2x + c\) ✓
02
Calculation
[3 marks]

Find \(\displaystyle\int \left(\dfrac{3}{x^{2}} + \sqrt{x}\right) dx\).

  • \(3x^{-2} + x^{\frac{1}{2}}\) integrates to \(\dfrac{3x^{-1}}{-1} + \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}\). ✓
  • \(= -\dfrac{3}{x} + \tfrac{2}{3}x^{\frac{3}{2}} + c\) ✓
03
Calculation
[2 marks]

Find \(\displaystyle\int (5 - 2x)^{3}\,dx\).

  • \(\dfrac{(5 - 2x)^{4}}{4 \times (-2)} + c\) ✓
  • \(= -\dfrac{(5 - 2x)^{4}}{8} + c\) ✓
04
Calculation
[3 marks]

Find \(\displaystyle\int \dfrac{x^{2} - 4}{x^{2}}\,dx\).

  • Split: \(\dfrac{x^{2} - 4}{x^{2}} = 1 - 4x^{-2}\). ✓
  • Integrate: \(x - \dfrac{4x^{-1}}{-1} + c\). ✓
  • \(= x + \dfrac{4}{x} + c\) ✓
05
Analysis
[4 marks]

A curve has \(\dfrac{dy}{dx} = 6x^{2} - 2x\) and passes through the point \((2, 9)\). Find the equation of the curve.

  • \(y = 2x^{3} - x^{2} + c\). ✓
  • At \((2, 9)\): \(9 = 16 - 4 + c\), so \(c = -3\). ✓
  • \(y = 2x^{3} - x^{2} - 3\) ✓
06
Analysis
[6 marks]

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?

  • (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. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.