This set has twelve original questions, ordered from easier to harder, covering all five lessons in integration methods. Questions 1 to 4 practise powers and constants, 5 to 8 add brackets and definite integrals, and 9 to 12 mix everything with checking.
Attempt each question on paper before opening the answer. Write the integrated expression on its own line, then substitute. Mark the ones you got wrong and use the routing list at the end.
Questions
1. Find ∫(4x³ − 6x² + 5) dx.
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Integrate each term: 4x³ gives x⁴, −6x² gives −2x³, and 5 gives 5x.
x⁴ − 2x³ + 5x + c
Check: differentiating gives 4x³ − 6x² + 5.
2. A curve has gradient dy/dx = 6x − 5 and passes through (2, 7). Find its equation.
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y = 3x² − 5x + c. Substitute (2, 7): 7 = 12 − 10 + c, so c = 5.
y = 3x² − 5x + 5
Check: at x = 2, 12 − 10 + 5 = 7.
3. Find ∫(x² + 3)/x² dx.
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Split first: (x² + 3)/x² = 1 + 3x−2. Integrating, 1 gives x and 3x−2 gives 3x−1/(−1) = −3/x.
x − 3/x + c
Check: differentiating gives 1 + 3x−2.
4. Find ∫(3√x − 2/√x) dx.
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Rewrite: 3x1/2 − 2x−1/2. First term: 3 × (2/3)x3/2 = 2x3/2. Second term: −2 × 2x1/2 = −4x1/2.
2x3/2 − 4√x + c
Check: differentiating gives 3x1/2 − 2x−1/2.
5. Find ∫(2x − 3)5 dx.
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Raise the power: (2x − 3)6/6. Divide by the inner coefficient 2, so the divisor is 12.
(2x − 3)6/12 + c
Check: 6(2x − 3)5 × 2 / 12 = (2x − 3)5.
6. Evaluate ∫ from 1 to 2 of (6x² − 2) dx.
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[2x³ − 2x] from 1 to 2. At x = 2: 16 − 4 = 12. At x = 1: 2 − 2 = 0.
12
7. Given f′(x) = 3x² − 4x and f(−1) = 1, find f(x).
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f(x) = x³ − 2x² + c. Substitute x = −1: (−1)³ − 2(1) + c = 1, so −1 − 2 + c = 1 and c = 4.
f(x) = x³ − 2x² + 4
Check: f(−1) = −1 − 2 + 4 = 1.
8. Evaluate ∫ from −1 to 3 of (x² − 2x) dx.
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[x³/3 − x²] from −1 to 3. At x = 3: 9 − 9 = 0. At x = −1: −1/3 − 1 = −4/3. Subtract: 0 − (−4/3) = 4/3.
4/3
9. Find ∫8e2x dx.
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The integral of e2x is ½e2x. Multiply by 8.
4e2x + c
Check: differentiating gives 4 × 2e2x = 8e2x.
10. Evaluate ∫ from 0 to 1 of (2x + 1)³ dx.
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Antiderivative: (2x + 1)⁴/8. At x = 1: 3⁴/8 = 81/8. At x = 0: 1/8. Subtract: 80/8 = 10.
10
Check: (2x + 1)³ is 1 at x = 0 and 27 at x = 1, so an answer between 1 and 27 is sensible.
11. A curve has gradient (2x − 1)² and passes through (1, 2). Find its equation.
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Integrate: y = (2x − 1)³/6 + c. Substitute (1, 2): 2 = 1/6 + c, so c = 11/6.
y = (2x − 1)³/6 + 11/6
Check: differentiating gives 3(2x − 1)² × 2 / 6 = (2x − 1)².
12. A student claims ∫4x(x² − 3) dx = (x² − 3)²/2 + c. Check it by differentiating, and correct it if needed.
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Differentiate: 2(x² − 3) × 2x / 2 = 2x(x² − 3), which is half of 4x(x² − 3). The claim is wrong.
The correct antiderivative is (x² − 3)² + c, because differentiating it gives 2(x² − 3) × 2x = 4x(x² − 3).
If you got these wrong
- Questions 1, 3, 4, 7: constants, negative powers or fractional powers went wrong. Revisit integrating a power with the correct constant.
- Questions 2, 7, 11: the point was not used correctly to find c. Revisit recovering a function from its derivative and a point.
- Questions 5, 9, 10, 11: the inner coefficient was missed. Revisit using a substitution-shaped reverse derivative.
- Questions 6, 8, 10: limits or negative values were mishandled. Revisit evaluating a definite integral with correct limits.
- Question 12 or any wrong answer you could not explain: revisit checking an antiderivative by differentiation.
The calculus shape and rate explorer, the non-calculator working trainer and the mistake log and retest queue help you keep track of the error types that return.
If the same questions keep catching you out, our teachers can look at your working in online one-to-one Additional Mathematics tuition.