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Admissions tests / ESAT / Maths 2 / Differentiation and integration

Demanding. 15 questions, 15 marks, about 30 minutes.

ESAT Mathematics 2: Differentiation and integration, set 3

The derivative as a gradient, differentiating powers of x, tangents, normals, stationary points, increasing and decreasing functions, indefinite and definite integration, and areas.

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

    A particle moves along a straight line so that its displacement from a fixed point O after t seconds is s = t^3 - 6t^2 + 9t (in metres). Find the particle's acceleration when t = 3.

    1. A 0 m/s^2
    2. B 18 m/s^2
    3. C 6 m/s^2
    4. D -12 m/s^2
  2. 21 mark

    The volume of a sphere of radius r is V = (4/3) pi r^3. Find dV/dr when r = 3, and identify what this value represents.

    1. A 36 pi cubic units per unit increase in radius: this is the rate at which the volume is increasing at the instant r = 3.
    2. B 12 pi cubic units per unit increase in radius: this is the rate at which the volume is increasing at the instant r = 3.
    3. C 36 pi cubic units: this is the total volume of the sphere when r = 3.
    4. D 24 pi cubic units per unit increase in radius: this is the rate at which the volume is increasing at the instant r = 3.
  3. 31 mark

    Given that y = (x^2 - 1)^2 / x^3 for x != 0, find dy/dx, simplifying your answer fully.

    1. A 1 - 2/x^2 - 3/x^4
    2. B 1 + 2/x^2 - 3/x^4
    3. C 1 + 2/x^2 + 3/x^4
    4. D x - 2/x + 1/x^3
  4. 41 mark

    Given that y = (2sqrt(x) - 3)^2 for x > 0, find dy/dx, giving your answer in the form a - b/sqrt(x).

    1. A 4
    2. B 4 - 6sqrt(x)
    3. C 4 - 12/sqrt(x)
    4. D 4 - 6/sqrt(x)
  5. 51 mark

    Find the coordinates of the point on the curve y = x^2 - 5x + 7 at which the tangent is parallel to the line y = 3x - 1.

    1. A (4, 3)
    2. B (4, -11)
    3. C (2, 1)
    4. D (4, 43)
  6. 61 mark

    The curve y = 2x^3 + 3x^2 - 12x + 7 has two stationary points. Find the x-coordinate of the one at which y has a local maximum.

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

    Find the set of values of x for which f(x) = x^3 - 3x^2 - 9x + 1 is an increasing function.

    1. A -1 < x < 3
    2. B x < -3 or x > 1
    3. C x < 0 or x > 2
    4. D x < -1 or x > 3
  8. 81 mark

    The curve y = f(x) crosses the x-axis at x = 3 within the interval 0 <= x <= 8. It lies above the x-axis for 0 < x < 3, and below the x-axis for 3 < x < 8. Given that the definite integral of f(x) dx from x = 0 to x = 3 equals 15, and the definite integral of f(x) dx from x = 3 to x = 8 equals -40, find the area enclosed between the curve and the x-axis for 0 <= x <= 8.

    1. A -25
    2. B 55
    3. C 25
    4. D 15
  9. 91 mark

    Find the indefinite integral of (sqrt(x) + 2)^2 with respect to x, giving your answer in terms of x.

    1. A x^2/2 + 4x + C
    2. B x^2/2 + 4x^(3/2) + 4x + C
    3. C x^2/2 + (8/3)x^(3/2) + 4x + C
    4. D x^2/2 + 6x^(3/2) + 4x + C
  10. 101 mark

    Evaluate the definite integral of (sqrt(x) - 3)^2 dx from x = 0 to x = 9.

    1. A 27/2
    2. B 243/2
    3. C -81/2
    4. D 459/2
  11. 111 mark

    The function g(x) is defined for x > 0 by g(x) equal to the definite integral of (3t^2 - 2) dt from t = 1 to t = x. Using the Fundamental Theorem of Calculus, find g'(x).

    1. A x^3 - 2x + 1
    2. B x^3 - 2x
    3. C 6x
    4. D 3x^2 - 2
  12. 121 mark

    Given that the definite integral of f(x) dx from x = 2 to x = 7 equals 15, and the definite integral of g(x) dx from x = 2 to x = 7 equals -6, find the definite integral of (2f(x) - 3g(x)) dx from x = 2 to x = 7.

    1. A 36
    2. B 48
    3. C 33
    4. D 12
  13. 131 mark

    Use the trapezium rule with 4 strips of width 1 to estimate the definite integral of (x^2 + 1) dx from x = 0 to x = 4.

    1. A 26
    2. B 17.5
    3. C 52
    4. D 22
  14. 141 mark

    The curve y = x^2 - 2x + 5 is used to estimate the definite integral of y dx from x = 0 to x = 3, using the trapezium rule. By finding d2y/dx2 for this curve, determine whether the trapezium rule estimate will be an overestimate or an underestimate of the true value.

    1. A Underestimate. d2y/dx2 = 2, which is positive, so the curve is concave up; each chord joining two points on a concave-up curve lies above the curve, so the trapezium bounded above by that chord leaves out a sliver of the true area beneath the curve.
    2. B Underestimate. Differentiating a second time with a sign slip gives d2y/dx2 = -2, so the curve appears concave down, and a concave-down curve's chords lie below it, causing the trapezium rule to underestimate the true value.
    3. C Overestimate. d2y/dx2 = 2, which is positive, so the curve is concave up over the whole interval; each chord lies above the curve, so the trapezium rule (whose top edge is that chord) covers more area than the true area under the curve.
    4. D Overestimate. The trapezium rule always overestimates a definite integral for any curve, regardless of its concavity.
  15. 151 mark

    A curve satisfies dy/dx = (x^3 - 8)/x^2 for x != 0, and passes through the point (2, 10). Find y in terms of x.

    1. A y = x^4/4 + 8/x + 2
    2. B y = x^2/2 + 8/x + 4
    3. C y = x^2/2 - 8/x + 12
    4. D y = x^2/2 + 8/x

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: C

    1. Acceleration is the second derivative of displacement with respect to time, so first find the velocity: ds/dt = 3t^2 - 12t + 9.
    2. Differentiate again to find the acceleration function: d2s/dt2 = 6t - 12.
    3. Substitute t=3: acceleration = 6(3) - 12 = 18 - 12 = 6.
    4. So the acceleration at t=3 is 6 m/s^2, and the answer is C.
    • Why not A: Correctly finds the velocity function ds/dt = 3t^2 - 12t + 9, but evaluates this first derivative at t=3 (giving 27-36+9=0) instead of the second derivative, mistaking velocity for acceleration.
    • Why not B: Correctly finds the acceleration function d2s/dt2 = 6t - 12, but when substituting t=3 forgets to subtract the constant term, computing only 6(3)=18 instead of 6(3)-12=6.
    • Why not D: Correctly finds the acceleration function d2s/dt2 = 6t - 12, but reports only the constant term of this expression, -12, rather than evaluating the whole expression at t=3.

  2. Question 2Answer: A

    1. Differentiate V = (4/3) pi r^3 with respect to r: bring down the power 3 and reduce it by one, giving dV/dr = (4/3) x 3 x pi r^2 = 4 pi r^2.
    2. Substitute r=3: dV/dr = 4 pi (3)^2 = 4 pi (9) = 36 pi.
    3. The derivative dV/dr measures how fast the volume changes for a small increase in radius, so at the instant r=3 the volume is increasing at a rate of 36 pi cubic units per unit increase in r.
    4. So the answer is A.
    • Why not B: Differentiates r^3 by reducing the power to r^2 but forgets to first multiply by the original power 3, so (4/3)pi r^3 differentiates to (4/3)pi r^2 instead of 4pi r^2; evaluating this wrong derivative at r=3 gives (4/3)pi(9) = 12pi.
    • Why not C: Correctly computes the derivative dV/dr = 4pi r^2 = 36pi at r=3, but then misreads this rate of change as if it were the volume itself, confusing dV/dr with V.
    • Why not D: Correctly differentiates to dV/dr = 4pi r^2, but squares r=3 incorrectly as 2 x 3 = 6 rather than 3^2 = 9, giving 4pi(6) = 24pi.

  3. Question 3Answer: B

    1. Expand the numerator: (x^2 - 1)^2 = x^4 - 2x^2 + 1.
    2. Divide each term by x^3 to simplify before differentiating: y = x^4/x^3 - 2x^2/x^3 + 1/x^3 = x - 2x^-1 + x^-3.
    3. Differentiate term by term: d/dx(x) = 1; d/dx(-2x^-1) = (-2)(-1)x^-2 = 2x^-2; d/dx(x^-3) = -3x^-4.
    4. Combining these gives dy/dx = 1 + 2x^-2 - 3x^-4 = 1 + 2/x^2 - 3/x^4, so the answer is B.
    • Why not A: Correctly expands and simplifies to y = x - 2x^-1 + x^-3, but when differentiating -2x^-1 keeps the sign unchanged instead of flipping it, treating d/dx(-2x^-1) as -2x^-2 rather than the correct (-2)(-1)x^-2 = +2x^-2.
    • Why not C: Expands (x^2-1)^2 with a sign slip as x^4 - 2x^2 - 1 instead of x^4 - 2x^2 + 1, and this flipped sign on the constant term carries all the way through to the sign of the final term of the derivative.
    • Why not D: Correctly simplifies y to x - 2x^-1 + x^-3, but stops there, giving the simplified expression for y itself instead of differentiating it to find dy/dx.

  4. Question 4Answer: D

    1. Expand the square: (2sqrt(x) - 3)^2 = (2sqrt(x))^2 - 2(2sqrt(x))(3) + 3^2 = 4x - 12sqrt(x) + 9.
    2. Rewrite sqrt(x) as x^(1/2), so y = 4x - 12x^(1/2) + 9.
    3. Differentiate term by term: d/dx(4x) = 4; d/dx(-12x^(1/2)) = -12 x (1/2) x^(-1/2) = -6x^(-1/2); d/dx(9) = 0.
    4. So dy/dx = 4 - 6x^(-1/2) = 4 - 6/sqrt(x), and the answer is D.
    • Why not A: Expands (2sqrt(x) - 3)^2 by dropping the cross term, treating it like (a-b)^2 = a^2 + b^2 instead of a^2 - 2ab + b^2, so 4x - 12sqrt(x) + 9 becomes just 4x + 9, whose derivative is the constant 4.
    • Why not B: Correctly expands to 4x - 12x^(1/2) + 9, but when differentiating -12x^(1/2) forgets to reduce the power by one, applying only the multiplication by the exponent and leaving x^(1/2) unchanged: -12 x (1/2) x^(1/2) = -6x^(1/2) = -6sqrt(x).
    • Why not C: Correctly expands to 4x - 12x^(1/2) + 9, but when differentiating -12x^(1/2) reduces the power correctly to -1/2 while forgetting to also multiply by the exponent 1/2 first, giving -12x^(-1/2) = -12/sqrt(x) instead of -6/sqrt(x).

  5. Question 5Answer: A

    1. Two lines are parallel exactly when they have the same gradient, so the tangent's gradient must equal the gradient of y = 3x - 1, which is 3.
    2. Differentiate the curve: dy/dx = 2x - 5.
    3. Set this equal to 3: 2x - 5 = 3, so 2x = 8 and x = 4.
    4. Find the corresponding y-value: y = 4^2 - 5(4) + 7 = 16 - 20 + 7 = 3.
    5. So the point is (4, 3), and the answer is A.
    • Why not B: Correctly solves 2x - 5 = 3 to find x = 4, but then substitutes back into y = x^2 - 5x + 7 with a sign slip, computing 16 - 20 - 7 = -11 instead of 16 - 20 + 7 = 3.
    • Why not C: Confuses the gradient of the line y = 3x - 1 with its y-intercept, setting the derivative equal to -1 instead of to the gradient 3, giving 2x - 5 = -1, x = 2, and then y = 4 - 10 + 7 = 1.
    • Why not D: Correctly solves for x = 4, but substitutes back into y = x^2 - 5x + 7 with the linear term's sign flipped, computing 16 + 20 + 7 = 43 instead of 16 - 20 + 7 = 3.

  6. Question 6Answer: C

    1. Differentiate: dy/dx = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1).
    2. Set dy/dx = 0: the stationary points are at x=-2 and x=1.
    3. Differentiate again: d2y/dx2 = 12x + 6.
    4. At x=-2: d2y/dx2 = 12(-2) + 6 = -18, which is negative, so x=-2 is a maximum.
    5. At x=1: d2y/dx2 = 12(1) + 6 = 18, which is positive, so x=1 is a minimum.
    6. So the local maximum is at x=-2, and the answer is C.
    • Why not A: When dividing 6x^2 + 6x - 12 = 0 by 6, correctly divides the x^2 and x terms but forgets to also divide the constant term, working with x^2 + x - 12 = 0 instead of x^2 + x - 2 = 0; this factors as (x+4)(x-3)=0, giving wrong stationary points x=-4 and x=3. Applying the second derivative test correctly to these wrong roots (12(-4)+6=-42<0) selects x=-4 as the maximum.
    • Why not B: Factorises 6(x^2 + x - 2) with a sign slip as 6(x-2)(x+1) instead of 6(x+2)(x-1), giving wrong stationary points x=2 and x=-1. Applying the second derivative test to these wrong roots (12(-1)+6=-6<0) selects x=-1 as the maximum.
    • Why not D: Correctly finds the stationary points x=-2 and x=1, and correctly computes the second derivative 12x+6, but reverses the second derivative test, treating the positive value at x=1 (12(1)+6=18) as indicating a maximum rather than a minimum.

  7. Question 7Answer: D

    1. Differentiate: f'(x) = 3x^2 - 6x - 9.
    2. Factorise: f'(x) = 3(x^2 - 2x - 3) = 3(x-3)(x+1).
    3. The critical points are x=3 and x=-1.
    4. Since the coefficient of x^2 in f'(x) is positive, f'(x) is negative between its roots and positive outside them, so f'(x) > 0 for x < -1 or x > 3.
    5. A function is increasing where its derivative is positive, so f(x) is increasing for x < -1 or x > 3, and the answer is D.
    • Why not A: Correctly finds the critical points x=-1 and x=3 by solving f'(x) = 3(x-3)(x+1) = 0, but since f'(x) is an upward-opening parabola in x, it is negative BETWEEN its roots and positive outside them; this option selects the interval between the roots, where the function is decreasing, not increasing.
    • Why not B: Differentiates -3x^2 with a sign slip as +6x instead of -6x, giving f'(x) = 3x^2 + 6x - 9 = 3(x+3)(x-1) with roots x=-3 and x=1, and selects the region outside these wrong roots.
    • Why not C: Drops the -9x term entirely when differentiating the four-term expression, leaving f'(x) = 3x^2 - 6x = 3x(x-2) with roots x=0 and x=2, and selects the region outside these wrong roots.

  8. Question 8Answer: B

    1. A definite integral is signed: positive where the curve is above the axis, negative where it is below. Area, however, must always be positive.
    2. Since the curve crosses the axis at x=3, the region must be split there, and each piece's contribution to the area must be taken as a positive magnitude before combining.
    3. The region 0 < x < 3 contributes an area of |15| = 15.
    4. The region 3 < x < 8 contributes an area of |-40| = 40.
    5. Total area = 15 + 40 = 55, so the answer is B.
    • Why not A: Adds the two signed definite integrals directly, 15 + (-40) = -25, and reports this raw signed total as if it were the area, without recognising that area must always be positive and that a curve crossing the axis needs each region measured separately.
    • Why not C: Adds the two signed integrals to get 15 + (-40) = -25 first, and only then takes the absolute value of this already-cancelled combined total, |-25| = 25, instead of taking the absolute value of each region before combining them.
    • Why not D: Only accounts for the region above the axis, 0 < x < 3, where the integral is 15, and forgets to include the contribution of the region below the axis, 3 < x < 8, entirely.

  9. Question 9Answer: C

    1. Expand the square: (sqrt(x) + 2)^2 = x + 4sqrt(x) + 4 = x + 4x^(1/2) + 4.
    2. Integrate term by term using the integral of x^n dx = x^(n+1)/(n+1) for n != -1.
    3. Integral of x = x^2/2.
    4. Integral of 4x^(1/2) = 4 x x^(3/2)/(3/2) = 4 x (2/3) x^(3/2) = (8/3)x^(3/2).
    5. Integral of 4 = 4x.
    6. Add the constant of integration: x^2/2 + (8/3)x^(3/2) + 4x + C, so the answer is C.
    • Why not A: Expands (sqrt(x) + 2)^2 by dropping the cross term, treating it like a^2 + b^2 instead of a^2 + 2ab + b^2, so x + 4sqrt(x) + 4 becomes just x + 4, which integrates to x^2/2 + 4x + C with no middle term at all.
    • Why not B: Correctly expands to x + 4x^(1/2) + 4, but when integrating 4x^(1/2) raises the power to 3/2 and forgets to also divide by that new power, giving 4x^(3/2) instead of 4 x (2/3) x^(3/2) = (8/3)x^(3/2).
    • Why not D: Correctly expands to x + 4x^(1/2) + 4, but when integrating 4x^(1/2) multiplies by the reciprocal of the new power upside down, using 3/2 instead of 2/3, giving 4 x (3/2) x^(3/2) = 6x^(3/2) instead of (8/3)x^(3/2).

  10. Question 10Answer: A

    1. Expand the square: (sqrt(x) - 3)^2 = x - 6sqrt(x) + 9 = x - 6x^(1/2) + 9.
    2. Find the antiderivative: F(x) = x^2/2 - 6 x (2/3) x^(3/2) + 9x = x^2/2 - 4x^(3/2) + 9x.
    3. Evaluate at the upper limit: F(9) = 81/2 - 4(27) + 81 = 40.5 - 108 + 81 = 13.5 = 27/2.
    4. Evaluate at the lower limit: F(0) = 0 - 0 + 0 = 0.
    5. The definite integral is F(9) - F(0) = 27/2 - 0 = 27/2, so the answer is A.
    • Why not B: Expands (sqrt(x) - 3)^2 by dropping the cross term, treating it like a^2 + b^2 instead of a^2 - 2ab + b^2, so x - 6sqrt(x) + 9 becomes just x + 9, giving an antiderivative of x^2/2 + 9x, which evaluates to 121.5 - 0 = 243/2.
    • Why not C: Correctly expands to x - 6x^(1/2) + 9, but when integrating -6x^(1/2) raises the power to 3/2 without dividing by it, using -6x^(3/2) instead of -4x^(3/2), giving an antiderivative that evaluates to -40.5 - 0 = -81/2.
    • Why not D: Correctly integrates but loses the negative sign on the middle term, using +4x^(3/2) instead of -4x^(3/2) in the antiderivative, which evaluates to 229.5 - 0 = 459/2.

  11. Question 11Answer: D

    1. The Fundamental Theorem of Calculus states that if g(x) = the definite integral of f(t) dt from a fixed lower limit to x, then g'(x) = f(x): differentiating an integral with a variable upper limit simply recovers the integrand, evaluated at x.
    2. Here the integrand is f(t) = 3t^2 - 2, so g'(x) = 3x^2 - 2 directly, with no need to evaluate the integral first.
    3. As a check: evaluating the integral gives g(x) = [t^3 - 2t] from 1 to x = (x^3 - 2x) - (1 - 2) = x^3 - 2x + 1, and differentiating this confirms g'(x) = 3x^2 - 2.
    4. So the answer is D.
    • Why not A: Correctly evaluates the integral to find g(x) = x^3 - 2x + 1 itself, but stops there, confusing this expression for g(x) with the derivative g'(x) that the question actually asks for.
    • Why not B: Evaluates the integral but forgets to include the contribution from the lower limit t=1 (which contributes the constant +1), giving g(x) = x^3 - 2x instead of x^3 - 2x + 1, and then, like the other error, reports this as if it were g'(x) rather than g(x).
    • Why not C: Correctly recognises that g'(x) relates to the integrand, but then differentiates the integrand (3t^2 - 2) a second time with respect to x, as if the Fundamental Theorem required one more differentiation, giving d/dx(3x^2 - 2) = 6x instead of stopping at 3x^2 - 2.

  12. Question 12Answer: B

    1. Integration is linear over a fixed range: the definite integral of (2f(x) - 3g(x)) dx equals 2 times the definite integral of f(x) dx, minus 3 times the definite integral of g(x) dx, over the same range.
    2. The integral of f(x) dx from 2 to 7 is 15, so 2 x 15 = 30.
    3. The integral of g(x) dx from 2 to 7 is -6, so 3 x (-6) = -18, and subtracting this term means subtracting -18, which is the same as adding 18.
    4. Combining: 30 - (-18) = 30 + 18 = 48, so the answer is B.
    • Why not A: Correctly scales the integral of f(x) by 2, giving 2(15)=30, but uses the integral of g(x) directly without multiplying it by its coefficient 3, computing 30 - (-6) = 36 instead of 30 - 3(-6) = 48.
    • Why not C: Uses the integral of f(x) directly without multiplying it by its coefficient 2, computing 15 - 3(-6) = 15 + 18 = 33 instead of 2(15) - 3(-6) = 48.
    • Why not D: Mishandles the subtraction sign in 2f(x) - 3g(x), treating it as though it were 2f(x) + 3g(x), computing 2(15) + 3(-6) = 30 - 18 = 12 instead of 2(15) - 3(-6) = 30 + 18 = 48.

  13. Question 13Answer: A

    1. Evaluate y = x^2 + 1 at the five strip boundaries x=0,1,2,3,4: y0=1, y1=2, y2=5, y3=10, y4=17.
    2. The trapezium rule for n strips of width h is: integral ~ h/2 x [y0 + yn + 2(y1 + ... + y_(n-1))].
    3. Here h=1, so the estimate = 0.5 x [1 + 17 + 2(2+5+10)] = 0.5 x [18 + 2(17)] = 0.5 x [18 + 34] = 0.5 x 52 = 26.
    4. So the answer is A.
    • Why not B: Correctly evaluates the five ordinates as 1, 2, 5, 10, 17, but forgets to double the three interior ordinates in the trapezium rule formula, computing 0.5 x (1+2+5+10+17) = 17.5 instead of doubling y1, y2 and y3.
    • Why not C: Correctly evaluates the ordinates and correctly doubles the interior ones (1 + 17 + 2(2+5+10) = 52), but forgets to multiply this total by h/2 = 1/2 at the end, leaving double the correct estimate.
    • Why not D: Evaluates the function incorrectly as y = x^2 at each strip boundary, dropping the '+1' constant term, giving ordinates 0, 1, 4, 9, 16 instead of 1, 2, 5, 10, 17, and applies the trapezium rule correctly to these wrong values.

  14. Question 14Answer: C

    1. Differentiate y = x^2 - 2x + 5 twice: dy/dx = 2x - 2, so d2y/dx2 = 2.
    2. Since d2y/dx2 = 2 is positive for every x, the curve is concave up over the whole interval 0 <= x <= 3.
    3. For a concave-up curve, the curve bends away below a straight line joining two of its points, so each chord lies above the curve.
    4. The trapezium rule approximates the area under the curve using these chords as the tops of trapezia, so each trapezium's area is more than the true area under the curve on that strip.
    5. So the trapezium rule overestimates the integral here, and the answer is C.
    • Why not A: States the correct geometric fact, that the chord lies above a concave-up curve, but then draws the opposite conclusion: a trapezium whose top edge sits above the curve covers MORE area than the curve, not less, so it overestimates rather than underestimating.
    • Why not B: Differentiates dy/dx = 2x - 2 correctly, but makes a sign error differentiating a second time, reporting d2y/dx2 = -2 instead of the correct +2; this wrongly signed value implies concave down, whose chords lie below the curve, leading to the wrong 'underestimate' conclusion.
    • Why not D: Assumes the trapezium rule has a fixed bias in every case, ignoring that whether it over- or under-estimates the true integral depends entirely on the concavity of the curve.

  15. Question 15Answer: B

    1. Simplify dy/dx before integrating: (x^3 - 8)/x^2 = x^3/x^2 - 8/x^2 = x - 8x^-2.
    2. Integrate term by term: integral of x dx = x^2/2; integral of -8x^-2 dx = -8 x x^-1/(-1) = 8x^-1 = 8/x.
    3. So y = x^2/2 + 8/x + C.
    4. Use the point (2, 10) to find C: substitute x=2, y=10: 10 = (2)^2/2 + 8/2 + C = 2 + 4 + C = 6 + C, so C = 4.
    5. The particular solution is y = x^2/2 + 8/x + 4, so the answer is B.
    • Why not A: Fails to simplify x^3/x^2 to x before integrating, and instead integrates the original unsimplified numerator term-by-term as if dy/dx were x^3 - 8x^-2, giving an antiderivative of x^4/4 + 8/x + C; using the point (2,10) with this wrong antiderivative gives C=2.
    • Why not C: Correctly simplifies dy/dx to x - 8x^-2, but makes a sign error integrating -8x^-2, forgetting to divide by the new negative power, and uses -8x^-1 instead of the correct +8x^-1; using the point (2,10) with this wrong antiderivative gives C=12.
    • Why not D: Correctly simplifies and integrates to reach y = x^2/2 + 8/x + C, but never uses the given point (2,10) to find C, effectively leaving the constant of integration as 0.

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