Mathematics · AQA 7357 · Topic H

Integration

Reversing differentiation to find areas and solve problems: standard integrals, substitution, integration by parts, partial fractions and differential equations.

  • 8 sub-topics
  • 1 in Year 12, 5 in Year 13
  • 129 practice questions
  • 8 animated lessons
Practice questions (129) Lessons Worksheets All AQA Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. H1Fundamental theorem of calculusY12
  2. H2Integrating standard functionsY13
  3. H3Definite integrals and areas between curvesY13
  4. H4Integration as the limit of a sumY13
  5. H5Integration by substitution and by partsY13
  6. H6Integration using partial fractionsY13
  7. H7First order separable differential equationsY13
  8. H8Interpret solutions of differential equations in contextY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 129 questions in this topic, free, no account needed.

Integration

Easy

A definite integral · Year 12 · 4 marks

  1. Evaluate ∫12(3x2+2x2)dx\displaystyle\int_1^2 \left(3x^2 + \frac{2}{x^2}\right)\mathrm{d}x.
    (4)
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  1. Write 2x2\frac{2}{x^2} as a power of xx so you can integrate it.∫(3x2+2x−2)dx\int \left(3x^2 + 2x^{-2}\right)\mathrm{d}x
  2. Add 1 to each power and divide by the new power.=[x3−2x−1]12= \left[x^3 - 2x^{-1}\right]_1^2
  3. Substitute the top limit, then subtract the value at the bottom limit.=(8−1)−(1−2)= (8 - 1) - (1 - 2)
  4. Take care with the double negative.=8= 8

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Integration

Medium

Area between a curve and a line · Year 12 · 5 marks

  1. Find the area enclosed by y=4x−x2y = 4x - x^2 and y=xy = x.
    (5)
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  1. Find where they meet.4x−x24x - x^2=x{}= x⇒x{}\Rightarrow x=0,x{}= 0, x=3{}= 3
  2. Top minus bottom: the curve is above the line here.∫03(4x−x2−x) dx\displaystyle\int_0^3 (4x - x^2 - x)\,\mathrm{d}x=∫03(3x−x2) dx{}= \int_0^3 (3x - x^2)\,\mathrm{d}x
  3. [32x2−13x3]03\left[\tfrac{3}{2}x^2 - \tfrac{1}{3}x^3\right]_0^3=13.5−9{}= 13.5 - 9
  4. =4.5= 4.5

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Integration

Not needed for Year 12Hard

Parts with a logarithm · Year 13 · 5 marks

  1. Find the exact value of ∫1exln⁡x dx\displaystyle\int_1^{\mathrm{e}} x\ln x\,\mathrm{d}x.
    (5)
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  1. Choose u=ln⁡xu = \ln x: we can differentiate ln⁡x\ln x but not easily integrate it.u=ln⁡xu = \ln x, v′=xv' = x: 12x2ln⁡x\tfrac{1}{2}x^2\ln x−∫12x dx{}- \displaystyle\int \tfrac{1}{2}x\,\mathrm{d}x
  2. =[12x2ln⁡x−14x2]1e= \left[\tfrac{1}{2}x^2\ln x - \tfrac{1}{4}x^2\right]_1^{\mathrm{e}}
  3. ln⁡e=1\ln \mathrm{e} = 1, ln⁡1=0\ln 1 = 0.=(12e2−14e2)= \left(\tfrac{1}{2}\mathrm{e}^2 - \tfrac{1}{4}\mathrm{e}^2\right)−(0−14){}- \left(0 - \tfrac{1}{4}\right)
  4. =14(e2+1)= \tfrac{1}{4}(\mathrm{e}^2 + 1)

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