Calculus 2 Quiz: Interpreting Graphs For Setup
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Interpreting Graphs For SetupQuestion 1 of 20

A region is bounded by the graphs of a downward-opening parabola with vertex (0, 4) that passes through (2, 0) and (-2, 0), and a line segment connecting (-2, 0) and (2, 0). The region is revolved about the line y = 5. Which integral represents the volume of the solid?

π ∫_-2^2 ( (5 - (4−x24-x^2))^2 - 5^2 ) dx
π ∫_-2^2 ( 5^2 - (1+x^2)^2 ) dx
π ∫_-2^2 ( 5^2 - (4-x^2)^2 ) dx
π ∫_-2^2 ( (4-x^2-5)^2 - (-5)^2 ) dx
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Calculus 2 Quiz

Calculus 2 Quiz: Interpreting Graphs For Setup

Practice Interpreting Graphs For Setup in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Interpreting Graphs For Setup, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

A region is bounded by the graphs of a downward-opening parabola with vertex (0, 4) that passes through (2, 0) and (-2, 0), and a line segment connecting (-2, 0) and (2, 0). The region is revolved about the line y = 5. Which integral represents the volume of the solid?

  1. π ∫_-2^2 ( (5 - (4−x24-x^2))^2 - 5^2 ) dx
  2. π ∫_-2^2 ( 5^2 - (1+x^2)^2 ) dx (correct answer)
  3. π ∫_-2^2 ( 5^2 - (4-x^2)^2 ) dx
  4. π ∫_-2^2 ( (4-x^2-5)^2 - (-5)^2 ) dx
Explanation: The parabola is y = 4 - x^2 and the line is y = 0. The region is revolved around y = 5. Using the washer method, the outer radius R(x) is the distance from the axis of rotation y = 5 to the farther curve y = 0, so R(x) = 5 - 0 = 5. The inner radius r(x) is the distance from y = 5 to the nearer curve y = 4 - x^2, so r(x) = 5 - (4 - x2x^2) = 1 + x^2. The volume is V = π ∫-2^2 (R(x)^2 - r(x)^2) dx = π ∫-2^2 (5^2 - (1 + x^2)^2) dx.

Question 2

A region R is bounded by the graph of a function y = f(x), the x-axis, and the lines x = a and x = b, with a < b. The graph of f(x) is entirely above the x-axis. The region is revolved about the line x = c, where c < a. Which integral represents the volume of the solid using the shell method?

  1. 2π ∫_a^b (x - c) f(x) dx (correct answer)
  2. 2π ∫_a^b (c - x) f(x) dx
  3. π ∫_a^b (f(x) - c)^2 dx
  4. 2π ∫_a^b x f(x) dx
Explanation: For the shell method with revolution around a vertical line, we integrate with respect to x. The height of a cylindrical shell at x is h = f(x). The axis of rotation is x = c. The radius of a shell at position x is the distance from x to c. Since x is in [a, b] and c < a, we have x > c. Thus, the radius is r = x - c. The volume of a shell is 2πrh Δx = 2π(x - c)f(x) Δx. The total volume is the integral of this expression: V = 2π ∫_a^b (x - c) f(x) dx.

Question 3

A region R is bounded on the right by the graph of a function x = f(y), on the left by the y-axis, and between y = c and y = d, where c < d. The graph of f(y) is strictly positive for y in [c, d]. This region R is rotated about the line x = -2. Which integral represents the volume of the resulting solid using the washer method?

  1. π ∫_c^d ( (f(y) + 2)^2 - 2^2 ) dy (correct answer)
  2. π ∫_c^d ( (f(y) - 2)^2 - 2^2 ) dy
  3. 2π ∫_c^d (y + 2) f(y) dy
  4. π ∫_c^d (f(y))^2 dy
Explanation: When rotating around a vertical line (x = -2), the washer method requires integrating with respect to y. The outer radius R(y) is the distance from the axis of rotation to the outer edge of the region, which is f(y). So, R(y) = f(y) - (-2) = f(y) + 2. The inner radius r(y) is the distance from the axis of rotation to the inner edge of the region, which is the y-axis (x=0). So, r(y) = 0 - (-2) = 2. The volume is given by V = π ∫_c^d (R(y)^2 - r(y)^2) dy = π ∫_c^d ( (f(y) + 2)^2 - 2^2 ) dy.

Question 4

Let R be the first-quadrant region bounded by the graph of a concave down function y = g(x) with g(0) > 0 and g(b) = 0 for some b > 0, and the coordinate axes. The region R is rotated about the y-axis. Which integral represents the volume of the solid using the shell method?

  1. 2π ∫_0^b x g(x) dx (correct answer)
  2. π ∫_0^b (g(x))^2 dx
  3. 2π ∫_0^{g(0)} y g^{-1}(y) dy
  4. π ∫_0^{g(0)} (g^{-1}(y))^2 dy
Explanation: Using the shell method to rotate a region about the y-axis, we integrate with respect to x. A typical cylindrical shell has radius r = x and height h = g(x). The volume of this shell is approximately 2πrh Δx = 2πx g(x) Δx. Integrating from the lower x-bound (x=0) to the upper x-bound (where the graph hits the x-axis, x=b) gives the total volume: V = 2π ∫_0^b x g(x) dx. Choice D represents the volume using the disk method, which would be correct. Choice C is an incorrect application of the shell method with respect to y. Choice B is the disk method for rotation about the x-axis.

Question 5

A region R is bounded by the graph of a polar curve r = f(θ) for θ in [α, β], where 0 ≤ α < β ≤ 2π. The graph shows that f(θ) is continuous and non-negative on this interval. Which integral represents the area of the region R?

  1. ∫_α^β f(θ) dθ
  2. (1/2) ∫_α^β (f'(θ))^2 dθ
  3. ∫_α^β √( (f(θ))^2 + (f'(θ))^2 ) dθ
  4. (1/2) ∫_α^β (f(θ))^2 dθ (correct answer)
Explanation: The formula for the area of a region bounded by a polar curve r = f(θ) from θ = α to θ = β is given by A = (1/2) ∫_α^β r^2 dθ. Substituting r = f(θ), we get A = (1/2) ∫_α^β (f(θ))^2 dθ. Choice A represents the area under the curve if f(θ) were plotted in a Cartesian system. Choice C is the formula for the arc length of a polar curve.

Question 6

The graphs of y = f(x) and y = g(x) bound a region. The graph of f(x) is always above the graph of g(x) on the interval [a, b]. The area of this region is given by A = ∫_a^b (f(x) - g(x)) dx. If this same region is used as the base of a solid whose cross-sections perpendicular to the x-axis are squares, what is the volume of the solid?

  1. ∫_a^b (f(x) - g(x))^2 dx (correct answer)
  2. ( ∫_a^b (f(x) - g(x)) dx )^2
  3. ∫_a^b ( (f(x))^2 - (g(x))^2 ) dx
  4. 2 ∫_a^b (f(x) - g(x)) dx
Explanation: The side length, s, of a square cross-section at a given x is the vertical distance between the two curves, which is s = f(x) - g(x). The area of this square cross-section is A(x) = s^2 = (f(x) - g(x))^2. The volume of the solid is the integral of the cross-sectional areas from x = a to x = b. Therefore, V = ∫_a^b (f(x) - g(x))^2 dx. Choice B incorrectly squares the total area. Choice C represents the difference in volumes of solids with square cross-sections generated from each function to the x-axis, which is not the same.

Question 7

The graphs of y = 1, y = f(x), and x = 0 bound a region in the first quadrant. The graph of f(x) is decreasing, with f(0) > 1, and it intersects y = 1 at x = k > 0. Which expression represents the area of this region as a single integral with respect to y?

  1. ∫_0^k (f(x) - 1) dx
  2. ∫_1^{f(0)} f^{-1}(y) dy (correct answer)
  3. ∫_0^{f(0)} f^{-1}(y) dy - k
  4. ∫_0^k f(x) dx - k
Explanation: To integrate with respect to y, we consider horizontal strips. The region is bounded by the y-axis (x=0) on the left and the curve x = f^{-1}(y) on the right. The y-values for the region range from y = 1 to the y-intercept of f(x), which is f(0). Therefore, the area is given by the integral of the horizontal width (right curve minus left curve) with respect to y. Area = ∫_1^{f(0)} (f^{-1}(y) - 0) dy = ∫_1^{f(0)} f^{-1}(y) dy. Choice A correctly represents the area as an integral with respect to x. Choice D is also the correct area with respect to x, expressed differently.

Question 8

The graph of a function y = f(x) is shown for the interval [0, L]. The average value of the function on this interval is k. A horizontal line y = k is drawn on the same graph. Which statement must be true about the area of the region under the graph of y=f(x) from x=0 to x=L, denoted A_f, and the area of the rectangle with vertices (0,0), (L,0), (L,k), and (0,k), denoted A_r?

  1. A_f > A_r
  2. A_f < A_r
  3. A_f = A_r (correct answer)
  4. The relationship cannot be determined without knowing f(x).
Explanation: The average value k of a function f(x) on an interval [a, b] is defined as k = (1/(b-a)) ∫_a^b f(x) dx. In this case, a=0 and b=L, so k = (1/L) ∫_0^L f(x) dx. The area under the graph of f(x) is A_f = ∫_0^L f(x) dx. From the average value formula, we can write A_f = k * L. The area of the rectangle A_r with height k and width L is also k * L. Therefore, A_f = A_r. This is the geometric interpretation of the average value of a function.

Question 9

The graph of a differentiable function f(x) is defined on [a, b]. Let L be the arc length of the curve y=f(x) from x=a to x=b, and let D be the straight-line distance between the points (a, f(a)) and (b, f(b)). Which statement correctly relates L and D?

  1. L ≥ D always (correct answer)
  2. L ≤ D always
  3. L = D only if f(x) is a constant function
  4. L > D always
Explanation: The arc length L is the length of the path along the curve y=f(x) from x=a to x=b. The distance D is the length of the straight line segment connecting the endpoints (a, f(a)) and (b, f(b)). The shortest distance between two points is a straight line. Therefore, the length of any other path, such as the arc of the curve, must be greater than or equal to the straight-line distance. Equality holds if and only if the function f(x) is a linear function on [a,b], as the curve itself is the straight line segment. So, L ≥ D is always true.

Question 10

The region R is in the first quadrant, bounded by the graphs of y = f(x) and y = g(x). The graphs intersect at the origin and at (k, h) where k>0, h>0. On the interval (0, k), the graph of f(x) is above the graph of g(x). The region R is revolved around the x-axis. Which integral gives the volume of the resulting solid?

  1. π ∫_0^k (f(x) - g(x))^2 dx
  2. 2π ∫_0^k x (f(x) - g(x)) dx
  3. π ∫_0^k ( (f(x))^2 - (g(x))^2 ) dx (correct answer)
  4. π ∫_0^h ( (g^{-1}(y))^2 - (f^{-1}(y))^2 ) dy
Explanation: The volume of a solid of revolution generated by rotating a region between two curves about the x-axis is found using the washer method. The outer radius is R(x) = f(x) (the farther curve from the axis) and the inner radius is r(x) = g(x) (the closer curve). The volume of a washer is π(R2R^2 - r2r^2)Δx. Integrating over the interval [0, k] gives the total volume V = π ∫_0^k ( (f(x))^2 - (g(x))^2 ) dx. Choice A is a common error, squaring the difference of the functions instead of differencing their squares. Choice B is the shell method formula for rotation about the y-axis.

Question 11

A function y = f(x) is graphed, where f(x) is positive and f'(x) is also positive for x > 0. The curve is rotated about the x-axis from x = a to x = b, where 0 < a < b. Which integral represents the surface area of the resulting shape?

  1. 2π ∫_a^b x √(1 + (f'(x))^2) dx
  2. 2π ∫_a^b f(x) √(1 + (f'(x))^2) dx (correct answer)
  3. ∫_a^b √(1 + (f(x))^2) dx
  4. π ∫_a^b (f(x))^2 dx
Explanation: The formula for the surface area of a solid generated by revolving a curve y = f(x) about the x-axis is S = 2π ∫_a^b r(x) dL, where r(x) is the radius of revolution and dL is the differential arc length. When revolving about the x-axis, the radius at a point x is the function value itself, so r(x) = f(x). The arc length differential is dL = √(1 + (f'(x))^2) dx. Combining these gives S = 2π ∫_a^b f(x) √(1 + (f'(x))^2) dx. Choice A is the formula for surface area when revolving about the y-axis. Choice D is the formula for volume.

Question 12

A continuous, positive, and decreasing function f(x) is defined for x ≥ 1. The series ∑_{n=2}^∞ f(n) is compared to the integral ∫_1^∞ f(x) dx. Visualizing the terms of the series as the areas of rectangles of unit width, which of the following inequalities can be established by interpreting the sum as a right-hand Riemann sum?

  1. ∑_{n=2}^∞ f(n) > ∫_1^∞ f(x) dx
  2. ∑_{n=2}^∞ f(n) < ∫_1^∞ f(x) dx (correct answer)
  3. ∑_{n=2}^∞ f(n) = ∫_2^∞ f(x) dx
  4. ∑_{n=2}^∞ f(n) < ∫_2^∞ f(x) dx
Explanation: The sum S₂ = ∑_{n=2}^∞ f(n) = f(2) + f(3) + ... can be visualized as a right-hand Riemann sum for the integral I₁ = ∫1^∞ f(x) dx. The rectangles have heights f(2), f(3), ... on the intervals [1,2], [2,3], ... respectively. Because f(x) is a decreasing function, the height of each rectangle (taken at the right endpoint) is the minimum value of the function on that interval. Therefore, the sum of the areas of the rectangles is less than the area under the curve. This means ∑{n=2}^∞ f(n) < ∫_1^∞ f(x) dx. Note that the sum is also less than ∫_2^∞ f(x) dx + f(2), but the direct comparison is an underestimate of the integral from 1 to infinity.

Question 13

A region is bounded by the graph of an odd function y = f(x) and the x-axis between x = -a and x = a. The graph of f(x) is above the x-axis for x in (0, a]. Which integral represents the volume of the solid formed by revolving this region about the y-axis?

  1. 0
  2. 2π ∫_{-a}^a x f(x) dx
  3. 4π ∫_0^a x f(x) dx (correct answer)
  4. π ∫_{-a}^a (f(x))^2 dx
Explanation: Using the shell method, the volume is V = 2π ∫{-a}^a x f(x) dx. Since f(x) is odd, we have f(-x) = -f(x). The integrand g(x) = x f(x) is even because g(-x) = (-x)f(-x) = (-x)(-f(x)) = xf(x) = g(x). For any even function, ∫{-a}^a g(x) dx = 2∫_0^a g(x) dx. Therefore, V = 2π · 2∫_0^a x f(x) dx = 4π ∫_0^a x f(x) dx.

Question 14

The region in the first quadrant bounded by the graph of a function y = f(x), the x-axis, and the line x = 4 is the base of a solid. The graph of f(x) is positive and decreasing for x in [0, 4]. For this solid, each cross-section perpendicular to the x-axis is a semicircle whose diameter lies in the xy-plane. Which integral represents the volume of the solid?

  1. (π/2) ∫_0^4 (f(x))^2 dx
  2. (π/8) ∫_0^4 (f(x))^2 dx (correct answer)
  3. (π/4) ∫_0^4 f(x) dx
  4. π ∫_0^4 f(x) dx
Explanation: The diameter of the semicircle at a given x is f(x). The radius is therefore r = f(x)/2. The area of a semicircle is (1/2)πr^2. So, the area of a cross-section is A(x) = (1/2)π(f(x)/2)^2 = (1/2)π(f(x)^2/4) = (π/8)(f(x))^2. The volume is the integral of the cross-sectional area: V = ∫_0^4 A(x) dx = (π/8) ∫_0^4 (f(x))^2 dx.

Question 15

Let R be the region in the first quadrant bounded by the graph of a decreasing function y = f(x), the line x = 1, and the coordinate axes. The graph of f intersects the y-axis at (0, k) and the x-axis at (1, 0). The region R is revolved around the line x = 2. Which integral represents the volume of the solid?

  1. π ∫_0^1 ( (f(x))^2 - 2^2 ) dx
  2. 2π ∫_0^1 (x-2) f(x) dx
  3. π ∫_0^k ( (2 - f^{-1}(y))^2 - 2^2 ) dy
  4. 2π ∫_0^1 (2-x) f(x) dx (correct answer)
Explanation: When finding volumes of solids of revolution, you need to choose the right method based on the axis of rotation. Since we're revolving around a vertical line (x = 2) that's outside our region, the shell method is most efficient. The shell method formula is V=2π∫ab(radius)(height) dxV = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx, where the radius is the distance from each point to the axis of rotation, and the height is the function value. Here, our region extends from x = 0 to x = 1, so we integrate from 0 to 1. The radius from any point at x-coordinate to the line x = 2 is |2 - x| = (2 - x) since x < 2 in our region. The height of each cylindrical shell is simply f(x). This gives us 2π∫01(2−x)f(x) dx2\pi \int_0^1 (2-x) f(x) \, dx, which is answer D. Let's examine why the other options fail: A uses the washer method formula but incorrectly subtracts 222^2 instead of the proper inner radius squared, and doesn't account for the shift from the y-axis. B has the wrong radius (x-2) which would be negative since x < 2, making this physically meaningless. C attempts the washer method with respect to y, but again uses 222^2 as the inner radius squared incorrectly—the inner radius should be the distance from the line x = 2 to the y-axis, which is 2, but this setup has other geometric issues. Remember: when the axis of rotation is parallel to but separate from the region's natural orientation, the shell method often provides the most straightforward setup. Always check that your radius expression gives positive values.

Question 16

The graph of a polar curve r = 1 + cos(θ) is a cardioid symmetric about the polar axis. It traces its full shape for θ in [0, 2π]. A second curve, the circle r = 3cos(θ), is also graphed. Which integral represents the area of the region that lies inside the circle r = 3cos(θ) and outside the cardioid r = 1 + cos(θ)?

  1. (1/2) ∫_{-π/2}^{π/2} ( (3cosθ)^2 - (1+cosθ)^2 ) dθ
  2. ∫_{-π/3}^{π/3} ( 3cosθ - (1+cosθ) ) dθ
  3. (1/2) ∫_{0}^{2π} ( (3cosθ)^2 - (1+cosθ)^2 ) dθ
  4. (1/2) ∫_{-π/3}^{π/3} ( (3cosθ)^2 - (1+cosθ)^2 ) dθ (correct answer)
Explanation: When you encounter polar area problems involving regions between curves, you need three key components: the correct area formula, the proper limits of integration, and identification of which curve is "outer" versus "inner." The area between two polar curves is given by 12∫αβ(router2−rinner2) dθ\frac{1}{2}\int_{\alpha}^{\beta} (r_{outer}^2 - r_{inner}^2) \, d\theta. Here, you want the region inside the circle r=3cos⁡θr = 3\cos\theta but outside the cardioid r=1+cos⁡θr = 1 + \cos\theta, so the circle is the outer curve and the cardioid is the inner curve. To find the integration limits, you need the intersection points where 3cos⁡θ=1+cos⁡θ3\cos\theta = 1 + \cos\theta. Solving: 2cos⁡θ=12\cos\theta = 1, so cos⁡θ=12\cos\theta = \frac{1}{2}, giving θ=±π3\theta = \pm\frac{\pi}{3}. The circle r=3cos⁡θr = 3\cos\theta only exists where cos⁡θ≥0\cos\theta \geq 0, which is θ∈[−π2,π2]\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}], and the relevant region is between the intersections. Therefore, the correct integral is 12∫−π/3π/3((3cos⁡θ)2−(1+cos⁡θ)2) dθ\frac{1}{2}\int_{-\pi/3}^{\pi/3} ((3\cos\theta)^2 - (1+\cos\theta)^2) \, d\theta, which is answer D. Answer A uses incorrect limits [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] that extend beyond the intersection points. Answer B omits the 12\frac{1}{2} factor and uses rr instead of r2r^2 in the integrand. Answer C uses limits [0,2π][0, 2\pi] where the circle doesn't even exist for much of the interval. Study tip: Always find intersection points first to determine integration limits, and remember polar area formulas use r2r^2, not rr.

Question 17

The graphs of f(x) and g(x) are continuous functions that intersect at x = a, x = b, and x = c, where a < b < c. The total area of the region enclosed between the graphs from x = a to x = c is desired. A graph reveals that f(x) ≥ g(x) on [a, b] and g(x) ≥ f(x) on [b, c]. Which expression represents this total area?

  1. ∫_a^c (f(x) - g(x)) dx
  2. ∫_a^b (f(x) - g(x)) dx - ∫_b^c (g(x) - f(x)) dx
  3. ∫_a^b (f(x) - g(x)) dx + ∫_b^c (g(x) - f(x)) dx (correct answer)
  4. ∫_a^c |f(x) - g(x)| dx
Explanation: To find the total area between two curves, one must integrate the absolute difference of the functions. This often requires splitting the integral at points where the functions cross. On [a, b], the area is ∫_a^b (f(x) - g(x)) dx since f(x) ≥ g(x). On [b, c], the area is ∫_b^c (g(x) - f(x)) dx since g(x) ≥ f(x). The total area is the sum of these two areas. Choice A would calculate the net signed area, which could be less than the total area. Choice B incorrectly subtracts the second area. Choice D is conceptually correct, but choice C correctly expresses the evaluation of that absolute value integral based on the graph's description.

Question 18

The region bounded by the graph of a function x=g(y), the y-axis, y=c, and y=d (with 0<c<d) is rotated about the x-axis. The graph of g(y) is positive for y in [c, d]. Which integral represents the volume of the resulting solid using the cylindrical shell method?

  1. π ∫_c^d (g(y))^2 dy
  2. 2π ∫_c^d y g(y) dy (correct answer)
  3. 2π ∫_c^d y √(1 + (g'(y))^2) dy
  4. π ∫_0^{max(g)} y^2 dy
Explanation: When revolving a region about the x-axis using the shell method, we integrate with respect to y. A typical cylindrical shell has a radius r, which is the distance from the x-axis, so r = y. The height of the shell, h, is the horizontal width of the region at that y-value, which is given by g(y). The volume of a shell is 2πrh Δy = 2πy g(y) Δy. Integrating over the y-interval [c, d] gives the total volume: V = 2π ∫_c^d y g(y) dy. Choice A represents the volume using the disk method if the region were rotated about the y-axis.

Question 19

The graphs of y = f(x), y = g(x), x=a, and x=b enclose a region, where f(x) > g(x) > 0 on [a,b]. This region is rotated about the y-axis. Which integral represents the volume of the solid generated, using the shell method?

  1. 2π ∫_a^b x (f(x) + g(x)) dx
  2. π ∫_a^b ( (f(x))^2 - (g(x))^2 ) dx
  3. 2π ∫_a^b (f(x) - g(x)) dx
  4. 2π ∫_a^b x (f(x) - g(x)) dx (correct answer)
Explanation: When you encounter a volume problem involving rotation about the y-axis, you need to choose between the shell method and washer method. The shell method is particularly useful when rotating about the y-axis because it naturally uses x as the variable of integration. The shell method formula for rotation about the y-axis is: V=2π∫ab(radius)×(height) dxV = 2\pi \int_a^b (\text{radius}) \times (\text{height}) \, dx For this problem, imagine taking a thin vertical strip at position x. When rotated about the y-axis, this strip forms a cylindrical shell with:
  • Radius = x (distance from the y-axis)
  • Height = f(x) - g(x) (difference between the upper and lower functions)
Therefore, the volume is 2π∫abx(f(x)−g(x))dx2\pi \int_a^b x(f(x) - g(x)) dx, which is answer D. Let's see why the other options are wrong: A) 2π∫abx(f(x)+g(x))dx2\pi \int_a^b x(f(x) + g(x)) dx incorrectly adds the functions instead of subtracting them. This would give you more than the actual enclosed region. B) π∫ab((f(x))2−(g(x))2)dx\pi \int_a^b ((f(x))^2 - (g(x))^2) dx is the washer method formula for rotation about the x-axis, not the y-axis. This completely misapplies the method. C) 2π∫ab(f(x)−g(x))dx2\pi \int_a^b (f(x) - g(x)) dx has the correct height but is missing the radius factor x, which is essential for the shell method. Remember: Shell method about the y-axis always includes the radius factor x in the integrand. Without it, you're not accounting for the varying distance from the axis of rotation.

Question 20

The graph of a continuous function f(x) on [1, ∞) is positive and decreasing. Let S be the sum of the series ∑_{n=1}^∞ f(n) and let I = ∫_1^∞ f(x) dx. Both S and I are convergent. A visualization of the series as a left-hand Riemann sum for the integral suggests a relationship between S and I. Which inequality is established by this visualization?

  1. S < I
  2. S = I
  3. S > I (correct answer)
  4. S < f(1)
Explanation: For a positive, decreasing function f(x), the left-hand Riemann sum overestimates the corresponding definite integral. The sum S = ∑_{n=1}^∞ f(n) = f(1) + f(2) + f(3) + ... can be visualized as the sum of areas of rectangles of width 1 and heights f(1), f(2), f(3), etc. The integral I = ∫_1^∞ f(x) dx is the area under the curve. The rectangles of the left-hand sum will completely contain the area under the curve from x=1 to infinity. Therefore, the sum of the areas of the rectangles is greater than the area under the curve, which means S > I.