Approximating Areas With Riemann Sums - AP Calculus BC

1. The sum $\sum_{k=1}^4 \frac{3}{4} \cdot \sin\left(\frac{3k}{4}\right)$ represents which type of Riemann sum approximation?
   Answer: Right Riemann sum for $\int_0^3 \sin x \, dx$ with $n = 4$ The sum has $\Delta x = \frac{3}{4}$ and evaluates at $\frac{3 \cdot 1}{4}, \frac{3 \cdot 2}{4}, \frac{3 \cdot 3}{4}, \frac{3 \cdot 4}{4} = \frac{3}{4}, \frac{3}{2}, \frac{9}{4}, 3$. With $n = 4$ on $[0,3]$, $\Delta x = \frac{3}{4}$, and these are the right endpoints of subintervals $[0,\frac{3}{4}], [\frac{3}{4},\frac{3}{2}], [\frac{3}{2},\frac{9}{4}], [\frac{9}{4},3]$. Choice A would start at $k=0$, choice C would use midpoints, choice D has wrong interval.
2. A right Riemann sum approximation of $\int_1^7 f(x) \, dx$ with $n = 3$ subintervals can be written as:
   Answer: $2[f(3) + f(5) + f(7)]$ With $n = 3$ on $[1,7]$, $\Delta x = 2$. The subintervals are $[1,3], [3,5], [5,7]$ with right endpoints $3, 5, 7$. The right sum is $\Delta x[f(3) + f(5) + f(7)] = 2[f(3) + f(5) + f(7)]$. Choice A uses left endpoints, choice C has wrong $\Delta x = \frac{2}{3}$, and choice D has wrong $\Delta x = 6$.
3. A function $f$ is increasing on $[a,b]$. If $L_n$, $R_n$, and $M_n$ represent left, right, and midpoint Riemann sums respectively with $n$ subintervals, then:
   Answer: $L_n \leq M_n \leq \int_a^b f(x) \, dx \leq R_n$ For an increasing function, the left sum underestimates (uses minimum values), the right sum overestimates (uses maximum values), and the midpoint sum is generally more accurate than both, falling between the left sum and the true integral value. Thus $L_n \leq M_n \leq \int_a^b f(x) \, dx \leq R_n$. The other choices incorrectly order these approximations.
4. A midpoint Riemann sum with $n = 5$ equal subintervals is used to approximate $\int_1^{11} \ln x \, dx$. The midpoints of the subintervals are:
   Answer: $2, 4, 6, 8, 10$ With $n = 5$ on $[1,11]$, $\Delta x = 2$. The subintervals are $[1,3], [3,5], [5,7], [7,9], [9,11]$ with midpoints $2, 4, 6, 8, 10$. Choice B gives left endpoints, choice C gives right endpoints, and choice D incorrectly includes both endpoints with an extra point.
5. For a decreasing function $g$ on $[0,4]$, which statement about Riemann sum approximations of $\int_0^4 g(x) \, dx$ is correct?
   Answer: The left Riemann sum provides an overestimate of the integral value For a decreasing function, the left endpoints give the maximum value on each subinterval, making the left Riemann sum an overestimate. The right endpoints give minimum values, making the right sum an underestimate. Choice A is backwards, choice B is backwards, and choice D is incorrect since midpoint sums are approximations, not exact values.
6. A right Riemann sum with $n = 8$ subintervals approximates $\int_4^{12} (x^2 - 3x) \, dx$. The width of each subinterval is:
   Answer: $1$ For any Riemann sum with $n$ equal subintervals on $[a,b]$, the width is $\Delta x = \frac{b-a}{n}$. Here, $\Delta x = \frac{12-4}{8} = \frac{8}{8} = 1$. The other choices result from incorrect calculations: choice A gives $\frac{4}{8}$, choice C gives $\frac{12}{8}$, and choice D gives $\frac{16}{8}$.
7. The left Riemann sum $\sum_{i=0}^{n-1} f(a + i\Delta x) \Delta x$ approximates $\int_a^b f(x) \, dx$ where $\Delta x = \frac{b-a}{n}$. If $a = 2$, $b = 10$, and $n = 4$, what are the $x$-values used in the sum?
   Answer: $2, 4, 6, 8$ With $a = 2$, $b = 10$, $n = 4$, we have $\Delta x = 2$. The left endpoints are $a + i\Delta x$ for $i = 0, 1, 2, 3$: $2 + 0(2) = 2$, $2 + 1(2) = 4$, $2 + 2(2) = 6$, $2 + 3(2) = 8$. Choice B gives right endpoints, choice C includes the right boundary, and choice D gives midpoints.
8. The trapezoidal rule approximation of $\int_0^2 e^{x^2} \, dx$ with $n = 2$ subintervals is:
   Answer: $\frac{1}{2}[e^0 + 2e^1 + e^4]$ With $n = 2$ on $[0,2]$, $\Delta x = 1$. Points are $0, 1, 2$ with function values $e^0, e^1, e^4$. The trapezoidal rule is $\frac{\Delta x}{2}[f(0) + 2f(1) + f(2)] = \frac{1}{2}[e^0 + 2e^1 + e^4]$. Choice A omits the factor of 2 for the middle term, choices C and D incorrectly separate out $\Delta x$ when it's already incorporated in the $\frac{1}{2}$ factor.
9. The trapezoidal approximation of $\int_0^4 (3x^2 + 1) \, dx$ using $n = 4$ subintervals equals:
   Answer: $1 \cdot \frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]$ With $n = 4$ on $[0,4]$, $\Delta x = 1$. Points are $0, 1, 2, 3, 4$ with function values $1, 4, 13, 28, 49$. The trapezoidal rule is $\frac{\Delta x}{2}[f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)] = \frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]$. Choice A omits $\Delta x$, choice C omits the factor of 2, choice D uses wrong $\Delta x$.
10. If $\int_0^8 f(x) \, dx$ is approximated using both a left Riemann sum and a right Riemann sum with $n = 4$ subintervals, and $f$ is a continuous increasing function, then:
    Answer: The right sum will be greater than the left sum by some positive amount For an increasing function, the left sum uses minimum values on each subinterval (underestimate) while the right sum uses maximum values (overestimate). Therefore, the right sum will be greater than the left sum. Choice A is wrong since Riemann sums are approximations, choice B has the inequality backwards, and choice D is incorrect since the midpoint sum position depends on the function's concavity, not just monotonicity.
11. A midpoint Riemann sum with $n = 3$ subintervals approximates $\int_0^3 e^x \, dx$. Which expression represents this approximation?
    Answer: $1 \cdot [e^{0.5} + e^{1.5} + e^{2.5}]$ With $n = 3$ on $[0,3]$, $\Delta x = 1$. The subintervals are $[0,1], [1,2], [2,3]$ with midpoints $0.5, 1.5, 2.5$. The midpoint sum is $\Delta x \sum f(\text{midpoints}) = 1[e^{0.5} + e^{1.5} + e^{2.5}]$. Choice B uses left endpoints, choice C uses right endpoints, and choice D has the wrong coefficient.
12. Using the midpoint rule with $n = 2$ subintervals, the approximation of $\int_1^5 \sqrt{x} \, dx$ is:
    Answer: $2(\sqrt{2} + \sqrt{4})$ With $n = 2$ on $[1,5]$, $\Delta x = 2$. The subintervals are $[1,3]$ and $[3,5]$ with midpoints $2$ and $4$. The approximation is $2[\sqrt{2} + \sqrt{4}]$. Choice B uses wrong midpoints $1$ and $3$, choice C has wrong $\Delta x = 4$, and choice D uses wrong midpoints $3$ and $5$.
13. The trapezoidal rule with $n = 4$ subintervals is used to approximate $\int_2^6 \frac{1}{x} \, dx$. What is the value of $\Delta x$ and how many function evaluations are needed?
    Answer: $\Delta x = 1$ and $5$ function evaluations are needed For $n = 4$ subintervals on $[2,6]$, $\Delta x = \frac{6-2}{4} = 1$. The trapezoidal rule requires evaluation at all endpoints: $x_0 = 2, x_1 = 3, x_2 = 4, x_3 = 5, x_4 = 6$, which is $5$ points total. Choice A has the wrong number of evaluations, while choices C and D have incorrect $\Delta x$ values.
14. Using the midpoint rule with $n = 4$ to approximate $\int_{-2}^2 x^3 \, dx$, the approximation equals:
    Answer: $1 \cdot [(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]$ With $n = 4$ on $[-2,2]$, $\Delta x = 1$. The subintervals are $[-2,-1], [-1,0], [0,1], [1,2]$ with midpoints $-1.5, -0.5, 0.5, 1.5$. The approximation is $1 \cdot [(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]$. Choice B uses left endpoints, choice C has wrong $\Delta x = 2$, choice D uses different points entirely.
15. If $f(x) = \sin x$ and we use a left Riemann sum with $n = 6$ equal subintervals to approximate $\int_0^\pi f(x) \, dx$, which of the following represents the approximation?
    Answer: $\frac{\pi}{6}[\sin(0) + \sin(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) + \sin(\frac{\pi}{2}) + \sin(\frac{2\pi}{3}) + \sin(\frac{5\pi}{6})]$ With $n = 6$ on $[0,\pi]$, $\Delta x = \frac{\pi}{6}$. Left endpoints are $0, \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}$. The left sum uses these six points. Choice B uses right endpoints, choice C has wrong $\Delta x = \frac{\pi}{3}$, and choice D has wrong $\Delta x = \pi$.

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