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: no calculator, exact answers, and no antiderivatives (use geometry and the properties of definite integrals). Notation: ∫ (a to b) f(x) dx means the definite integral of f from x = a to x = b. This set is for both Calculus AB and Calculus BC students.
Question 1 (multiple choice · foundation)
f is continuous, ∫ (1 to 6) f(x) dx = 8 and ∫ (1 to 4) f(x) dx = 11. What is ∫ (6 to 4) 2f(x) dx?
- (A) −6
- (B) 6
- (C) 3
- (D) 38
Answer and explanation
Answer: (B). Adjacent intervals: ∫ (4 to 6) f = ∫ (1 to 6) f − ∫ (1 to 4) f = 8 − 11 = −3. Reverse the limits: ∫ (6 to 4) f = 3. Take out the constant: ∫ (6 to 4) 2f = 2 × 3 = 6.
- (A) is 2 ∫ (4 to 6) f. It forgets that the limits run from 6 down to 4.
- (C) is ∫ (6 to 4) f. It forgets the factor 2.
- (D) adds the two given integrals instead of subtracting: 2(8 + 11).
Question 2 (multiple choice · core)
What is ∫ (−4 to 0) [√(16 − x²) + 3] dx?
- (A) 4π + 3
- (B) 4π + 12
- (C) 8π + 12
- (D) 16π + 12
Answer and explanation
Answer: (B). Split the sum. On [−4, 0], y = √(16 − x²) is a quarter circle of radius 4, area (1/4)π(4²) = 4π. ∫ (−4 to 0) 3 dx is a rectangle 3 high and 4 wide, area 12. Total: 4π + 12.
- (A) adds 3 instead of 3 × (width 4).
- (C) uses a semicircle. From −4 to 0 is only a quarter of the circle.
- (D) uses the whole circle.
Question 3 (multiple choice · core)
Let f(x) = 2 for x < 3 and f(x) = x for x ≥ 3. What is ∫ (0 to 5) f(x) dx?
- (A) 10
- (B) 25/2
- (C) 14
- (D) It does not exist, because f is not continuous on [0, 5].
Answer and explanation
Answer: (C). f has a jump at x = 3, so split there. ∫ (0 to 3) 2 dx = 6 (a rectangle). ∫ (3 to 5) x dx is a trapezoid with parallel sides 3 and 5 and width 2: (1/2)(3 + 5)(2) = 8. Total: 14.
- (A) uses f = 2 on the whole interval.
- (B) uses f = x on the whole interval: a triangle with area (1/2)(5)(5).
- (D) is false. A bounded function with a jump discontinuity still has a definite integral; you split at the jump.
Question 4 (multiple choice · core)
f and g are continuous on an interval containing a, b and c. Which statement must be true?
- (A) ∫ (a to b) f(x)g(x) dx = (∫ (a to b) f(x) dx)(∫ (a to b) g(x) dx)
- (B) ∫ (a to b) [f(x) + 3] dx = ∫ (a to b) f(x) dx + 3
- (C) ∫ (a to b) f(x) dx + ∫ (b to c) f(x) dx = ∫ (a to c) f(x) dx
- (D) ∫ (a to b) |f(x)| dx = |∫ (a to b) f(x) dx|
Answer and explanation
Answer: (C). This is the adjacent-interval property. It holds for any order of a, b and c.
- (A) fails for f(x) = g(x) = x on [0, 2]: the left side is 8/3, the right side is 2 × 2 = 4. There is no product rule for integrals.
- (B) fails for f(x) = 0 on [0, 2]: the left side is ∫ (0 to 2) 3 dx = 6, the right side is 3. The constant must be multiplied by the width.
- (D) fails for f(x) = x on [−1, 1]: total area 1, but net area 0.
Question 5 (graph · core)
The graph of f on 0 ≤ x ≤ 8 is shown above. It is a semicircle on [0, 4), a horizontal segment on [4, 6) and part of the line y = x − 7 on [6, 8], except that f(7) = 3.
(a) Find ∫ (0 to 8) f(x) dx. (b) Find ∫ (8 to 4) f(x) dx. (c) Find ∫ (0 to 6) [3f(x) − 1] dx. (d) f is not continuous on [0, 8]. Explain why the integral in (a) still exists, and why the value f(7) = 3 does not affect it.
Worked solution and suggested Marlbridge rubric
Signed areas of the pieces:
- [0, 4]: semicircle of radius 2, area (1/2)π(2²) = 2π.
- [4, 6]: rectangle 2 wide and 2 deep, below the axis: −4.
- [6, 7]: triangle with base 1 and height 1, below: −1/2.
- [7, 8]: triangle with base 1 and height 1, above: +1/2.
(a) 2π − 4 − 1/2 + 1/2 = 2π − 4.
(b) ∫ (4 to 8) f = −4 − 1/2 + 1/2 = −4. Reversing the limits: ∫ (8 to 4) f = 4.
(c) ∫ (0 to 6) f = 2π − 4. So ∫ (0 to 6) [3f − 1] = 3(2π − 4) − 1 × 6 = 6π − 18.
(d) f is bounded and has only finitely many discontinuities, all jumps (at 4 and 6) or a removable one (at 7). Split the integral at 4 and 6 and add the pieces. At x = 7, f differs from the line only at a single point. A point has zero width, so it adds no area.
| Point | What earns it |
|---|---|
| 1 | Correct signed areas of all four pieces, including 2π for the semicircle |
| 1 | (a) 2π − 4 and (b) 4, with the sign change from reversed limits |
| 1 | (c) 6π − 18, multiplying the constant 1 by the width 6 |
| 1 | (d) Splitting at the jumps, and a single point having zero width (no area) |
Question 6 (constructed response · core)
f and g are continuous, with
∫ (0 to 3) f(x) dx = 4, ∫ (0 to 7) f(x) dx = −2, ∫ (3 to 7) g(x) dx = 5.
(a) Find ∫ (3 to 7) f(x) dx. (b) Find ∫ (3 to 7) [4f(x) − g(x)] dx. (c) Find ∫ (7 to 3) [g(x) + 2] dx. (d) Find the constant k for which ∫ (0 to 7) [f(x) + k] dx = 12.
Worked solution and suggested Marlbridge rubric
(a) ∫ (3 to 7) f = ∫ (0 to 7) f − ∫ (0 to 3) f = −2 − 4 = −6.
(b) 4(−6) − 5 = −29.
(c) Reverse the limits: −∫ (3 to 7) [g + 2] = −(5 + 2 × 4) = −13. (The width from 3 to 7 is 4.)
(d) ∫ (0 to 7) f + k(7 − 0) = −2 + 7k. Set −2 + 7k = 12, so k = 2.
| Point | What earns it |
|---|---|
| 1 | (a) −6 using adjacent intervals |
| 1 | (b) −29 using the constant-multiple and difference properties |
| 1 | (c) −13: reversed limits and the constant 2 multiplied by the width 4 |
| 1 | (d) −2 + 7k = 12 and k = 2 |
Question 7 (constructed response · stretch)
Let p(x) = |x − 2| − 1.
(a) Find ∫ (−1 to 5) p(x) dx. (b) Find the total area between the graph of p and the x-axis for −1 ≤ x ≤ 5. Explain why it differs from (a). (c) Let s(x) = p(x) for x ≠ 2, and s(2) = 10. Find ∫ (−1 to 5) s(x) dx, and justify. (d) Find the value of c, with c > 3, for which ∫ (1 to c) p(x) dx = 0.
Worked solution and suggested Marlbridge rubric
The graph of p is a V with its point at (2, −1). It crosses the axis where |x − 2| = 1, at x = 1 and x = 3.
(a) Three triangles:
- [−1, 1]: above the axis, base 2, height p(−1) = 2, area +2.
- [1, 3]: below the axis, base 2, depth 1, area −1.
- [3, 5]: above the axis, base 2, height p(5) = 2, area +2.
So ∫ (−1 to 5) p = 2 − 1 + 2 = 3. (Alternative: ∫ |x − 2| = 9/2 + 9/2 = 9, and ∫ 1 dx over width 6 is 6; 9 − 6 = 3.)
(b) Total area counts every region as positive: 2 + 1 + 2 = 5. The integral in (a) is smaller because the region from 1 to 3 lies below the axis and is subtracted.
(c) s differs from p only at the single point x = 2. A point has zero width, so it contributes no area, and ∫ (−1 to 5) s = 3.
(d) ∫ (1 to 3) p = −1. For c > 3, the region from 3 to c is a triangle above the axis with base c − 3 and height c − 3, area (c − 3)²/2. Adjacent intervals: −1 + (c − 3)²/2 = 0, so (c − 3)² = 2 and c = 3 + √2 (about 4.41).
| Point | What earns it |
|---|---|
| 1 | (a) 3, from signed areas or from the alternative split |
| 1 | (b) Total area 5, with the reason that the integral subtracts the region below the axis |
| 1 | (c) 3, because changing one point does not change the integral |
| 1 | (d) −1 + (c − 3)²/2 = 0 and c = 3 + √2 (the root 3 − √2 is rejected because c > 3) |
How did you do?
- Q1 or Q6 wrong: redo Worked example 1 in the study guide, and watch reversed limits and constants.
- Q2 or Q7(a) wrong: reread “Integrals from geometry” and Worked example 3.
- Q3, Q5(d) or Q7(c) wrong: reread “Integrals of functions with jumps and holes” and Worked example 2.
- Q4 wrong: reread “What is not a property”.
Then tick off the topic checklist.