AP Calculus AB Quiz: Approximating Areas With Riemann Sums

What this quiz covers

This quiz focuses on Approximating Areas With Riemann Sums, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $R_n$ is an overestimate of the integral.
2. $R_n$ is an underestimate of the integral. (correct answer)
3. $R_n$ is equal to the value of the integral.
4. The relationship cannot be determined from the given information.

Explanation: For a strictly decreasing function on a given subinterval, the function's value at the right endpoint is the minimum value on that subinterval. Therefore, a right Riemann sum, which uses the function values at the right endpoints, will be an underestimate of the true value of the integral.

Question 2

1. $T_n < I < L_n$ (correct answer)
2. $L_n < I < T_n$
3. $I < T_n < L_n$
4. $T_n < L_n < I$

Explanation: Since $f'(x) < 0$, the function $f$ is decreasing. For a decreasing function, the left Riemann sum $L_n$ is an overestimate of the integral $I$. Since $f''(x) < 0$, the function is concave down. For a concave down function, the trapezoidal sum $T_n$ is an underestimate of the integral $I$. Combining these facts, we have $T_n < I < L_n$.

Question 3

1. $\frac{A_L + A_R}{2}$ (correct answer)
2. $A_L + A_R$
3. $\sqrt{A_L A_R}$
4. $A_R - A_L$

Explanation: For equal subintervals of width $\Delta x$, the left Riemann sum is $A_L = \Delta x \sum_{i=0}^{n-1} f(x_i)$ and the right Riemann sum is $A_R = \Delta x \sum_{i=1}^{n} f(x_i)$. The trapezoidal sum is $A_T = \frac{\Delta x}{2} \sum_{i=1}^{n} (f(x_{i-1}) + f(x_i)) = \frac{\Delta x}{2} [ (f(x_0)+f(x_1)) + ... + (f(x_{n-1})+f(x_n)) ] = \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)] = \frac{1}{2} [\Delta x(f(x_0)+...+f(x_{n-1})) + \Delta x(f(x_1)+...+f(x_n))] = \frac{A_L+A_R}{2}$.

Question 4

1. 200 cubic meters
2. 400 cubic meters (correct answer)
3. 420 cubic meters
4. 540 cubic meters

Explanation: The total volume is approximated by the integral of the rate, $\int_0^6 R(t) dt$. The interval is $[0, 6]$ with 3 subintervals, so the width is $\Delta t = \frac{6-0}{3}=2$ hours. A left Riemann sum uses the left endpoints $t=0, 2, 4$. The approximation is $\Delta t [R(0) + R(2) + R(4)] = 2[50+70+80] = 2[200] = 400$. The units are (hours) * (cubic meters/hour) = cubic meters.

Question 5

1. 26
2. 36
3. 41 (correct answer)
4. 46

Explanation: The interval of integration is $[0, 3]$ with $n=3$ subintervals, so the width of each subinterval is $\Delta x = \frac{3-0}{3} = 1$. The subintervals are $[0,1], [1,2], [2,3]$. A right Riemann sum uses the right endpoints $x=1, 2, 3$. The approximation is $\Delta x [f(1) + f(2) + f(3)] = 1[8 + 13 + 20] = 41$.

Question 6

1. 14 (correct answer)
2. 21
3. 22
4. 30

Explanation: The interval is $[0, 4]$ with $n=4$ subintervals, so the width of each subinterval is $\Delta x = \frac{4-0}{4} = 1$. The subintervals are $[0,1], [1,2], [2,3], [3,4]$. For a left Riemann sum, we use the left endpoints of these subintervals, which are $0, 1, 2, 3$. The approximation is $\Delta x [f(0) + f(1) + f(2) + f(3)] = 1[0^2 + 1^2 + 2^2 + 3^2] = 1[0+1+4+9] = 14$.

Question 7

1. 50
2. 60
3. 70 (correct answer)
4. 35

Explanation: The interval is $[1, 7]$ with $n=3$ subintervals, so the width of each subinterval is $\Delta x = \frac{7-1}{3} = 2$. The subintervals are $[1,3], [3,5], [5,7]$. For a right Riemann sum, we use the right endpoints $3, 5, 7$. The approximation is $\Delta x [g(3) + g(5) + g(7)] = 2[8 + 12 + 15] = 2[35] = 70$.

Question 8

1. 1.0
2. 3.5 (correct answer)
3. 5.0
4. 9.0

Explanation: The interval is $[0, 2]$ with $n=2$ subintervals, so the width is $\Delta x = \frac{2-0}{2} = 1$. The subintervals are $[0,1]$ and $[1,2]$. The midpoints are $0.5$ and $1.5$. The midpoint Riemann sum is $\Delta x [f(0.5) + f(1.5)] = 1[(0.5)^3 + (1.5)^3] = 1[0.125 + 3.375] = 3.5$.

Question 9

1. $L_n < \int_a^b f(x) dx < R_n$ (correct answer)
2. $R_n < \int_a^b f(x) dx < L_n$
3. $L_n < R_n < \int_a^b f(x) dx$
4. $\int_a^b f(x) dx < L_n < R_n$

Explanation: For a strictly increasing function on a given subinterval, the minimum value occurs at the left endpoint and the maximum value occurs at the right endpoint. Therefore, a left Riemann sum ($L_n$) will be an underestimate of the true integral, and a right Riemann sum ($R_n$) will be an overestimate. This leads to the inequality $L_n < \int_a^b f(x) dx < R_n$.

Question 10

1. $T_4$ is an underestimate of $\int_0^4 h(x) dx$.
2. $T_4$ is an overestimate of $\int_0^4 h(x) dx$. (correct answer)
3. $T_4$ is equal to $\int_0^4 h(x) dx$.
4. The relationship cannot be determined without knowing if $h(x)$ is increasing or decreasing.

Explanation: The condition $h''(x) > 0$ means that the graph of $h(x)$ is concave up. For a concave up function, the secant line segment connecting the endpoints of any subinterval lies above the curve. Since the trapezoidal rule uses these secant line segments to form the tops of the trapezoids, the area of each trapezoid is greater than the area under the curve on that subinterval. Thus, the trapezoidal sum $T_4$ is an overestimate.

Question 11

1. $2(f(0)+f(2)+f(4))$
2. $3(f(2)+f(4)+f(6))$
3. $2(f(2)+f(4)+f(6))$ (correct answer)
4. $3(f(0)+f(3)+f(6))$

Explanation: The interval is $[0, 6]$ with $n=3$ subintervals of equal width. The width of each subinterval is $\Delta x = \frac{6-0}{3} = 2$. The subintervals are $[0,2], [2,4], [4,6]$. For a right Riemann sum, we use the function values at the right endpoints of these subintervals, which are $x=2, x=4, x=6$. The approximation is given by the sum of the areas of the three rectangles: $f(2) \cdot \Delta x + f(4) \cdot \Delta x + f(6) \cdot \Delta x = \Delta x (f(2)+f(4)+f(6)) = 2(f(2)+f(4)+f(6))$.

Question 12

1. $\frac{1}{4}(e^0 + 2e^{0.25} + 2e^1 + 2e^{2.25} + e^4)$ (correct answer)
2. $\frac{1}{2}(e^0 + e^{0.25} + e^1 + e^{2.25})$
3. $\frac{1}{2}(e^{0.25} + e^1 + e^{2.25} + e^4)$
4. $\frac{1}{4}(e^{0.25} + e^1 + e^{2.25} + e^4)$

Explanation: The interval is $[0, 2]$ with $n=4$ subintervals, so the width is $\Delta x = \frac{2-0}{4} = 0.5$. The endpoints are $0, 0.5, 1, 1.5, 2$. The trapezoidal rule formula is $\frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)]$. Plugging in the values: $\frac{0.5}{2}[e^{0^2} + 2e^{0.5^2} + 2e^{1^2} + 2e^{1.5^2} + e^{2^2}]$. This simplifies to $\frac{1}{4}[e^0 + 2e^{0.25} + 2e^1 + 2e^{2.25} + e^4]$.

Question 13

1. The estimate is an overestimate because the function is strictly increasing. (correct answer)
2. The estimate is an underestimate because the function is strictly increasing.
3. The estimate is exact because the function is differentiable.
4. The relationship cannot be determined without knowing the concavity of the function.

Explanation: The distance traveled is the integral of the speed function, $\int_0^4 s(t) dt$. Since the speed function $s(t)$ is strictly increasing, the value at the right endpoint of any subinterval is the maximum value on that subinterval. Therefore, a right Riemann sum will use the maximum value for each rectangle's height, resulting in an overestimate of the actual distance traveled.

Question 14

1. 1
2. 2 (correct answer)
3. 5
4. 10

Explanation: The width of each subinterval (rectangle) in a Riemann sum with equal subintervals is given by the formula $\Delta x = \frac{b-a}{n}$, where $[a, b]$ is the interval of integration and $n$ is the number of subintervals. In this case, $a=2, b=12$, and $n=5$. Therefore, the width is $\Delta x = \frac{12-2}{5} = \frac{10}{5} = 2$.

Question 15

1. $f(x)$ is increasing
2. $f(x)$ is decreasing (correct answer)
3. $f(x)$ is concave up
4. $f(x)$ is concave down

Explanation: If a left Riemann sum is an overestimate, it means the function's value at the left endpoint of each subinterval is the maximum value in that subinterval. This occurs when a function is decreasing. If a right Riemann sum is an underestimate, it means the function's value at the right endpoint is the minimum value, which also occurs when a function is decreasing. Therefore, $f(x)$ must be a decreasing function on the interval.