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Admissions tests / TMUA / Paper 1 / Integration

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

TMUA Paper 1: Integration, set 1

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

    Find the total area enclosed between the curve y = x^2 - 4, the x-axis, and the lines x = 0 and x = 3.

    1. A 23/3
    2. B -3
    3. C 3
    4. D 7/3
  2. 21 mark

    Find the value of the integral from 1 to 4 of (3 sqrt(x) - 2/x^2) dx.

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

    Let g(x) = integral from 0 to x of (2t^3 - 5t) dt. Find g'(3).

    1. A -39
    2. B 3
    3. C 39
    4. D 54
  4. 41 mark

    Given that the integral from 1 to 6 of f(x) dx = 20, and the integral from 4 to 6 of f(x) dx = 9, find the integral from 1 to 4 of f(x) dx.

    1. A 29
    2. B 9
    3. C -11
    4. D 11
  5. 51 mark

    Use the trapezium rule with strips of width 1 to estimate the integral from 0 to 4 of x^2 dx, using the five ordinates at x = 0, 1, 2, 3, 4.

    1. A 44
    2. B 22
    3. C 15
    4. D 32
  6. 61 mark

    A curve satisfies dy/dx = 4x^3 - 6x, and y = 2 when x = 1. Find the value of y when x = 2.

    1. A 4
    2. B 53
    3. C 8
    4. D 6
  7. 71 mark

    Find the total area enclosed between the curve y = x^3 - 4x and the x-axis, for -1 <= x <= 2.

    1. A -9/4
    2. B 9/4
    3. C 7/4
    4. D 23/4
  8. 81 mark

    Find the value of the integral from 1 to 4 of (x^2 - 5)/sqrt(x) dx.

    1. A 12/5
    2. B 112/5
    3. C -5/2
    4. D -12/5
  9. 91 mark

    F(x) is an antiderivative of f(x) = 2x - 3, and F(2) = 1. Find F(5).

    1. A 12
    2. B -11
    3. C 13
    4. D 31
  10. 101 mark

    Given that the integral from 0 to 3 of f(x) dx = 8 and the integral from 0 to 3 of g(x) dx = -5, find the integral from 0 to 3 of (2f(x) - 3g(x)) dx.

    1. A 1
    2. B -3
    3. C 34
    4. D 31
  11. 111 mark

    The curve y = sqrt(x) is concave (it curves downward, like the inside of a dome) on the interval 1 <= x <= 9. Which of the following correctly describes a trapezium-rule estimate of the integral from 1 to 9 of sqrt(x) 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 an overestimate, because the chords joining points on the curve lie above the curve.
    3. C The trapezium rule gives the exact value, because sqrt(x) is a smooth curve with no straight sections.
    4. D It is impossible to say whether it is an overestimate or an underestimate without calculating both values numerically.
  12. 121 mark

    A curve has gradient function dy/dx = 3x^2 + 2, and passes through the point (0, -1). Find the value of y when x = -2.

    1. A -12
    2. B -13
    3. C 3
    4. D -9

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: A

    1. The curve y = x^2 - 4 crosses the x-axis where x^2 - 4 = 0, so at x = -2 and x = 2. Only x = 2 lies inside [0, 3].
    2. For 0 <= x <= 2 the curve is below the axis (y <= 0), and for 2 <= x <= 3 it is above the axis (y >= 0), so the region splits into two pieces at x = 2.
    3. An antiderivative of x^2 - 4 is x^3/3 - 4x.
    4. Area of the first piece = integral from 0 to 2 of (4 - x^2) dx = [4x - x^3/3] from 0 to 2 = (8 - 8/3) - 0 = 16/3.
    5. Area of the second piece = integral from 2 to 3 of (x^2 - 4) dx = [x^3/3 - 4x] from 2 to 3 = (9 - 12) - (8/3 - 8) = -3 + 16/3 = 7/3.
    6. Total area = 16/3 + 7/3 = 23/3.
    • Why not B: This is the value of the plain definite integral from 0 to 3, which is not the same thing as the area: the curve dips below the x-axis on part of the interval, so the signed integral undercounts the true area.
    • Why not C: This is the absolute value of the definite integral from 0 to 3 taken as a whole, rather than the sum of the absolute values of the two pieces either side of the root at x = 2.
    • Why not D: This is only the area of the piece from x = 2 to x = 3, where the curve is above the axis; it ignores the region from x = 0 to x = 2, where the curve is below the axis.

  2. Question 2Answer: B

    1. Rewrite the integrand using index notation: 3 sqrt(x) - 2/x^2 = 3x^(1/2) - 2x^-2.
    2. Integrate term by term: the antiderivative of 3x^(1/2) is 3 * x^(3/2)/(3/2) = 2x^(3/2), and the antiderivative of -2x^-2 is -2 * x^-1/(-1) = 2x^-1 = 2/x.
    3. So F(x) = 2x^(3/2) + 2/x.
    4. F(4) = 2 * 4^(3/2) + 2/4 = 2*8 + 1/2 = 33/2.
    5. F(1) = 2*1 + 2/1 = 4.
    6. The integral equals F(4) - F(1) = 33/2 - 4 = 25/2.
    • Why not A: This comes from a sign error integrating -2x^-2: writing its antiderivative as -2x^-1 instead of the correct +2x^-1, which flips the sign of that term throughout.
    • Why not C: This comes from forgetting to raise the power of x when integrating 3x^(1/2): using 3x^(1/2)/(3/2) instead of the correct 3x^(3/2)/(3/2).
    • Why not D: This is the negative of the correct answer, from evaluating F(1) - F(4) instead of F(4) - F(1).

  3. Question 3Answer: C

    1. By the Fundamental Theorem of Calculus, if g(x) = integral from a to x of f(t) dt, then g'(x) = f(x), so no integration is actually needed here.
    2. With f(t) = 2t^3 - 5t, this gives g'(x) = 2x^3 - 5x directly.
    3. Substitute x = 3: g'(3) = 2*(3^3) - 5*3 = 2*27 - 15 = 54 - 15.
    4. g'(3) = 39.
    • Why not A: This comes from confusing d/dx of the integral from a to x of f(t) dt (which equals f(x)) with d/dx of the integral from x to b of f(t) dt (which equals -f(x)), introducing an unwanted minus sign.
    • Why not B: This comes from misreading the integrand's power, using x^2 in place of x^3, giving 2*3^2 - 5*3 = 3 instead of 2*3^3 - 5*3.
    • Why not D: This comes from dropping the -5x term after substituting x = 3, leaving only 2*3^3 = 54.

  4. Question 4Answer: D

    1. Because 1, 4 and 6 are in order, the interval [1, 6] splits into the contiguous pieces [1, 4] and [4, 6].
    2. So the integral from 1 to 6 of f(x) dx = the integral from 1 to 4 of f(x) dx + the integral from 4 to 6 of f(x) dx.
    3. Substitute the given values: 20 = (integral from 1 to 4) + 9.
    4. So the integral from 1 to 4 of f(x) dx = 20 - 9 = 11.
    • Why not A: This adds the two given integrals (20 + 9) instead of recognising that the integral from 4 to 6 is already part of the integral from 1 to 6, so it must be subtracted, not added.
    • Why not B: This simply restates the given value of the integral from 4 to 6, mistaking it for the answer instead of using it to find the integral from 1 to 4.
    • Why not C: This subtracts in the wrong order (9 - 20 instead of 20 - 9), giving the negative of the correct value.

  5. Question 5Answer: B

    1. The ordinates are y0 = 0^2 = 0, y1 = 1^2 = 1, y2 = 2^2 = 4, y3 = 3^2 = 9, y4 = 4^2 = 16, with strip width h = 1.
    2. The trapezium rule states: integral is approximately (h/2) * [y0 + 2(y1 + y2 + y3) + y4].
    3. Substitute: (1/2) * [0 + 2(1 + 4 + 9) + 16] = (1/2) * [0 + 28 + 16] = (1/2) * 44.
    4. The estimate is 22.
    • Why not A: This omits the factor of 1/2 at the front of the trapezium rule formula, using y0 + 2(y1+y2+y3) + y4 without halving it.
    • Why not C: This forgets to double the interior ordinates y1, y2 and y3 in the formula, using (1/2)(y0+y1+y2+y3+y4) instead of (1/2)(y0 + 2(y1+y2+y3) + y4).
    • Why not D: This treats the whole interval as a single trapezium using only the two end ordinates, (1/2)(4)(y0+y4), instead of using all four strips.

  6. Question 6Answer: C

    1. Integrate dy/dx = 4x^3 - 6x term by term: y = x^4 - 3x^2 + C.
    2. Use the condition y = 2 when x = 1: 2 = 1^4 - 3*(1^2) + C = 1 - 3 + C = -2 + C, so C = 4.
    3. The curve is y = x^4 - 3x^2 + 4.
    4. At x = 2: y = 2^4 - 3*(2^2) + 4 = 16 - 12 + 4.
    5. y = 8.
    • Why not A: This omits the constant of integration entirely, so the condition y = 2 when x = 1 is never used to pin down y = x^4 - 3x^2 + C.
    • Why not B: This comes from applying the power rule incorrectly to 4x^3, integrating it as 4x^4 instead of the correct x^4.
    • Why not D: This assumes the constant of integration equals the given y-value, C = 2, instead of substituting x = 1, y = 2 into y = x^4 - 3x^2 + C and solving for C.

  7. Question 7Answer: D

    1. Factorise: x^3 - 4x = x(x^2 - 4) = x(x-2)(x+2), with roots at x = -2, 0, 2. Only x = 0 lies inside [-1, 2].
    2. Check the sign: at x = -0.5, y = (-0.5)^3 - 4*(-0.5) = -0.125 + 2 = 1.875 > 0, so the curve is above the axis on [-1, 0]. At x = 1, y = 1 - 4 = -3 < 0, so the curve is below the axis on [0, 2].
    3. An antiderivative is F(x) = x^4/4 - 2x^2. Then F(-1) = 1/4 - 2 = -7/4, F(0) = 0, F(2) = 4 - 8 = -4.
    4. Area of the first piece = F(0) - F(-1) = 0 - (-7/4) = 7/4.
    5. The second piece has signed value F(2) - F(0) = -4, so its area is the absolute value, 4.
    6. Total area = 7/4 + 4 = 7/4 + 16/4 = 23/4.
    • Why not A: This is the plain definite integral from -1 to 2, computed without noticing that the curve crosses the axis at x = 0 inside the interval, so positive and negative regions have cancelled.
    • Why not B: This takes the absolute value of the overall definite integral from -1 to 2, rather than splitting the region at the root x = 0 and taking the absolute value of each piece separately.
    • Why not C: This only accounts for the region from x = -1 to x = 0, where the curve is above the axis, and ignores the region from x = 0 to x = 2, where the curve dips below.

  8. Question 8Answer: A

    1. Simplify the integrand first: (x^2 - 5)/sqrt(x) = x^2/x^(1/2) - 5/x^(1/2) = x^(3/2) - 5x^-(1/2), subtracting the exponents.
    2. Integrate term by term: the antiderivative of x^(3/2) is x^(5/2)/(5/2) = (2/5)x^(5/2), and the antiderivative of -5x^-(1/2) is -5 * x^(1/2)/(1/2) = -10x^(1/2).
    3. So F(x) = (2/5)x^(5/2) - 10x^(1/2).
    4. F(4) = (2/5)*32 - 10*2 = 64/5 - 20 = -36/5. (Using 4^(5/2) = (sqrt(4))^5 = 2^5 = 32.)
    5. F(1) = (2/5)*1 - 10*1 = 2/5 - 10 = -48/5.
    6. The integral equals F(4) - F(1) = -36/5 - (-48/5) = 12/5.
    • Why not B: This comes from a sign error integrating the -5x^-(1/2) term, obtaining +10x^(1/2) instead of the correct -10x^(1/2).
    • Why not C: This comes from simplifying x^2/sqrt(x) by multiplying the exponents (2 x 1/2) instead of subtracting them, treating it as x^1 rather than the correct x^(3/2).
    • Why not D: This is the negative of the correct answer, from evaluating F(1) - F(4) instead of F(4) - F(1).

  9. Question 9Answer: C

    1. By the Fundamental Theorem of Calculus, F(5) - F(2) = integral from 2 to 5 of f(x) dx.
    2. An antiderivative of f(x) = 2x - 3 is x^2 - 3x.
    3. Evaluate: at x = 5, x^2 - 3x = 25 - 15 = 10. At x = 2, x^2 - 3x = 4 - 6 = -2.
    4. So the integral from 2 to 5 equals 10 - (-2) = 12.
    5. F(5) = F(2) + 12 = 1 + 12 = 13.
    • Why not A: This reports the value of the integral from 2 to 5 of f(x) dx itself, forgetting to add the given F(2) = 1 to recover F(5).
    • Why not B: This subtracts the integral from F(2) instead of adding it, computing F(2) - 12 instead of F(2) + 12.
    • Why not D: This comes from a sign error finding the antiderivative of 2x - 3, using x^2 + 3x instead of the correct x^2 - 3x.

  10. Question 10Answer: D

    1. Integration is linear: the integral of (2f(x) - 3g(x)) dx over an interval equals 2 times the integral of f(x) dx minus 3 times the integral of g(x) dx, over that same interval.
    2. Substitute the given values: 2*8 - 3*(-5).
    3. 2*8 = 16, and -3*(-5) = 15.
    4. 16 + 15 = 31.
    • Why not A: This uses the magnitude of the given integral for g, +5, instead of the given value -5, so the sign of that contribution is lost.
    • Why not B: This combines the two given values first, 8 + (-5) = 3, then multiplies by (2-3) = -1, instead of scaling each integral separately before combining them.
    • Why not C: This swaps the coefficients, applying 3 to the integral of f and 2 to the integral of g instead of the other way round.

  11. Question 11Answer: A

    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 = 1, y = 1, and at x = 9, y = 3. The chord between these points passes through x = 4 at height 1 + (3-1)/(9-1) * (4-1) = 1.75.
    3. The true curve at x = 4 gives y = sqrt(4) = 2, which is higher than the chord's value of 1.75.
    4. So the curve lies above its chords across the interval, meaning the true area under the curve is larger than the area under the chords.
    5. The trapezium rule (which uses the area under the chords) therefore underestimates the true integral.
    • Why not B: This has the direction backwards: a chord joining two points on a curve that bends downward (like sqrt(x)) lies below the curve, not above it. Chords lying above the curve is what happens for a curve that bends upward, such as y = x^2.
    • Why not C: This wrongly assumes any smooth curve is approximated exactly by straight-line chords; that is only true when the curve is itself a straight line.
    • Why not D: This is too pessimistic: while the exact size of the error does need a calculation, the direction of the error (over- or underestimate) is determined by the curve's shape alone, before any numbers are found.

  12. Question 12Answer: B

    1. Integrate dy/dx = 3x^2 + 2 term by term: y = x^3 + 2x + C.
    2. Use the point (0, -1): -1 = 0^3 + 2*0 + C, so C = -1.
    3. The curve is y = x^3 + 2x - 1.
    4. At x = -2: y = (-2)^3 + 2*(-2) - 1 = -8 - 4 - 1.
    5. y = -13.
    • Why not A: This omits the constant of integration, so the condition that the curve passes through (0, -1) is never used to find C.
    • Why not C: This comes from evaluating (-2)^3 as +8 instead of -8, dropping the sign when cubing a negative number.
    • Why not D: This comes from integrating the constant term 2 as simply '2' instead of '2x', forgetting that the integral of a constant with respect to x carries a factor of x.

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