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Admissions tests / TMUA / Paper 1 / Integration

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Integration, set 2

Indefinite and definite integration, areas between a curve and an axis, areas between two curves, and integration as the reverse of differentiation.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    The curve y = f(x) satisfies f(x) <= 0 for 1 <= x <= 3, and f(x) >= 0 for 3 <= x <= 8. Given that the integral from 1 to 3 of f(x) dx = -6, and the integral from 3 to 8 of f(x) dx = 15, find the area enclosed between the curve and the x-axis for 1 <= x <= 8.

    1. A 9
    2. B 21
    3. C -9
    4. D 15
  2. 21 mark

    Find the total area enclosed between the curve y = x^2 - 4x and the x-axis, between x = 0 and x = 5.

    1. A -25/3
    2. B 25/3
    3. C 7/3
    4. D 13
  3. 31 mark

    Find integral of (6x^2 - 4/sqrt(x)) dx.

    1. A 2x^3 - 8 sqrt(x) + C
    2. B 2x^3 - 2 sqrt(x) + C
    3. C 6x^3 - 8 sqrt(x) + C
    4. D 2x^3 + 8 sqrt(x) + C
  4. 41 mark

    Find the value of the integral from 1 to 4 of (x - 4)/sqrt(x) dx.

    1. A 38/3
    2. B -14
    3. C -10/3
    4. D 10/3
  5. 51 mark

    Let k(x) = integral from x to 4 of (t^2 + 3) dt. Find k'(1).

    1. A 4
    2. B -19
    3. C -1
    4. D -4
  6. 61 mark

    It is known that the integral from 0 to x of f(t) dt = x^4 - 3x^2 for every value of x. Find f(2).

    1. A 4
    2. B 20
    3. C -4
    4. D 44
  7. 71 mark

    Given that the integral from 0 to 3 of f(x) dx = 5, the integral from 3 to 7 of f(x) dx = -2, and the integral from 7 to 10 of f(x) dx = 6, find the integral from 0 to 10 of f(x) dx.

    1. A 13
    2. B 11
    3. C 9
    4. D 3
  8. 81 mark

    Given that the integral from 2 to 9 of f(x) dx = 15, find the integral from 9 to 2 of 4f(x) dx.

    1. A -60
    2. B 60
    3. C -15
    4. D 15
  9. 91 mark

    Use the trapezium rule with strips of width 2 to estimate the integral from 0 to 6 of (x^2 + 1) dx, using the four ordinates at x = 0, 2, 4, 6.

    1. A 164
    2. B 82
    3. C 60
    4. D 114
  10. 101 mark

    The curve y = x^2 is convex (it curves upward, like the inside of a bowl) on the interval 2 <= x <= 6. Which of the following correctly describes a trapezium-rule estimate of the integral from 2 to 6 of x^2 dx?

    1. A The trapezium rule gives an underestimate, because the chords joining points on the curve lie below the curve.
    2. B The trapezium rule gives the exact value, because x^2 is a polynomial and the rule is exact for the integral of any polynomial.
    3. C It is impossible to say whether it is an overestimate or an underestimate without calculating both values numerically.
    4. D The trapezium rule gives an overestimate, because the chords joining points on the curve lie above the curve.
  11. 111 mark

    A curve satisfies dy/dx = 6x^2 - 4x + 1, and y = 5 when x = 2. Find the value of y when x = -1.

    1. A -10
    2. B -5
    3. C -4
    4. D 0
  12. 121 mark

    A curve satisfies dy/dx = 3 sqrt(x) - 2, and y = 1 when x = 4. Find the value of y when x = 9.

    1. A 36
    2. B 48
    3. C 29
    4. D 49

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: B

    1. The curve is below the x-axis on [1,3] (since f(x) <= 0 there) and above it on [3,8] (since f(x) >= 0 there), so the region splits into two pieces at x = 3.
    2. The area of a piece where the curve is below the axis equals the absolute value of its signed integral: the piece on [1,3] contributes |-6| = 6.
    3. The area of a piece where the curve is above the axis equals its signed integral directly, since that value is already positive: the piece on [3,8] contributes |15| = 15.
    4. Total area = 6 + 15 = 21.
    • Why not A: Adds the two given integrals directly (-6 + 15 = 9) without first taking the absolute value of the negative piece: the signed net change across the whole interval is not the same thing as the area, because the region below the axis must contribute positively to an area.
    • Why not C: This negates the combined signed total (-(-6+15) = -9) rather than splitting the interval at the sign change and summing the absolute value of each piece separately; an area can never be negative in the first place.
    • Why not D: This uses only the integral over [3, 8], where the curve lies above the axis, and discards the integral over [1, 3] entirely instead of including the magnitude of that piece too.

  2. Question 2Answer: D

    1. The curve y = x^2 - 4x = x(x-4) crosses the x-axis where x = 0 and x = 4. Only x = 4 lies strictly inside (0, 5).
    2. For 0 <= x <= 4 the curve is below the axis (check x = 2: y = 4 - 8 = -4 < 0), and for 4 <= x <= 5 it is above the axis (check x = 4.5: y = 20.25 - 18 = 2.25 > 0), so the region splits into two pieces at x = 4.
    3. An antiderivative of x^2 - 4x is x^3/3 - 2x^2.
    4. Area of the first piece = |integral from 0 to 4 of (x^2-4x) dx| = |(64/3 - 32) - 0| = |-32/3| = 32/3.
    5. Area of the second piece = |integral from 4 to 5 of (x^2-4x) dx| = |(125/3 - 50) - (64/3 - 32)| = |-25/3 - (-32/3)| = |7/3| = 7/3.
    6. Total area = 32/3 + 7/3 = 39/3 = 13.
    • Why not A: This is the value of the plain definite integral from 0 to 5, which is not the same thing as the area: the curve dips below the x-axis for most of the interval, so the signed integral does not equal the area, and a genuine area cannot come out negative in any case.
    • Why not B: This takes the absolute value of the overall signed integral from 0 to 5, rather than splitting the region at the root x = 4 and summing the absolute value of each piece separately.
    • Why not C: This only accounts for the region from x = 4 to x = 5, where the curve is above the axis, and ignores the larger region from x = 0 to x = 4, where the curve dips below the axis.

  3. Question 3Answer: A

    1. Rewrite the integrand using index notation: 6x^2 - 4/sqrt(x) = 6x^2 - 4x^-(1/2).
    2. Integrate term by term: the antiderivative of 6x^2 is 6*x^3/3 = 2x^3, and the antiderivative of -4x^-(1/2) is -4 * x^(1/2)/(1/2) = -8x^(1/2) = -8 sqrt(x).
    3. Combine the two results and add the constant of integration, since this is an indefinite integral.
    4. integral of (6x^2 - 4/sqrt(x)) dx = 2x^3 - 8 sqrt(x) + C.
    • Why not B: This integrates -4x^-(1/2) by multiplying by the new exponent 1/2 instead of dividing by it, giving -4*(1/2)x^(1/2) = -2 sqrt(x) instead of the correct -4*2*x^(1/2) = -8 sqrt(x).
    • Why not C: This integrates 6x^2 by raising the power without dividing the coefficient by the new power, giving 6x^3 instead of the correct 6x^3/3 = 2x^3.
    • Why not D: This drops the negative sign carried by the term -4x^-(1/2), giving +8 sqrt(x) instead of the correct -8 sqrt(x).

  4. Question 4Answer: C

    1. Simplify the integrand first by dividing each term by sqrt(x): (x-4)/sqrt(x) = x/sqrt(x) - 4/sqrt(x) = x^(1/2) - 4x^-(1/2).
    2. Integrate term by term: the antiderivative of x^(1/2) is x^(3/2)/(3/2) = (2/3)x^(3/2), and the antiderivative of -4x^-(1/2) is -4 * x^(1/2)/(1/2) = -8x^(1/2).
    3. So F(x) = (2/3)x^(3/2) - 8x^(1/2).
    4. F(4) = (2/3)*8 - 8*2 = 16/3 - 16 = -32/3. (Using 4^(3/2) = (sqrt(4))^3 = 2^3 = 8.)
    5. F(1) = (2/3)*1 - 8*1 = 2/3 - 8 = -22/3.
    6. The integral equals F(4) - F(1) = -32/3 - (-22/3) = -10/3.
    • Why not A: This mishandles the negative power when integrating -4x^-(1/2): dividing by the OLD exponent (-1/2) instead of the new exponent (+1/2) turns the term into +8 sqrt(x) instead of the correct -8 sqrt(x), flipping its contribution.
    • Why not B: This simplifies (x-4)/sqrt(x) incorrectly as -3x^(1/2) (treating -4/sqrt(x) as -4x^(1/2) rather than -4x^-(1/2)), losing the sign on the exponent when dividing the constant term by sqrt(x).
    • Why not D: This is the negative of the correct answer, from evaluating F(1) - F(4) instead of F(4) - F(1).

  5. Question 5Answer: D

    1. Because x is the LOWER limit here, rewrite the integral so x sits in the upper limit: k(x) = integral from x to 4 of (t^2+3) dt = -integral from 4 to x of (t^2+3) dt.
    2. By the Fundamental Theorem of Calculus, d/dx of integral from a to x of f(t) dt = f(x), so d/dx of -integral from 4 to x of (t^2+3) dt = -(x^2+3).
    3. So k'(x) = -(x^2+3).
    4. k'(1) = -(1^2+3) = -(1+3) = -4.
    • Why not A: This uses +(x^2+3) in place of the correct -(x^2+3), missing the extra minus sign that appears because x is the LOWER limit of the integral here rather than the upper limit.
    • Why not B: This substitutes x = 4 (the fixed upper limit of the integral) into the derivative expression, instead of x = 1, the point at which k'(x) is actually being evaluated.
    • Why not C: This differentiates only the t^2 part of the integrand and forgets the constant term +3, giving -(x^2) instead of the correct -(x^2+3).

  6. Question 6Answer: B

    1. Let g(x) = integral from 0 to x of f(t) dt = x^4 - 3x^2. By the Fundamental Theorem of Calculus, g'(x) = f(x), so f(x) is found by differentiating the given formula.
    2. g'(x) = 4x^3 - 6x.
    3. So f(x) = 4x^3 - 6x.
    4. f(2) = 4*(2^3) - 6*2 = 4*8 - 12 = 32 - 12 = 20.
    • Why not A: This evaluates the given formula for the integral itself at x = 2 (2^4 - 3*2^2 = 16 - 12 = 4), mistaking the antiderivative for f(2) instead of differentiating it first, as the Fundamental Theorem of Calculus requires.
    • Why not C: This differentiates x^4 incorrectly as x^3 instead of the correct 4x^3, forgetting to bring the power down as a coefficient.
    • Why not D: This differentiates -3x^2 with the wrong sign, as +6x instead of the correct -6x.

  7. Question 7Answer: C

    1. Because 0, 3, 7 and 10 are in order, the interval [0, 10] splits into the three contiguous pieces [0,3], [3,7] and [7,10].
    2. The additive property of the integral says the integral over the whole interval equals the sum of the integrals over the contiguous pieces.
    3. Substitute the given values: integral from 0 to 10 of f(x) dx = 5 + (-2) + 6.
    4. 5 + (-2) + 6 = 9.
    • Why not A: This treats the given value -2 as though it were positive, computing 5 + 2 + 6 = 13 and losing the sign of the middle piece.
    • Why not B: This adds only the two outer integrals, 5 + 6 = 11, and leaves out the middle contiguous interval [3, 7] entirely.
    • Why not D: This averages the three given values, (5 + (-2) + 6)/3 = 3, instead of adding them: contiguous integrals combine by summation, not by averaging.

  8. Question 8Answer: A

    1. Reversing the limits of a definite integral negates its value: the integral from 9 to 2 of f(x) dx = -(the integral from 2 to 9 of f(x) dx) = -15.
    2. Integration is linear, so the integral from 9 to 2 of 4f(x) dx = 4 * (the integral from 9 to 2 of f(x) dx).
    3. Substitute: 4 * (-15) = -60.
    • Why not B: This treats the integral from 9 to 2 as equal to the integral from 2 to 9 without applying the sign flip that reversing the limits requires, computing 4*15 = 60 instead of 4*(-15).
    • Why not C: This correctly reverses the sign for the swapped limits, getting -15, but then forgets to scale by the coefficient 4 in front of f(x), leaving the answer as -15 instead of multiplying through to -60.
    • Why not D: This neither reverses the sign for the swapped limits nor scales by the coefficient 4, simply restating the given value 15 unchanged.

  9. Question 9Answer: B

    1. The ordinates are y0 = 0^2+1 = 1, y1 = 2^2+1 = 5, y2 = 4^2+1 = 17, y3 = 6^2+1 = 37, with strip width h = 2.
    2. The trapezium rule states: integral is approximately (h/2) * [y0 + 2(y1 + y2) + y3].
    3. Substitute: (2/2) * [1 + 2(5+17) + 37] = 1 * [1 + 44 + 37] = 1 * 82.
    4. The estimate is 82.
    • Why not A: This omits the factor of 1/2 at the front of the trapezium rule formula, using h[y0 + 2(y1+y2) + y3] instead of (h/2)[y0 + 2(y1+y2) + y3], doubling the true estimate.
    • Why not C: This forgets to double the interior ordinates (at x=2 and x=4), using (h/2)(y0+y1+y2+y3) instead of (h/2)[y0 + 2(y1+y2) + y3].
    • Why not D: This treats the whole interval [0,6] as a single trapezium using only the two end ordinates, (6/2)(y0+y3), instead of using all three strips.

  10. Question 10Answer: D

    1. The trapezium rule replaces the curve between each pair of ordinates with a straight chord, and adds up the areas of the resulting trapezia.
    2. Check with an example: at x = 2, y = 4, and at x = 6, y = 36. The chord between these points has gradient (36-4)/(6-2) = 8, so at x = 4 it passes through height 4 + 8*2 = 20.
    3. The true curve at x = 4 gives y = 4^2 = 16, which is lower than the chord's value of 20.
    4. So the curve lies below its chords across the interval, meaning the true area under the curve is smaller than the area under the chords.
    5. The trapezium rule, which uses the area under the chords, therefore overestimates the true integral.
    • Why not A: This has the direction backwards: for a curve that bends upward (convex), such as x^2, the chord joining two points on the curve lies above the curve, not below it. Chords lying below the curve is what happens for a curve that bends downward (concave), such as sqrt(x).
    • Why not B: This wrongly assumes the trapezium rule is exact for any polynomial; it is only exact when the curve itself is a straight line (degree 1 or lower), since only then does a chord coincide exactly with the curve between the two points.
    • Why not C: This is too pessimistic: while the exact size of the error does need a calculation, the direction of the error (over- or underestimate) is decided by the curve's shape (convex or concave) alone, before any numbers are found.

  11. Question 11Answer: A

    1. Integrate dy/dx = 6x^2 - 4x + 1 term by term: y = 2x^3 - 2x^2 + x + C.
    2. Use the condition y = 5 when x = 2: 5 = 2*(2^3) - 2*(2^2) + 2 + C = 16 - 8 + 2 + C = 10 + C, so C = -5.
    3. The curve is y = 2x^3 - 2x^2 + x - 5.
    4. At x = -1: y = 2*(-1)^3 - 2*(-1)^2 + (-1) - 5 = -2 - 2 - 1 - 5.
    5. y = -10.
    • Why not B: This omits the constant of integration entirely, so the condition y = 5 when x = 2 is never used to pin down y = 2x^3 - 2x^2 + x + C.
    • Why not C: This integrates -4x incorrectly as -4x^2 instead of the correct -2x^2, forgetting to divide by the new power when integrating.
    • Why not D: This assumes the constant of integration equals the given y-value, C = 5, instead of substituting x = 2, y = 5 into y = 2x^3 - 2x^2 + x + C and solving for C.

  12. Question 12Answer: C

    1. Integrate dy/dx = 3x^(1/2) - 2 term by term: the antiderivative of 3x^(1/2) is 3 * x^(3/2)/(3/2) = 2x^(3/2), and the antiderivative of -2 is -2x.
    2. So y = 2x^(3/2) - 2x + C.
    3. Use the condition y = 1 when x = 4: 1 = 2*(4^(3/2)) - 2*4 + C = 2*8 - 8 + C = 8 + C, so C = -7. (Using 4^(3/2) = (sqrt(4))^3 = 2^3 = 8.)
    4. The curve is y = 2x^(3/2) - 2x - 7.
    5. At x = 9: y = 2*(9^(3/2)) - 2*9 - 7 = 2*27 - 18 - 7 = 54 - 25.
    6. y = 29.
    • Why not A: This omits the constant of integration entirely, so the condition y = 1 when x = 4 is never used to pin down y = 2x^(3/2) - 2x + C.
    • Why not B: This integrates 3x^(1/2) incorrectly as 3x^(3/2) instead of the correct 2x^(3/2) = 3*(2/3)x^(3/2), forgetting to multiply the coefficient by the reciprocal of the new power.
    • Why not D: This integrates the constant term -2 with the wrong sign, as +2x instead of the correct -2x.

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