AP Calculus AB Practice Quiz: Approximating Areas with Riemann Sums
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 11 questions to check your progress.
Question 1 of 11
All Questions (11)
A) The derivative of a function at a point
B) A definite integral
C) An indefinite integral
D) The limit of a function
Correct Answer: B
A definite integral represents the net signed area under a curve. A Riemann sum approximates this area by summing the areas of geometric shapes, such as rectangles, over partitions of the interval. [cite: 2625]
A) 46
B) 56
C) 69
D) 70
Correct Answer: A
The width of each of the three subintervals is (6-0)/3 = 2. A left Riemann sum uses the function value at the left endpoint of each subinterval. The approximation is `Δx * [f(0) + f(2) + f(4)] = 2 * [5 + 7 + 11] = 2 * 23 = 46`. [cite: 2626, 2627]
A) 8
B) 10
C) 12
D) 14
Correct Answer: C
The interval [1, 5] is divided into four subintervals, so the width of each is (5-1)/4 = 1. The subintervals are [1, 2], [2, 3], [3, 4], and [4, 5]. A right Riemann sum uses the right endpoints: x=2, 3, 4, 5. From the graph, f(2)=2, f(3)=3, f(4)=4, and f(5)=3. The approximation is `1*f(2) + 1*f(3) + 1*f(4) + 1*f(5) = 1*2 + 1*3 + 1*4 + 1*3 = 12`. [cite: 2626, 2627]
A) 155 ft
B) 165 ft
C) 170 ft
D) 180 ft
Correct Answer: B
The distance is the integral of velocity. The trapezoidal sum is calculated for each subinterval. The subintervals have widths 3-0=3, 8-3=5, and 10-8=2. The approximation is `(1/2)(v(0)+v(3))*3 + (1/2)(v(3)+v(8))*5 + (1/2)(v(8)+v(10))*2 = (1/2)(10+20)*3 + (1/2)(20+18)*5 + (1/2)(18+12)*2 = (1/2)(30)*3 + (1/2)(38)*5 + (1/2)(30)*2 = 45 + 95 + 30 = 165` feet. [cite: 2627]
A) 10
B) 17
C) 20
D) 34
Correct Answer: C
With two subintervals on [0, 4], the width of each is (4-0)/2 = 2. The subintervals are [0, 2] and [2, 4]. The midpoints are x=1 and x=3. The function is f(x) = x^2. The midpoint Riemann sum approximation is `width * (f(midpoint1) + f(midpoint2)) = 2 * (f(1) + f(3)) = 2 * (1^2 + 3^2) = 2 * (1 + 9) = 2 * 10 = 20`. [cite: 2626, 2627]
A) It is an overestimate.
B) It is an underestimate.
C) It is exactly equal to the definite integral.
D) It is impossible to determine without knowing the concavity.
Correct Answer: A
For a decreasing function, the left endpoint of any subinterval is the highest point in that subinterval. Therefore, the rectangle formed using the left endpoint's height will have an area greater than the area under the curve for that subinterval, resulting in an overall overestimate of the definite integral. [cite: 2629]
A) An overestimate.
B) An underestimate.
C) Exactly equal to the true area.
D) Dependent on the concavity of `g(x)`.
Correct Answer: A
For an increasing function, the right endpoint of any subinterval is the highest point in that subinterval. The rectangle formed using the right endpoint's height will have an area greater than the area under the curve for that subinterval, leading to an overall overestimate. [cite: 2629]
A) The function `f` is increasing.
B) The function `f` is decreasing.
C) The graph of `f` is concave up.
D) The graph of `f` is concave down.
Correct Answer: C
The top of each trapezoid in a trapezoidal sum is a secant line connecting two points on the curve. If the function is concave up, the curve lies below its secant lines. Therefore, the area of each trapezoid will be greater than the area under the curve on that subinterval, resulting in an overestimate. [cite: 2629]
A) 10 * [R(0) + R(10) + R(20) + R(30) + R(40) + R(50)]
B) 10 * [R(10) + R(20) + R(30) + R(40) + R(50) + R(60)]
C) 6 * [R(0) + R(10) + R(20) + R(30) + R(40) + R(50) + R(60)]
D) 6 * [R(10) + R(20) + R(30) + R(40) + R(50) + R(60)]
Correct Answer: B
The total interval is from 0 to 60 minutes. With 6 subintervals of equal width, the width of each subinterval is Δt = (60-0)/6 = 10 minutes. A right Riemann sum uses the right endpoint of each subinterval. The right endpoints are t = 10, 20, 30, 40, 50, and 60. The sum is the width of the subintervals (10) multiplied by the sum of the function values at these right endpoints. [cite: 2626, 2627]
A) L < T < I < R
B) L < I < T < R
C) R < I < T < L
D) T < I < L < R
Correct Answer: A
Since `f` is increasing, the left sum (L) is an underestimate and the right sum (R) is an overestimate (`L < I < R`). Since `f` is concave down, the secant lines for the trapezoidal sum (T) lie below the curve, making T an underestimate (`T < I`). For any subinterval, the area of the trapezoid is greater than the area of the left rectangle because the function is increasing. Thus, `L < T`. Combining these inequalities gives the final order: `L < T < I < R`. [cite: 2629]
A) These methods always provide the exact value of the integral.
B) They are only applicable to functions represented by a table of values.
C) The antiderivative of the integrand cannot be found or is difficult to find in terms of elementary functions.
D) They are used to find the derivative of the function.
Correct Answer: C
The Fundamental Theorem of Calculus allows for exact evaluation of definite integrals, but it requires finding an antiderivative. For many functions, an elementary antiderivative does not exist or is very difficult to find. Numerical methods provide a way to approximate the value of the definite integral in such cases, or when the function is only defined by discrete data points. [cite: 2628]