Approximating Areas With Riemann Sums
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AP Calculus BC › Approximating Areas With Riemann Sums
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?
$$\Delta x = 1$$ and $$5$$ function evaluations are needed
$$\Delta x = 2$$ and $$5$$ function evaluations are needed
$$\Delta x = 1$$ and $$4$$ function evaluations are needed
$$\Delta x = 2$$ and $$4$$ function evaluations are needed
Explanation
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.
A right Riemann sum with $$n = 8$$ subintervals approximates $$\int_4^{12} (x^2 - 3x) , dx$$. The width of each subinterval is:
$$0.5$$
$$1.5$$
$$1$$
$$2$$
Explanation
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 $$\frac{16}{8}$$.
Using the midpoint rule with $$n = 2$$ subintervals, the approximation of $$\int_1^5 \sqrt{x} , dx$$ is:
$$2(\sqrt{2} + \sqrt{4})$$
$$4(\sqrt{2} + \sqrt{4})$$
$$2(\sqrt{3} + \sqrt{5})$$
$$2(\sqrt{1} + \sqrt{3})$$
Explanation
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 $$5$$.
The trapezoidal approximation of $$\int_0^4 (3x^2 + 1) , dx$$ using $$n = 4$$ subintervals equals:
$$1 \cdot \frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]$$
$$2 \cdot \frac{1}{2}[1 + 4 + 13 + 28 + 49]$$
$$\frac{1}{2}[1 + 4 + 13 + 28 + 49]$$
$$\frac{1}{2}[1 + 2(4) + 2(13) + 2(28) + 49]$$
Explanation
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 $$\Delta x$$.
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?
$$2, 4, 6, 8, 10$$
$$3, 5, 7, 9$$
$$2, 4, 6, 8$$
$$4, 6, 8, 10$$
Explanation
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.
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:
$$1, 3, 5, 7, 9$$
$$3, 5, 7, 9, 11$$
$$2, 4, 6, 8, 10$$
$$1, 3, 5, 7, 9, 11$$
Explanation
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.
For a decreasing function $$g$$ on $$0,4$$, which statement about Riemann sum approximations of $$\int_0^4 g(x) , dx$$ is correct?
The left Riemann sum provides an underestimate of the integral value
The left Riemann sum provides an overestimate of the integral value
The right Riemann sum provides an overestimate of the integral value
The midpoint sum is always exactly equal to the integral value
Explanation
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.
Using the midpoint rule with $$n = 4$$ to approximate $$\int_{-2}^2 x^3 , dx$$, the approximation equals:
$$1 \cdot[(-2)^3 + (-1)^3 + (0)^3 + (1)^3]$$
$$2 \cdot[(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]$$
$$1 \cdot[(-1.5)^3 + (-0.5)^3 + (0.5)^3 + (1.5)^3]$$
$$1 \cdot[(-1)^3 + (0)^3 + (1)^3 + (2)^3]$$
Explanation
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.
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:
$$L_n \leq M_n \leq R_n \leq \int_a^b f(x) , dx$$
$$M_n \leq L_n \leq \int_a^b f(x) , dx \leq R_n$$
$$L_n \leq M_n \leq \int_a^b f(x) , dx \leq R_n$$
$$L_n \leq \int_a^b f(x) , dx \leq M_n \leq R_n$$
Explanation
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.
The sum $$\sum_{k=1}^4 \frac{3}{4} \cdot \sin\left(\frac{3k}{4}\right)$$ represents which type of Riemann sum approximation?
Left Riemann sum for $$\int_0^4 \sin x , dx$$ with $$n = 3$$
Right Riemann sum for $$\int_0^3 \sin x , dx$$ with $$n = 4$$
Left Riemann sum for $$\int_0^3 \sin x , dx$$ with $$n = 4$$
Midpoint Riemann sum for $$\int_0^3 \sin x , dx$$ with $$n = 4$$
Explanation
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.