TMUA Integration — Practice Questions by Topic

These are SummitPapers original questions, not official past paper questions. Official TMUA questions sorted by topic are on TMUA past papers by topic.

MM7Integration
114 questions78 on Paper 1 · 36 on Paper 2
9 free solutionsThe rest show the correct letter only

Covers: definite integrals, areas between a curve and a line, the trapezium rule, recognising a derivative inside an integral.

Paper 1 (78)Paper 2 (36)Practise this topic

Back to the syllabus · Integration · 1 of 114

Question 1Integration

A curve has gradient dy/dx = 2xy − 6x and passes through (0, 5). Express y in terms of x.

What this topic tests

Integration here means definite integrals, the area between a curve and a line, the trapezium rule, and recognising a derivative inside an integral.

How it is assessed

This is Section 1, on both papers in this set. Each question has five options. A paper is 20 questions in 75 minutes, with no calculator.

This note is written for SummitPapers. The official list of what can be examined is the specification, together with the Notes on Mathematics and, for Paper 2, the Notes on Logic and Proof.

Key methods

A definite integral is a signed area

Where the curve is below the axis, the integral is negative. An area asked for as a region is the absolute value of those pieces, added. A question that says “integral” wants the signed value.

The trapezium rule

With n strips of width h, the rule weights the first and last heights once and the heights in between twice, then multiplies by h/2. Missing the factor 1/2, or the double weight, is a different number.

Common mistakes

Forgetting the chain

The integral of f'(g(x)) g'(x) is f(g(x)). Integrating f'(g(x)) without the inner derivative leaves a missing factor.

Limits in the wrong order

Swapping the limits changes the sign. An area between a curve and a line uses the difference of the functions, with the upper curve first on each piece.

Worked example

This question is also in the list below, with the solution folded. It is opened here so the method is on the page before the other questions.

Find the area of the finite region bounded by the curve y = x² + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.

A. 26/3
B. 8/3
C. 20
D. 6
E. 14

Answer: A. 26/3

Worked solution. On 0 ≤ x ≤ 2 the curve is on or above the x-axis, so the area is the integral of x² + 3x from 0 to 2. An antiderivative is x³/3 + (3/2)x². At x = 2 this is 8/3 + 6 = 8/3 + 18/3 = 26/3, and the value at 0 is 0. The area is 26/3.

Why the other options look right. B integrates x² and drops 3x, leaving 8/3. C multiplies the value of the curve at x = 2 by the width 2, giving (4 + 6) × 2 = 20. D integrates 3x and drops x², leaving (3/2) × 4 = 6. E integrates x² as x³, so the value at 2 is 8 + 6 = 14.

Paper 1 questions

Paper 1 is Applications of Mathematical Knowledge. Calculators are not allowed.

Q1. A curve has gradient dy/dx = 2xy − 6x and passes through (0, 5). Express y in terms of x.
Free · worked solution included
Answer and worked solution
Answer: A. 2e^x² + 3
Worked solution. Factorise the gradient: dy/dx = 2x(y − 3). The curve passes through y = 5, so y ≠ 3 and the variables can be separated: the integral of 1/(y − 3) with respect to y equals the integral of 2x with respect to x, so ln|y − 3| = x² + C and y − 3 = A e^x². Substituting x = 0 and y = 5 gives A + 3 = 5, so A = 2. Therefore y = 2e^x² + 3. Check: dy/dx = 4x e^x² and 2x(y − 3) = 2x × 2e^x² = 4x e^x².
Why the other options look right. B uses the y-value 5 at x = 0 as the coefficient of the exponential instead of solving A + 3 = 5, giving 5e^x² + 3, which passes through (0, 8). C writes e^{x² + C} as e^x² + C, so y − 3 = e^x² + C; the point (0, 5) gives C = 1 and y = e^x² + 4. D factorises 2xy − 6x as 2x(y + 3), so y + 3 = A e^x²; the point (0, 5) gives A = 8 and y = 8e^x² − 3. E integrates 2x as 2x² instead of x², so y − 3 = A e^2x² and the point (0, 5) gives y = 2e^2x² + 3.
Q2. Water enters a tank at 2t² − t litres per minute, where t is the number of minutes after a valve is opened. The rate is positive for t ≥ 1. How many litres enter between t = 1 and t = 4?
Free · worked solution included
Answer and worked solution
Answer: D. 69/2
Worked solution. The amount is the definite integral of 2t² − t from 1 to 4. An antiderivative is (2/3)t³ − (1/2)t². At t = 4 this is 128/3 − 8 = 104/3, and at t = 1 it is 2/3 − 1/2 = 1/6. The difference is 104/3 − 1/6 = 207/6 = 69/2.
Why the other options look right. A evaluates the antiderivative at 4 and subtracts the value at 0, leaving 104/3. B integrates −t as −t², and [(2/3)t³ − t²] from 1 to 4 equals 27. C changes the sign of the integral of −t, and [(2/3)t³ + (1/2)t²] from 1 to 4 equals 99/2. E integrates 2t² as t³, and [t³ − (1/2)t²] from 1 to 4 equals 111/2.
Q3. Find the area of the finite region bounded by the curve y = x² + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.
Free · worked solution included
Answer and worked solution
Answer: A. 26/3
Worked solution. On 0 ≤ x ≤ 2 the curve is on or above the x-axis, so the area is the integral of x² + 3x from 0 to 2. An antiderivative is x³/3 + (3/2)x². At x = 2 this is 8/3 + 6 = 8/3 + 18/3 = 26/3, and the value at 0 is 0. The area is 26/3.
Why the other options look right. B integrates x² and drops 3x, leaving 8/3. C multiplies the value of the curve at x = 2 by the width 2, giving (4 + 6) × 2 = 20. D integrates 3x and drops x², leaving (3/2) × 4 = 6. E integrates x² as x³, so the value at 2 is 8 + 6 = 14.
Q4. The tangent to the curve y = x³ at the point (1, 1) meets the curve again at a second point. What is the area of the region enclosed by the curve and this tangent?
Free · worked solution included
Answer and worked solution
Answer: E. 27/4
Worked solution. The gradient is dy/dx = 3x², which is 3 at x = 1, so the tangent is y − 1 = 3(x − 1), that is y = 3x − 2. The curve and the tangent meet where x³ − 3x + 2 = 0, which factorises as (x − 1)²(x + 2) = 0, so the second point is at x = −2. For −2 < x < 1 the difference x³ − (3x − 2) = (x − 1)²(x + 2) is positive, so the curve lies above the tangent. The area is the integral from −2 to 1 of (x³ − 3x + 2) dx = [x⁴/4 − 3x²/2 + 2x] from −2 to 1 = (1/4 − 3/2 + 2) − (4 − 6 − 4) = 3/4 + 6 = 27/4.
Why the other options look right. A integrates the tangent minus the curve, (3x − 2) − x³, from −2 to 1 and reports the negative signed value as the area. B integrates x³ − 3x + 2 only from 0 to 1. C takes the second meeting point as x = 2, from a sign slip in the factor x + 2, and integrates from 1 to 2. D finds the area between the curve y = x³ and the x-axis for −2 ≤ x ≤ 1, which is 4 + 1/4, instead of the area between the curve and the tangent.
Q5. What is the value of the integral from 0 to 1 of x/(x + 1)² dx?
Free · worked solution included
Answer and worked solution
Answer: D. ln 2 − 1/2
Worked solution. The derivative of ln(x + 1) + 1/(x + 1) is 1/(x + 1) − 1/(x + 1)² = x/(x + 1)². Evaluating from 0 to 1 gives (ln 2 + 1/2) − (0 + 1) = ln 2 − 1/2.
Why the other options look right. A evaluates only ln(x + 1) from 0 to 1. B ignores the numerator x and integrates 1/(x + 1)² from 0 to 1, which gives 1 − 1/2 = 1/2. C forgets to subtract the lower-limit value 1, leaving ln 2 + 1/2. E reverses the limits.
Q6. What is the value of the integral from 0 to π of x sin x dx?
Free · worked solution included
Answer and worked solution
Answer: C. π
Worked solution. Let I be the integral. Replacing x by π − x gives I = the integral from 0 to π of (π − x) sin x dx, because sin(π − x) = sin x. Expanding, I = π times the integral of sin x, minus I. The integral of sin x from 0 to π is 2, so 2I = 2π and I = π.
Why the other options look right. A treats the contributions on either side of π/2 as cancelling. B is the integral of sin x from 0 to π, with the factor x omitted. D is 2I, the value obtained before dividing by 2. E takes that integral of sin x to be the length π, so the reflection step becomes 2I = π × π and I = π²/2.
Q7. What is the value of the integral from 0 to 2 of x / (x² + 1)² dx?
Free · worked solution included
Answer and worked solution
Answer: C. 2/5
Worked solution. Let u = x² + 1, so du = 2x dx. The limits change from x = 0 to x = 2 into u = 1 to u = 5. The integral becomes (1/2) times the integral of u⁻² from 1 to 5, which is (1/2) [ −1/u ] from 1 to 5 = (1/2)(−1/5 + 1) = (1/2)(4/5) = 2/5. The same antiderivative in x is −1/(2(x² + 1)), and evaluating it from 0 to 2 again gives −1/10 − (−1/2) = 2/5.
Why the other options look right. A drops the lower-limit term and evaluates only the antiderivative at the upper limit, −1/(2 × 5) = −1/10. B puts the factor 1/2 in twice, writing the antiderivative as −1/(4(x² + 1)), which changes by −1/20 + 1/4 = 1/5 from 0 to 2. D drops the upper-limit term and computes 0 − (−1/2) = 1/2. E uses −1/(x² + 1) as the antiderivative, omitting the factor 1/2, and that changes by 4/5 from 0 to 2.
Q8. In the expansion of (x + 2/x)⁶, let a be the largest coefficient. The line y = (a/120)x and the curve y = x² enclose one finite region. What is the area of that region?
Free · worked solution included
Answer and worked solution
Answer: B. 4/3
Worked solution. The general term of (x + 2/x)⁶ is C(6, k) x^{6 − k} (2/x)^k = C(6, k) 2^k x^{6 − 2k}. The coefficients C(6, k) 2^k for k = 0 to 6 are 1, 12, 60, 160, 240, 192 and 64, so a = 240. The line is y = 2x. It meets y = x² where x² = 2x, so x = 0 or x = 2. Between 0 and 2 the line is above the curve, and this is the only finite enclosed region. The area is the integral from 0 to 2 of (2x − x²) dx = [x² − x³/3] from 0 to 2 = 4 − 8/3 = 4/3.
Why the other options look right. A takes the line to meet the curve at x = 1 instead of x = 2, and integrates 2x − x² from 0 to 1, which gives 1 − 1/3 = 2/3. C reports the x-coordinate 2 of the second intersection point instead of the area. D is the integral of x² alone from 0 to 2. E is the integral of the line alone, which is [x²] from 0 to 2 = 4.
Q9. Find the area of the region enclosed by the curve y = x² − 4, the x-axis and the lines x = −3 and x = 3.
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Answer
Answer: C. 46/3
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Q10. A curve satisfies dy/dx = 6x² − 4x⁻³ + 2x⁻² for x not equal to 0, and y = 9 when x = 1. Which expression gives y?
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Answer
Answer: C. 2x³ + 2x⁻² − 2x⁻¹ + 7
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Q11. The function f satisfies f(x) > 0 for every real x, and the integral from 1 to 4 of f(x) dx equals K. Which one of the following must be true?
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Answer
Answer: B. The integral from 0 to 3 of [f(x + 1) + 2] dx equals K + 6
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Q12. Evaluate the integral from 1 to 9 of (4 − 3x) / (x√x) dx.
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Answer
Answer: C. −20/3
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Q13. The curve y = f(x) is given by f(x) = x(x − p)(x − q)(r − x), where 0 < p < q < r. The total area enclosed by the curve and the x-axis for 0 ≤ x ≤ r is 24. The integral of f from 0 to r is 2, and the integral from 0 to q is −5. What is the integral of f from p to r?
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Answer
Answer: B. −4
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Q14. The trapezium rule with two strips of equal width is used to estimate the integral from 0 to 2 of g(x) dx. For which of the following functions g is the estimate exactly equal to the integral?
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Answer
Answer: C. g(x) = (x − 1)³
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Q15. Find the area enclosed between the curves y = 8√x and y = x³/4.
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Answer
Answer: D. 80/3
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Q16. For each x, let m(x) be the larger of x² and 3x − 2. What is the integral from 0 to 2 of m(x) dx?
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Answer
Answer: E. 17/6
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Q17. It is given that dy/dx = 1/(√x(1 + √x)²) for x > 0, and that y = 2 when x = 1. What is y when x = 9?
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Answer
Answer: D. 5/2
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Q18. Suppose that 3 times the integral from 0 to 1 of f(x) dx, plus 2 times the integral from 1 to 2 of f(x) dx, equals 5. Also, the integral from 0 to 1 of f(x + 1) dx equals 4. What is the integral from 0 to 2 of f(x) dx?
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Answer
Answer: C. 3
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Q19. The curve y = x² − c, where 0 < c < 9, crosses the positive x-axis at x = √c. Let R be the area between the curve and the x-axis from x = 0 to x = √c, and let S be the area between the curve and the x-axis from x = √c to x = 3. For which value of c are these two areas equal?
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Answer
Answer: A. 3
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Q20. For m > 0, the line y = mx and the curve y = x³ enclose a region in the first quadrant. This region has the same area as the triangle bounded by the line y = mx, the x-axis and the line x = 1. What is the value of m?
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Answer
Answer: C. 2
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Q21. A continuous function f satisfies f(x) + f(4 − x) = x² − 4x + 10 for 0 ≤ x ≤ 4. What is the value of the integral of f(x) from 0 to 4?
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Answer
Answer: A. 44/3
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Q22. The function f is odd, so f(−x) = −f(x) for every x. The integral from 0 to 2 of f(x) dx equals 3, and the integral from 0 to 2 of (f(x))² dx equals 6. What is the integral from −2 to 2 of (f(x) + 1)(f(−x) + 4) dx?
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Answer
Answer: E. 4
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Q23. Use the trapezium rule with 3 strips to estimate the integral from 0 to 3/2 of 8x² dx.
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Answer
Answer: C. 19/2
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Q24. For every integer n ≥ 0, the integral from n to n + 1 of f(x) dx equals 2n + 1. Evaluate the sum from r = 1 to 6 of the integral from 0 to r of f(x) dx.
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Answer
Answer: B. 91
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Q25. Find the finite area enclosed by the curve y = x² − |x| − 6 and the x-axis.
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Answer
Answer: A. 27
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Q26. The integrals from 0 to 2 of (px + q) dx and of x(px + q) dx are 10 and 12 respectively. Find 2p + q.
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Answer
Answer: D. 8
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Q27. For every integer n ≥ 0, the integral of f from n to n + 1 equals 3n + 1. Evaluate the sum of the integrals of f from 0 to 2, from 1 to 2, from 3 to 2, and from 4 to 2.
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Answer
Answer: E. −15
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Q28. The trapezium rule with 2 strips is used to estimate the integral from 0 to 4 of √x dx. What is the positive difference between the estimate and the exact value?
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Answer
Answer: A. 10/3 − 2√2
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Q29. The function F satisfies dF/dx = x|x| for every real x, and F(0) = 0. Which expression is F(x)?
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Answer
Answer: D. |x|³/3
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Q30. As the real number a varies, what is the smallest possible value of the integral from 0 to 1 of (x² − a)² dx?
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Answer
Answer: A. 4/45
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Q31. Evaluate the integral from 0 to 2 of (x² + 1) dx.
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Answer
Answer: D. 14/3
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Q32. Evaluate ∫ from 1 to 3 of 2x dx.
Free · worked solution included
Answer and worked solution
Answer: C. 8
Worked solution. An antiderivative of 2x is x². Evaluating from 1 to 3 gives 3² − 1² = 9 − 1 = 8.
Why the other options look right. A evaluates 2x at the limits without integrating, giving 2(3) − 2(1) = 6 − 2 = 4. B is just the integrand's value at the upper limit, 2 × 3 = 6, instead of a difference of antiderivative values. D is the value of x² at the upper limit only, 3² = 9, ignoring the lower limit. E takes the antiderivative of 2x as 2x², forgetting to divide by the new power 2, so 2(3²) − 2(1²) = 18 − 2 = 16.
Q33. For a positive integer n, let floor(x) be the greatest integer less than or equal to x. The integral from 0 to n of [3ⁿ − log base 3 of ((floor(x) + 1)²)] dx equals which expression?
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Answer
Answer: D. n × 3ⁿ − 2 log base 3 of n!
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Q34. Which of the following integrals has the greatest value?
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Answer
Answer: C. the integral from π/6 to π/3 of 6 sin x dx
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Q35. A function f satisfies the following three equations. The integral from 0 to 7 of f(x) dx equals 9. The integral from −3 to 7 of f(x) dx equals 15. The integral from −7 to 4 of f(−x) dx equals 2. What is the integral from 0 to −4 of f(x) dx?
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Answer
Answer: E. 7
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Q36. The constant k is greater than 1, and the integral from 1 to k of (2x − 3) dx equals 6. Find k.
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Answer
Answer: E. 4
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Q37. Evaluate the definite integral of x(2 − x)² from 0 to 1.
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Answer
Answer: D. 11/12
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Q38. Find the value of the integral from −1 to 1 of (x³ + 6x²) dx.
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Answer
Answer: D. 4
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Q39. Find the value of the integral from 0 to 1 of x²(1 − x)² dx.
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Answer
Answer: C. 1/30
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Q40. Evaluate the integral from −3 to 3 of (x⁵ − 4x³ + 2x²) dx.
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Answer
Answer: C. 36
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Q41. Evaluate the integral from 1 to 2 of x(2 − x)³ dx.
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Answer
Answer: A. 3/10
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Q42. Evaluate the integral of (x³ + 2)/x² with respect to x, from x = 1 to x = 2.
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Answer
Answer: E. 5/2
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Q43. Find the value of the integral from x = 1 to x = 2 of (2x − 1)² dx.
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Answer
Answer: C. 13/3
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Q44. Find the integral of x³ + x² between x = −3 and x = 3.
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Answer
Answer: A. 18
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Q45. Find the integral of x²(1 − x)³ between x = 0 and x = 1.
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Answer
Answer: E. 1/60
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Q46. Evaluate the integral of (2x − 1)(x + 2) from x = 1 to x = 3.
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Answer
Answer: B. 76/3
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Q47. Evaluate the integral of (2x + 1)³ from x = 0 to x = 1.
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Answer
Answer: E. 10
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Q48. Evaluate the integral from −1 to 1 of (x⁵ + 9x²) dx.
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Answer
Answer: C. 6
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Q49. Evaluate the integral from 0 to 1 of x²(1 − x)⁴ dx.
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Answer
Answer: D. 1/105
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Q50. What is the value of the integral from 1 to 4 of (6√x − 5/√x) dx?
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Answer
Answer: B. 18
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Q51. Which of these five integrals has the greatest value?
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Answer
Answer: D. the integral from 0 to 2 of (8 − x) dx
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Q52. The line y = c, where 0 < c < 4, divides the region under y = 4x − x² and above the x-axis into two regions of equal area. What is c?
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Answer
Answer: C. 4 − 4^2/3
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Q53. Find the value of the integral from 1 to 4 of (6√x + 32/x³) dx.
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Answer
Answer: C. 43
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Q54. Evaluate the integral from 0 to 6 of x|x − 3| dx.
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Answer
Answer: A. 27
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Q55. The function f satisfies f(x) > 0 for every real x. The integral of f from 1 to 3 equals P, and the integral of f from 3 to 7 equals Q. Find the integral from −1 to 5 of (f(x + 2) + 3) dx, in terms of P and Q.
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Answer
Answer: C. P + Q + 18
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Q56. What is the value of the integral from −2 to 2 of (x³ + 5x) / (x² + 1) dx?
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Answer
Answer: A. 0
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Q57. Let n be a positive integer, and write floor(t) for the greatest integer less than or equal to t. What is the integral from 1 to 2ⁿ of floor(log₂ x) dx?
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Answer
Answer: D. (n − 2)2ⁿ + 2
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Q58. Which of these integrals has the largest value?
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Answer
Answer: E. the integral from 0 to π/3 of 1 dx
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Q59. A function f satisfies the integral from 0 to 4 of f(x) dx = 10, the integral from −2 to 0 of f(x) dx = 3, and the integral from 2 to 4 of f(−x) dx = 5. What is the integral from −4 to 0 of f(x) dx?
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Answer
Answer: C. 8
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Q60. Find the area of the region bounded by the curve y = √x, the line y = x − 6 and the x-axis.
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Answer
Answer: D. 27/2
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Q61. Find the area of the finite region between the curve y = 2x² − 3 and the line y = x.
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Answer: D. 125/24
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Q62. Find the area of the finite region between the curves y = x² and y = 4 − x².
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Answer
Answer: E. 16√2/3
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Q63. Find the area of the finite region bounded by the curve y = 8/x², the line y = x and the line x = 4.
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Answer: C. 4
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Q64. A line is tangent to the curve y = √x at the point (a², a), where a > 0. Find the area of the finite region bounded by the curve, the tangent and the x-axis.
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Answer: E. a³/3
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Q65. The area of the finite region between the parabolas y = x² + 3ax + a and y = a − 2x² equals 4. Find the possible values of a.
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Answer: A. a = ±2
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Q66. Find the area of the finite region between the curves y = 12 − x² and y = |x|.
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Answer: C. 45
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Q67. Evaluate ∫ from −1 to 2 of [3(x + |x|) − 2x|x|] dx.
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Answer: E. 22/3
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Q68. The positive number k satisfies ∫ from 0 to k of (√x + x²) dx = 8/3. Find k.
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Answer: B. 2^2/3
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Q69. Let f(x) = ∫ from 0 to x of 2t dt, and let g(x) = ∫ from 0 to 1 of xt dt. Which statement is true?
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Answer: D. g(f(p)) > f(g(p)) for every p > 0
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Q70. For 0 ≤ t ≤ 3, define f(t) = ∫ from 0 to 3 of |x − t| dx. Find the minimum value of f(t).
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Answer: C. 9/4
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Q71. The trapezium rule with three trapezia is used to estimate ∫ from 0 to 6 of |x(x − 2)(x − 5)| dx. What value does the rule give?
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Answer: B. 40
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Q72. Find the total area of the two finite regions enclosed between the curve y = x(x − a)(x − 3a), where a > 0, and the x-axis.
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Answer: A. 37a⁴/12
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Q73. The function f satisfies f(x) + 2f(−x) = 1 + x² ∫ from −1 to 1 of f(u) du for every x. Find ∫ from −1 to 1 of f(x) dx.
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Answer: A. 6/7
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Q74. Place the following integrals in order of size, from smallest to largest. P = ∫ from 0 to 1 of e^x dx, Q = ∫ from 0 to 1 of e^x/2 dx, R = ∫ from 0 to 1 of e^2x dx.
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Answer: C. Q < P < R
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Q75. Let f(x) = | |x − 1| − |x − 5| |, the absolute value of |x − 1| − |x − 5|. What is the integral from 0 to 6 of f(x) dx?
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Answer: D. 16
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Q76. The continuous function f satisfies f(x) = f(6 − x) for every real x. Suppose (the integral from 2 to 3 of f)² + (the integral from 4 to 6 of f)² + (the integral from 2 to 4 of f)(the integral from 0 to 2 of f) − 5(the integral from 3 to 6 of f) + 6 = 0. What is the sum of the possible values of the integral from 0 to 3 of f?
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Answer: C. 5
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Q77. Let {x} denote the fractional part of x. What is the integral from 0 to 3 of {2x} dx?
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Answer: C. 3/2
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Q78. The integral from 0 to a of (2x + 4) dx equals 12, and a > 0. What is a?
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Answer: B. 2
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Paper 2 questions

Paper 2 is Mathematical Reasoning. It can test this same topic. Argument, proof, and identifying errors are the topics that appear on Paper 2 only.

Q1. Evaluate the integral from 1 to 2 of (x² − 3/x²)² dx.
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Answer: C. 113/40
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Q2. The function f is increasing and f(0) = 0. The positive constants a and b satisfy a < b. The area of the region enclosed by y = f(x), the x-axis and the lines x = a and x = b is R. Define g(x) = f(x) + 3f(b). What is the area enclosed by y = g(x), the x-axis and the lines x = a and x = b?
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Answer: C. R + 3(b − a)f(b)
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Q3. The trapezium rule with 6 strips is used to approximate each of the following integrals. Which approximation is an overestimate?
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Answer: B. the integral from π/4 to π/2 of cos² x dx
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Q4. The function f is continuous and strictly increasing for 0 ≤ x ≤ 2. Which of the following must be true? I: the integral from 0 to 1 of f(x) dx is less than the integral from 1 to 2 of f(x) dx. II: the integral from 0 to 2 of f(x) dx is at most 2f(1). III: 2f(0) < the integral from 0 to 2 of f(x) dx < 2f(2).
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Answer: A. I and III only
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Q5. The function f is continuous for all real x, and F(x) = the integral from 0 to x of f(t) dt. Which completion is correct? F has a local maximum at x = 2 ...
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Answer: E. ... only if f(2) = 0.
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Q6. A continuous function f satisfies the integral from 0 to 2 of (f(x))² dx, plus 2, equals twice the integral from 0 to 2 of f(x) dx. Which of the following must be true? I: f(x) = 1 for every x with 0 ≤ x ≤ 2. II: the integral from 0 to 2 of f(x) dx equals 2. III: the integral from 0 to 2 of (f(x))² dx equals 4.
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Answer: A. I and II only
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Q7. Find the value of the integral from 1 to 4 of (√x − 2)²/√x dx.
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Answer: A. 2/3
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Q8. Which of the following statements about polynomials f and g are true? I: If the integral from 0 to 1 of f(t) dt is at least the integral from 0 to 1 of g(t) dt, then f(x) ≥ g(x) for some x with 0 ≤ x ≤ 1. II: If f(0) = g(0) and f'(x) ≥ g'(x) for all x ≥ 0, then f(x) ≥ g(x) for all x ≥ 0. III: If f(x) ≥ g(x) for all x ≥ 0, then f(x)² ≥ g(x)² for all x ≥ 0.
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Answer: E. I and II only
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Q9. A sequence of functions is defined by f₁(x) = 1 and f_n+1(x) = 1 + (the integral from 0 to x of f_n(t) dt) for n ≥ 1. What is the integral from 0 to 2 of f₄(x) dx?
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Answer: B. 6
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Q10. Let P be the integral from 0 to 1 of 2^x² dx, Q the integral from 0 to 1 of 2^x dx, and R the integral from 0 to 1 of (1 + x) dx. Which statement is correct?
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Answer: A. P < Q < R
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Q11. Evaluate the integral from 1 to 4 of (x + 1/√x)² dx, minus the integral from 1 to 4 of (x − 1/√x)² dx.
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Answer: B. 56/3
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Q12. The ceiling of x, written ceil(x), is x rounded up to the nearest integer. For example, ceil(π) = 4 and ceil(8) = 8. What is the value of the integral from 0 to 6 of 3^ceil(x) dx?
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Answer: A. 1092
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Q13. Let f be a polynomial with real coefficients. For p < q, define J(p, q) to be the integral from p to q of (f(x) + f(−x)) dx. Which of these statements must be true? 1: If f is odd, then J(p, q) = 0 whenever p < q. 2: If f(x) ≥ 0 for every x ≥ 0, then J(p, q) ≥ 0 whenever 0 ≤ p < q. 3: J(p, q) = 0 for every p < q only if f is odd.
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Answer: E. 1 and 3 only
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Q14. The integral of (1 − |x|)(2 − x) with respect to x, from −1 to 1, equals
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Answer: D. 2
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Q15. The trapezium rule with four strips is used to estimate the integral from 1 to 5 of |x(x − 3)(x − 5)| dx. The estimate equals
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Answer: B. 14
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Q16. Let floor(t) denote the greatest integer less than or equal to t. The value of the integral from 0 to 6 of (x + floor(x/3)) dx is
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Answer: C. 21
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Q17. Evaluate the integral from −1 to 3 of (|x| + |x − 2|) dx.
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Answer: D. 10
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Q18. The trapezium rule with 3 strips of equal width is applied on the interval [0, 3] to the integral of x² + k, where k is a constant, and gives the estimate 31/2. What is the exact value of the integral from 0 to 3 of (x² + k) dx?
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Answer: A. 15
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Q19. Evaluate the integral from 0 to 6 of |x − 2| dx.
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Answer: D. 10
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Q20. A function f has f(0) = 5, f(0.5) = 4, f(1) = 2, f(1.5) = 2 and f(2) = 1. Using only these values, the trapezium rule with 4 strips of equal width is applied to the integral from 0 to 2 of x f(x) dx. What estimate does it give?
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Answer: A. 4
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Q21. Evaluate the integral from 0 to 3 of |x² − 4| dx.
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Answer: B. 23/3
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Q22. For each positive integer n, let T_n be the estimate of the integral from 0 to 1 of x² dx given by the trapezium rule with n strips of equal width. What is the smallest n for which T_n − 1/3 < 1/100?
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Answer: C. 5
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Q23. What is the exact value of the integral of |x² − 2x| from x = 0 to x = 4?
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Answer: C. 8
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Q24. A function f takes the values f(0) = 3, f(1) = 5, f(2) = 4, f(3) = 6 and f(4) = 2. The integral of f(x) from 0 to 4 is estimated twice with the trapezium rule: once with 4 strips of equal width and once with 2 strips of equal width. By how much does the 4-strip estimate exceed the 2-strip estimate?
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Answer: E. 9/2
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Q25. Find the value of the integral from 0 to 5 of |x − 1| dx.
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Answer: C. 17/2
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Q26. The trapezium rule with 4 strips of equal width is used to approximate the integral of 2^x from 0 to 2. What value does the rule give?
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Answer: A. (9 + 6√2)/4
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Q27. Evaluate the integral of |x − 4| from x = 0 to x = 7.
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Answer: B. 25/2
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Q28. The trapezium rule with two strips of equal width is used to estimate the integral of √(x³ + 1) over the interval from 0 to 2. What estimate does it give?
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Answer: E. 2 + √2
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Q29. Evaluate the integral from −1 to 6 of |x − 1| dx.
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Answer: C. 29/2
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Q30. The trapezium rule with 4 strips of equal width is used to estimate the integral from 0 to π of sin x dx. What value does the rule give?
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Answer: A. π(1 + √2)/4
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Q31. The integral from 0 to x of (|t − 1| − 2) dt is zero for exactly one positive value of x. What is that value?
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Answer: A. 3 + √7
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Q32. The function f satisfies the integral from 0 to 2 of f(x) dx = 5, the integral from 1 to 4 of f(x) dx = 9, and the integral from 2 to 4 of f(x) dx = 6. What is the integral from 0 to 1 of f(x) dx?
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Answer: B. 2
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Q33. Which of these integrals has the largest value? You do not need to find the exact value of every integral.
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Answer: C. the integral from 0 to π of (2 + cos x) dx
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Q34. What is the value of the integral from 0 to 3 of |x − 1|(x − 2) dx?
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Answer: E. −1/6
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Q35. The trapezium rule with four trapezia is applied to the integral from 0 to 4 of |x² − 2x − 3| dx. What value does it give?
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Answer: C. 11
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Q36. Let floor(t) denote the greatest integer less than or equal to t. What is the integral from 0 to 2 of floor(x²) dx?
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Answer: C. 5 − √2 − √3
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A curve has gradient dy/dx = 2xy − 6x and passes through (0, 5). Express y in terms of x.

What this topic tests

How it is assessed

Key methods

A definite integral is a signed area

The trapezium rule

Common mistakes

Forgetting the chain

Limits in the wrong order

Worked example

Find the area of the finite region bounded by the curve y = x2 + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.

Paper 1 questions

Q1. A curve has gradient dy/dx = 2xy − 6x and passes through (0, 5). Express y in terms of x.

Q2. Water enters a tank at 2t2 − t litres per minute, where t is the number of minutes after a valve is opened. The rate is positive for t ≥ 1. How many litres enter between t = 1 and t = 4?

Q3. Find the area of the finite region bounded by the curve y = x2 + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.

Q4. The tangent to the curve y = x3 at the point (1, 1) meets the curve again at a second point. What is the area of the region enclosed by the curve and this tangent?

Q5. What is the value of the integral from 0 to 1 of x/(x + 1)2 dx?

Q6. What is the value of the integral from 0 to π of x sin x dx?

Q7. What is the value of the integral from 0 to 2 of x / (x2 + 1)2 dx?

Q8. In the expansion of (x + 2/x)6, let a be the largest coefficient. The line y = (a/120)x and the curve y = x2 enclose one finite region. What is the area of that region?

Q9. Find the area of the region enclosed by the curve y = x2 − 4, the x-axis and the lines x = −3 and x = 3.

Q10. A curve satisfies dy/dx = 6x2 − 4x−3 + 2x−2 for x not equal to 0, and y = 9 when x = 1. Which expression gives y?

Q11. The function f satisfies f(x) > 0 for every real x, and the integral from 1 to 4 of f(x) dx equals K. Which one of the following must be true?

Q12. Evaluate the integral from 1 to 9 of (4 − 3x) / (x√x) dx.

Q13. The curve y = f(x) is given by f(x) = x(x − p)(x − q)(r − x), where 0 < p < q < r. The total area enclosed by the curve and the x-axis for 0 ≤ x ≤ r is 24. The integral of f from 0 to r is 2, and the integral from 0 to q is −5. What is the integral of f from p to r?

Q14. The trapezium rule with two strips of equal width is used to estimate the integral from 0 to 2 of g(x) dx. For which of the following functions g is the estimate exactly equal to the integral?

Q15. Find the area enclosed between the curves y = 8√x and y = x³/4.

Q16. For each x, let m(x) be the larger of x² and 3x − 2. What is the integral from 0 to 2 of m(x) dx?

Q17. It is given that dy/dx = 1/(√x(1 + √x)²) for x > 0, and that y = 2 when x = 1. What is y when x = 9?

Q18. Suppose that 3 times the integral from 0 to 1 of f(x) dx, plus 2 times the integral from 1 to 2 of f(x) dx, equals 5. Also, the integral from 0 to 1 of f(x + 1) dx equals 4. What is the integral from 0 to 2 of f(x) dx?

Q19. The curve y = x2 − c, where 0 < c < 9, crosses the positive x-axis at x = √c. Let R be the area between the curve and the x-axis from x = 0 to x = √c, and let S be the area between the curve and the x-axis from x = √c to x = 3. For which value of c are these two areas equal?

Q20. For m > 0, the line y = mx and the curve y = x3 enclose a region in the first quadrant. This region has the same area as the triangle bounded by the line y = mx, the x-axis and the line x = 1. What is the value of m?

Q21. A continuous function f satisfies f(x) + f(4 − x) = x2 − 4x + 10 for 0 ≤ x ≤ 4. What is the value of the integral of f(x) from 0 to 4?

Q22. The function f is odd, so f(−x) = −f(x) for every x. The integral from 0 to 2 of f(x) dx equals 3, and the integral from 0 to 2 of (f(x))² dx equals 6. What is the integral from −2 to 2 of (f(x) + 1)(f(−x) + 4) dx?

Q23. Use the trapezium rule with 3 strips to estimate the integral from 0 to 3/2 of 8x2 dx.

Q24. For every integer n ≥ 0, the integral from n to n + 1 of f(x) dx equals 2n + 1. Evaluate the sum from r = 1 to 6 of the integral from 0 to r of f(x) dx.

Q25. Find the finite area enclosed by the curve y = x2 − |x| − 6 and the x-axis.

Q26. The integrals from 0 to 2 of (px + q) dx and of x(px + q) dx are 10 and 12 respectively. Find 2p + q.

Q27. For every integer n ≥ 0, the integral of f from n to n + 1 equals 3n + 1. Evaluate the sum of the integrals of f from 0 to 2, from 1 to 2, from 3 to 2, and from 4 to 2.

Q28. The trapezium rule with 2 strips is used to estimate the integral from 0 to 4 of √x dx. What is the positive difference between the estimate and the exact value?

Q29. The function F satisfies dF/dx = x|x| for every real x, and F(0) = 0. Which expression is F(x)?

Q30. As the real number a varies, what is the smallest possible value of the integral from 0 to 1 of (x2 − a)2 dx?

Q31. Evaluate the integral from 0 to 2 of (x² + 1) dx.

Q32. Evaluate ∫ from 1 to 3 of 2x dx.

Q33. For a positive integer n, let floor(x) be the greatest integer less than or equal to x. The integral from 0 to n of [3n − log base 3 of ((floor(x) + 1)2)] dx equals which expression?

Q34. Which of the following integrals has the greatest value?

Q35. A function f satisfies the following three equations. The integral from 0 to 7 of f(x) dx equals 9. The integral from −3 to 7 of f(x) dx equals 15. The integral from −7 to 4 of f(−x) dx equals 2. What is the integral from 0 to −4 of f(x) dx?

Q36. The constant k is greater than 1, and the integral from 1 to k of (2x − 3) dx equals 6. Find k.

Q37. Evaluate the definite integral of x(2 − x)² from 0 to 1.

Q38. Find the value of the integral from −1 to 1 of (x³ + 6x²) dx.

Q39. Find the value of the integral from 0 to 1 of x²(1 − x)² dx.

Q40. Evaluate the integral from −3 to 3 of (x⁵ − 4x³ + 2x²) dx.

Q41. Evaluate the integral from 1 to 2 of x(2 − x)³ dx.

Q42. Evaluate the integral of (x³ + 2)/x² with respect to x, from x = 1 to x = 2.

Q43. Find the value of the integral from x = 1 to x = 2 of (2x − 1)² dx.

Q44. Find the integral of x³ + x² between x = −3 and x = 3.

Q45. Find the integral of x²(1 − x)³ between x = 0 and x = 1.

Q46. Evaluate the integral of (2x − 1)(x + 2) from x = 1 to x = 3.

Q47. Evaluate the integral of (2x + 1)³ from x = 0 to x = 1.

Q48. Evaluate the integral from −1 to 1 of (x⁵ + 9x²) dx.

Q49. Evaluate the integral from 0 to 1 of x²(1 − x)⁴ dx.

Q50. What is the value of the integral from 1 to 4 of (6√x − 5/√x) dx?

Q51. Which of these five integrals has the greatest value?

Q52. The line y = c, where 0 < c < 4, divides the region under y = 4x − x2 and above the x-axis into two regions of equal area. What is c?

Q53. Find the value of the integral from 1 to 4 of (6√x + 32/x3) dx.

Q54. Evaluate the integral from 0 to 6 of x|x − 3| dx.

Q55. The function f satisfies f(x) > 0 for every real x. The integral of f from 1 to 3 equals P, and the integral of f from 3 to 7 equals Q. Find the integral from −1 to 5 of (f(x + 2) + 3) dx, in terms of P and Q.

Q56. What is the value of the integral from −2 to 2 of (x3 + 5x) / (x2 + 1) dx?

Q57. Let n be a positive integer, and write floor(t) for the greatest integer less than or equal to t. What is the integral from 1 to 2n of floor(log2 x) dx?

Q58. Which of these integrals has the largest value?

Q59. A function f satisfies the integral from 0 to 4 of f(x) dx = 10, the integral from −2 to 0 of f(x) dx = 3, and the integral from 2 to 4 of f(−x) dx = 5. What is the integral from −4 to 0 of f(x) dx?

Q60. Find the area of the region bounded by the curve y = √x, the line y = x − 6 and the x-axis.

Q61. Find the area of the finite region between the curve y = 2x² − 3 and the line y = x.

Q62. Find the area of the finite region between the curves y = x² and y = 4 − x².

Q63. Find the area of the finite region bounded by the curve y = 8/x², the line y = x and the line x = 4.

Q64. A line is tangent to the curve y = √x at the point (a², a), where a > 0. Find the area of the finite region bounded by the curve, the tangent and the x-axis.

Q65. The area of the finite region between the parabolas y = x² + 3ax + a and y = a − 2x² equals 4. Find the possible values of a.

Q66. Find the area of the finite region between the curves y = 12 − x² and y = |x|.

Q67. Evaluate ∫ from −1 to 2 of [3(x + |x|) − 2x|x|] dx.

Q68. The positive number k satisfies ∫ from 0 to k of (√x + x²) dx = 8/3. Find k.

Find the area of the finite region bounded by the curve y = x² + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.

Q2. Water enters a tank at 2t² − t litres per minute, where t is the number of minutes after a valve is opened. The rate is positive for t ≥ 1. How many litres enter between t = 1 and t = 4?

Q3. Find the area of the finite region bounded by the curve y = x² + 3x, the x-axis and the line x = 2. Take the region where 0 ≤ x ≤ 2.

Q4. The tangent to the curve y = x³ at the point (1, 1) meets the curve again at a second point. What is the area of the region enclosed by the curve and this tangent?

Q5. What is the value of the integral from 0 to 1 of x/(x + 1)² dx?

Q7. What is the value of the integral from 0 to 2 of x / (x² + 1)² dx?

Q8. In the expansion of (x + 2/x)⁶, let a be the largest coefficient. The line y = (a/120)x and the curve y = x² enclose one finite region. What is the area of that region?

Q9. Find the area of the region enclosed by the curve y = x² − 4, the x-axis and the lines x = −3 and x = 3.

Q10. A curve satisfies dy/dx = 6x² − 4x⁻³ + 2x⁻² for x not equal to 0, and y = 9 when x = 1. Which expression gives y?

Q19. The curve y = x² − c, where 0 < c < 9, crosses the positive x-axis at x = √c. Let R be the area between the curve and the x-axis from x = 0 to x = √c, and let S be the area between the curve and the x-axis from x = √c to x = 3. For which value of c are these two areas equal?

Q20. For m > 0, the line y = mx and the curve y = x³ enclose a region in the first quadrant. This region has the same area as the triangle bounded by the line y = mx, the x-axis and the line x = 1. What is the value of m?

Q21. A continuous function f satisfies f(x) + f(4 − x) = x² − 4x + 10 for 0 ≤ x ≤ 4. What is the value of the integral of f(x) from 0 to 4?

Q23. Use the trapezium rule with 3 strips to estimate the integral from 0 to 3/2 of 8x² dx.

Q25. Find the finite area enclosed by the curve y = x² − |x| − 6 and the x-axis.

Q30. As the real number a varies, what is the smallest possible value of the integral from 0 to 1 of (x² − a)² dx?

Q33. For a positive integer n, let floor(x) be the greatest integer less than or equal to x. The integral from 0 to n of [3ⁿ − log base 3 of ((floor(x) + 1)²)] dx equals which expression?

Q52. The line y = c, where 0 < c < 4, divides the region under y = 4x − x² and above the x-axis into two regions of equal area. What is c?

Q53. Find the value of the integral from 1 to 4 of (6√x + 32/x³) dx.

Q56. What is the value of the integral from −2 to 2 of (x³ + 5x) / (x² + 1) dx?

Q57. Let n be a positive integer, and write floor(t) for the greatest integer less than or equal to t. What is the integral from 1 to 2ⁿ of floor(log₂ x) dx?

Q74. Place the following integrals in order of size, from smallest to largest. P = ∫ from 0 to 1 of e^x dx, Q = ∫ from 0 to 1 of e^x/2 dx, R = ∫ from 0 to 1 of e^2x dx.

Q1. Evaluate the integral from 1 to 2 of (x² − 3/x²)² dx.

Q7. Find the value of the integral from 1 to 4 of (√x − 2)²/√x dx.

Q9. A sequence of functions is defined by f₁(x) = 1 and f_n+1(x) = 1 + (the integral from 0 to x of f_n(t) dt) for n ≥ 1. What is the integral from 0 to 2 of f₄(x) dx?

Q10. Let P be the integral from 0 to 1 of 2^x² dx, Q the integral from 0 to 1 of 2^x dx, and R the integral from 0 to 1 of (1 + x) dx. Which statement is correct?

Q12. The ceiling of x, written ceil(x), is x rounded up to the nearest integer. For example, ceil(π) = 4 and ceil(8) = 8. What is the value of the integral from 0 to 6 of 3^ceil(x) dx?

Q22. For each positive integer n, let T_n be the estimate of the integral from 0 to 1 of x² dx given by the trapezium rule with n strips of equal width. What is the smallest n for which T_n − 1/3 < 1/100?

Q26. The trapezium rule with 4 strips of equal width is used to approximate the integral of 2^x from 0 to 2. What value does the rule give?

Q35. The trapezium rule with four trapezia is applied to the integral from 0 to 4 of |x² − 2x − 3| dx. What value does it give?

Q36. Let floor(t) denote the greatest integer less than or equal to t. What is the integral from 0 to 2 of floor(x²) dx?