These are original Marlbridge practice questions, not past exam questions. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Assumptions for every question: angles in radians; no calculator except in Question 6(c), where decimals are given to 3 decimal places; exact answers elsewhere. Notation: ∫ (a to b) f(x) dx is the definite integral from a to b, and [F(x)] (a to b) means F(b) − F(a). The context in Question 6 is invented. This set is for both Calculus AB and Calculus BC students; Question 7(d) is for BC students only.
Question 1 (multiple choice · foundation)
∫ (4x + 6)/(x² + 3x + 1) dx =
- (A) 2 ln|x² + 3x + 1| + C
- (B) ln|x² + 3x + 1| + C
- (C) (4x + 6) ln|x² + 3x + 1| + C
- (D) (2x² + 6x)/(x³/3 + 3x²/2 + x) + C
Answer and explanation
Answer: (A). The signal: the derivative of the bottom is 2x + 3, and the top is 4x + 6 = 2(2x + 3). Let u = x² + 3x + 1, so du = (2x + 3) dx and (4x + 6) dx = 2 du. Then ∫ 2/u du = 2 ln|u| + C = 2 ln|x² + 3x + 1| + C.
- (B) misses the factor 2. Its derivative is (2x + 3)/(x² + 3x + 1), half the integrand.
- (C) keeps the top outside the logarithm. Only constants can come outside an integral, and the 4x + 6 is used up by du.
- (D) integrates the top and the bottom separately. ∫ f/g dx is not (∫ f dx)/(∫ g dx).
Question 2 (multiple choice · core)
∫ 1/(x² − 8x + 17) dx =
- (A) ln|x² − 8x + 17| + C
- (B) arctan(x − 4) + C
- (C) −1/(x − 4) + C
- (D) (1/(2x − 8)) ln|x² − 8x + 17| + C
Answer and explanation
Answer: (B). The signal: a quadratic denominator with nothing on top. Its discriminant is 64 − 68 = −4 < 0, so it does not factor. Complete the square: x² − 8x + 17 = (x − 4)² + 1. With u = x − 4, du = dx, the integral is ∫ 1/(u² + 1) du = arctan u + C = arctan(x − 4) + C.
- (A) uses the ln pattern without the derivative of the bottom on top. Its derivative is (2x − 8)/(x² − 8x + 17), not the integrand.
- (C) drops the + 1 and treats the bottom as (x − 4)². Its derivative is 1/(x − 4)².
- (D) divides by du/dx = 2x − 8 to “fix” the missing factor. You cannot divide by a variable expression this way, and differentiating (D) does not give back the integrand.
Question 3 (multiple choice · core)
For which integral is the substitution u = x² + 4 the most effective first step?
- (A) ∫ 1/(x² + 4) dx
- (B) ∫ x(x² + 4)⁵ dx
- (C) ∫ (x² + 4)⁵ dx
- (D) ∫ x²/(x² + 4) dx
Answer and explanation
Answer: (B). du = 2x dx, so the integrand needs a factor x (up to a constant). Only (B) has it: x dx = ½ du, and ∫ ½ u⁵ du = u⁶/12 + C = (x² + 4)⁶/12 + C.
- (A) has no x on top. It is an arctan form: ½ arctan(x/2) + C.
- (C) has no factor x either. Expand (x² + 4)⁵ and use the power rule term by term.
- (D) has top degree equal to bottom degree, so divide first: x²/(x² + 4) = 1 − 4/(x² + 4), giving x − 2 arctan(x/2) + C.
Question 4 (multiple choice · core)
Exactly one of these indefinite integrals cannot be written using powers, roots, exponentials, logarithms, trig and inverse trig functions. Which one?
- (A) ∫ x²√(x³ + 4) dx
- (B) ∫ √(x³ + 4) dx
- (C) ∫ cos²x sin x dx
- (D) ∫ cos(2x) dx
Answer and explanation
Answer: (B). √(x³ + 4) has no antiderivative in familiar functions. No substitution helps, because the derivative of x³ + 4, which is 3x², is not present, and there is nothing to expand or divide. A definite integral of √(x³ + 4) still exists; you would find its value with technology.
- (A) has the factor x²: with u = x³ + 4, x² dx = (1/3) du, so the answer is (2/9)(x³ + 4)^(3/2) + C.
- (C) has sin x, the derivative of cos x up to a sign: with u = cos x, the answer is −cos³x/3 + C.
- (D) is a basic rule with a linear inside: ½ sin(2x) + C.
Question 5 (constructed response · core)
For each integral, state the signal you see and the technique you will use, then find the integral.
(a) ∫ (x³ − 2x + 1)/x² dx (b) ∫ eˣ/(eˣ + 3) dx (c) ∫ (x² + 2x)/(x + 1) dx
(d) Check your answer to (c) by differentiating it.
Worked solution and suggested Marlbridge rubric
(a) Signal: a fraction over a single term. Technique: split. (x³ − 2x + 1)/x² = x − 2/x + x⁻². So the integral is x²/2 − 2 ln|x| − 1/x + C.
(b) Signal: the derivative of eˣ + 3 is eˣ, which is the top. Technique: substitution with u = eˣ + 3, du = eˣ dx. ∫ (1/u) du = ln|u| + C = ln(eˣ + 3) + C. No absolute value is needed, because eˣ + 3 > 0.
(c) Signal: top degree 2 > bottom degree 1. Technique: long division. x² + 2x = (x + 1)(x + 1) − 1, so (x² + 2x)/(x + 1) = x + 1 − 1/(x + 1). The integral is x²/2 + x − ln|x + 1| + C.
(d) d/dx [x²/2 + x − ln|x + 1|] = x + 1 − 1/(x + 1) = ((x + 1)² − 1)/(x + 1) = (x² + 2x)/(x + 1). This is the integrand, so (c) is correct.
| Point | What earns it |
|---|---|
| 1 | (a) Splits the fraction into powers and integrates all three terms correctly, including ln |
| 1 | (b) Names substitution with u = eˣ + 3 and obtains ln(eˣ + 3) + C |
| 1 | (c) Divides first and obtains x²/2 + x − ln |
| 1 | (d) Differentiates and shows the result simplifies to the original integrand |
Acceptable alternative for (c): substitute w = x + 1 so x² + 2x = w² − 1, then ∫ (w − 1/w) dw = w²/2 − ln|w| + C. This differs from the answer above only by a constant. Omitting + C throughout costs one point in total (take it from (a)), not one per part.
Question 6 (constructed response · core)
Invented context. Water flows into a garden tank through three inlets. Time t is measured in minutes, and each rate is in litres per minute.
- Inlet P: p(t) = 40t/(t² + 16)
- Inlet Q: q(t) = 40/(t² + 16)
- Inlet R: r(t) = 5e^(−0.04t²)
(a) Without a calculator, find the exact volume that flows through inlet P from t = 0 to t = 6. Give your answer in the form k ln m. (b) Without a calculator, find the exact volume that flows through inlet Q from t = 0 to t = 4. (c) A student tries to find ∫ (0 to 6) r(t) dt by substitution and gets stuck. Explain why, then use a calculator to find the volume through inlet R from t = 0 to t = 6.
Worked solution and suggested Marlbridge rubric
(a) Signal: the derivative of t² + 16 is 2t, and the top is 40t = 20(2t). Let u = t² + 16, du = 2t dt, so 40t dt = 20 du. New limits: t = 0 gives u = 16; t = 6 gives u = 52. ∫ (16 to 52) 20/u du = 20 [ln u] (16 to 52) = 20 ln(52/16) = 20 ln(13/4) litres (about 23.573 litres).
(b) Signal: no t on top, and t² + 16 = t² + 4² has no real roots. Technique: arctan form. ∫ (0 to 4) 40/(t² + 4²) dt = 40 × (1/4) [arctan(t/4)] (0 to 4) = 10(arctan 1 − arctan 0) = 10 × π/4 = 5π/2 litres (about 7.854 litres).
(c) Substitution with u = −0.04t² needs du = −0.08t dt, but there is no factor t in r(t). No other technique works either: e^(−0.04t²) has no antiderivative in familiar functions. The definite integral still exists, so use technology: ∫ (0 to 6) 5e^(−0.04t²) dt ≈ 20.169 litres.
| Point | What earns it |
|---|---|
| 1 | (a) Uses substitution with u = t² + 16 and changes the limits to 16 and 52 (or returns to t before substituting) |
| 1 | (a) Obtains 20 ln(13/4) litres, or an equivalent such as 20 ln 52 − 20 ln 16 |
| 1 | (b) Uses the arctan form with k = 4 and obtains 5π/2 litres |
| 1 | (c) Explains that the factor t needed for substitution is missing / there is no closed-form antiderivative, and gives 20.169 litres |
Units are required in at least one answer. Using u-limits 0 and 6 after substituting in (a) loses the second point.
Question 7 (constructed response · stretch)
Let I = ∫ (0 to 1) (2x³ + 3x² + 2x + 5)/(x² + 1) dx.
(a) A student says, “The bottom is x² + 1, so I will substitute u = x² + 1.” Explain why this is not a good first step. (b) Use long division to write the integrand as a polynomial plus a proper fraction. (c) Find the exact value of I. (d) BC only. Find ∫ (0 to 1) 1/(x² + 7x + 12) dx. Name the technique, and say why neither substitution nor arctan applies.
Worked solution and suggested Marlbridge rubric
(a) The top has degree 3 and the bottom degree 2, so the fraction is top-heavy. Also du = 2x dx, and the top is not a constant multiple of x. Substitution leaves a mixture of x and u. Long division is the right first step.
(b) Divide 2x³ + 3x² + 2x + 5 by x² + 1: (x² + 1)(2x + 3) = 2x³ + 3x² + 2x + 3, which leaves a remainder of 2. So
(2x³ + 3x² + 2x + 5)/(x² + 1) = 2x + 3 + 2/(x² + 1)
(c) Integrate each piece: ∫ (2x + 3) dx = x² + 3x, and ∫ 2/(x² + 1) dx = 2 arctan x. I = [x² + 3x + 2 arctan x] (0 to 1) = (1 + 3 + 2 × π/4) − 0 = 4 + π/2 (about 5.571).
(d) BC only. There is no x on top, so substitution has nothing to work with. The discriminant is 49 − 48 = 1 > 0, so the bottom factors and the arctan form does not apply: x² + 7x + 12 = (x + 3)(x + 4). Technique: linear partial fractions. 1 = A(x + 4) + B(x + 3); x = −3 gives A = 1 and x = −4 gives B = −1. So the integral is [ln|x + 3| − ln|x + 4|] (0 to 1) = (ln 4 − ln 5) − (ln 3 − ln 4) = ln(16/15) (about 0.065).
| Point | What earns it (AB: out of 4; BC: out of 5) |
|---|---|
| 1 | (a) Notes the top’s degree is at least the bottom’s, or that du = 2x dx does not match the top |
| 1 | (b) Correct quotient 2x + 3 and remainder 2 |
| 1 | (c) Correct antiderivative x² + 3x + 2 arctan x |
| 1 | (c) Correct value 4 + π/2, using arctan 1 = π/4 |
| 1 | (d) BC only: partial fractions 1/(x + 3) − 1/(x + 4), a reason for rejecting substitution and arctan, and the value ln(16/15) |
Acceptable equivalent forms for (d): 2 ln 4 − ln 3 − ln 5, or ln 16 − ln 15.
How did you do?
- Q1, Q3 or Q6(a) wrong: revisit the substitution signal in “The toolkit and the signal for each tool” in the study guide.
- Q2 or Q6(b) wrong: see Worked example 1(d) on completing the square and the arctan form.
- Q5 or Q7 wrong: compare your work with the lookalike table and Worked example 2; divide or split before anything else.
- Q4 or Q6(c) wrong: reread “When there is no formula”.
- BC students, Q7(d) wrong: see Worked example 3 and the linear partial fractions guide.
Then tick off the topic checklist.