Calculus 1 Quiz: Area Between Curves X Functions
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Area Between Curves X FunctionsQuestion 1 of 20

Let R be the region bounded by y=x2−2xy=x^2-2x and y=xy=x. Which of the following integrals represents the area of R?

∫03(x2−3x)dx\int_{0}^{3} (x^2-3x) dx
∫02(x2−2x)dx+∫23xdx\int_{0}^{2} (x^2-2x) dx + \int_{2}^{3} x dx
∫−13(x−(x2−2x))dx\int_{-1}^{3} (x - (x^2-2x)) dx
∫03(3x−x2)dx\int_{0}^{3} (3x-x^2) dx
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Calculus 1 Quiz

Calculus 1 Quiz: Area Between Curves X Functions

Practice Area Between Curves X Functions in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Area Between Curves X Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

Let R be the region bounded by y=x2−2xy=x^2-2x and y=xy=x. Which of the following integrals represents the area of R?

  1. ∫03(x2−3x)dx\int_{0}^{3} (x^2-3x) dx
  2. ∫02(x2−2x)dx+∫23xdx\int_{0}^{2} (x^2-2x) dx + \int_{2}^{3} x dx
  3. ∫−13(x−(x2−2x))dx\int_{-1}^{3} (x - (x^2-2x)) dx
  4. ∫03(3x−x2)dx\int_{0}^{3} (3x-x^2) dx (correct answer)
Explanation: When finding the area between two curves, you need to determine which function is on top, find their intersection points, and set up the integral as ∫ab[top function−bottom function] dx\int_a^b [\text{top function} - \text{bottom function}] \, dx. First, find where y=x2−2xy = x^2 - 2x and y=xy = x intersect by setting them equal: x2−2x=xx^2 - 2x = x, which gives x2−3x=0x^2 - 3x = 0, so x(x−3)=0x(x-3) = 0. The curves intersect at x=0x = 0 and x=3x = 3. Next, determine which function is on top in the interval [0,3][0,3]. Test a point like x=1x = 1: the line gives y=1y = 1, while the parabola gives y=1−2=−1y = 1 - 2 = -1. Since 1>−11 > -1, the line y=xy = x is above the parabola y=x2−2xy = x^2 - 2x throughout this interval. The area is therefore ∫03[x−(x2−2x)] dx=∫03(3x−x2) dx\int_0^3 [x - (x^2 - 2x)] \, dx = \int_0^3 (3x - x^2) \, dx, which matches answer choice D. Looking at the wrong answers: A uses the wrong integrand x2−3xx^2 - 3x (parabola minus line instead of line minus parabola). B incorrectly splits the integral and adds areas of individual functions rather than their difference. C has the correct integrand but wrong limits—it uses −1-1 to 3 instead of 0 to 3. Study tip: Always sketch the curves or test a point between intersections to determine which function is on top. The area between curves is always "top minus bottom," and getting this backwards is the most common error.

Question 2

Let f(x)=x3f(x) = x^3. Let LL be the line that is tangent to the graph of ff at x=1x = 1. What is the area of the region enclosed by the graph of ff, the line LL, and the line x=−1x = -1?

  1. 2
  2. 4 (correct answer)
  3. 6
  4. 8
Explanation: First, find the equation of the tangent line LL. f(x)=x3⇒f′(x)=3x2f(x) = x^3 \Rightarrow f'(x) = 3x^2. At x=1x = 1, f(1)=1f(1) = 1 and the slope is m=f′(1)=3m = f'(1) = 3. The tangent line is y−1=3(x−1)⇒y=3x−2y - 1 = 3(x - 1) \Rightarrow y = 3x - 2. The region is bounded by y=x3y = x^3, y=3x−2y = 3x - 2, and x=−1x = -1. The tangent point is at x=1x=1. The area is A=∫−11∣x3−(3x−2)∣dxA = \int_{-1}^{1} |x^3 - (3x - 2)| dx. On the interval [−1,1][-1, 1], the cubic function y=x3y=x^3 is above the tangent line y=3x−2y=3x-2. For instance, at x=0x=0, 0>−20 > -2. At x=−1x=-1, −1>−5-1 > -5. So the integrand is x3−3x+2x^3 - 3x + 2. A=∫−11(x3−3x+2)dxA = \int_{-1}^{1} (x^3 - 3x + 2) dx. The antiderivative is [x44−3x22+2x]−11[\frac{x^4}{4} - \frac{3x^2}{2} + 2x]_{-1}^{1}. Evaluating: (14−32+2)−(14−32−2)=(1−6+84)−(1−6−84)=34−(−134)=164=4(\frac{1}{4} - \frac{3}{2} + 2) - (\frac{1}{4} - \frac{3}{2} - 2) = (\frac{1-6+8}{4}) - (\frac{1-6-8}{4}) = \frac{3}{4} - (-\frac{13}{4}) = \frac{16}{4} = 4.

Question 3

Find the area of the region bounded by y=12−x2y = 12 - x^2 and y=x2−6y = x^2 - 6.

  1. 36
  2. 72 (correct answer)
  3. 108
  4. 144
Explanation: First, find the points of intersection: 12−x2=x2−6⇒18=2x2⇒x2=9⇒x=±312 - x^2 = x^2 - 6 \Rightarrow 18 = 2x^2 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3. The upper curve is y=12−x2y = 12 - x^2 and the lower curve is y=x2−6y = x^2 - 6. The area is A=∫−33((12−x2)−(x2−6))dx=∫−33(18−2x2)dxA = \int_{-3}^{3} ((12 - x^2) - (x^2 - 6)) dx = \int_{-3}^{3} (18 - 2x^2) dx. By symmetry, A=2∫03(18−2x2)dx=2[18x−2x33]03=2((18(3)−2(33)3)−0)=2(54−543)=2(54−18)=2(36)=72A = 2 \int_{0}^{3} (18 - 2x^2) dx = 2[18x - \frac{2x^3}{3}]_{0}^{3} = 2((18(3) - \frac{2(3^3)}{3}) - 0) = 2(54 - \frac{54}{3}) = 2(54 - 18) = 2(36) = 72.

Question 4

What is the area of the finite region bounded by the curve y=x2y = x^2, its tangent line at x=1x = 1, and the y-axis?

  1. 1/61/6
  2. 1/31/3 (correct answer)
  3. 1/21/2
  4. 2/32/3
Explanation: First, find the equation of the tangent line. Let f(x)=x2f(x) = x^2. The derivative is f′(x)=2xf'(x) = 2x. At x=1x = 1, the y-coordinate is f(1)=12=1f(1) = 1^2 = 1 and the slope of the tangent line is m=f′(1)=2(1)=2m = f'(1) = 2(1) = 2. The equation of the tangent line is y−1=2(x−1)y - 1 = 2(x - 1), which simplifies to y=2x−1y = 2x - 1. The region is bounded by y=x2y = x^2, y=2x−1y = 2x - 1, and the y-axis (x=0x=0). The intersection is at x=1x=1. We need to find the area between these curves from x=0x = 0 to x=1x = 1. For x∈[0,1]x \in [0, 1], x2≥2x−1x^2 \ge 2x - 1 because x2−(2x−1)=(x−1)2≥0x^2 - (2x-1) = (x-1)^2 \ge 0. The area is A=∫01(x2−(2x−1))dx=∫01(x2−2x+1)dxA = \int_{0}^{1} (x^2 - (2x - 1)) dx = \int_{0}^{1} (x^2 - 2x + 1) dx. Evaluating the integral: [x33−x2+x]01=(13−1+1)−0=13[\frac{x^3}{3} - x^2 + x]_{0}^{1} = (\frac{1}{3} - 1 + 1) - 0 = \frac{1}{3}.

Question 5

What is the area of the region bounded by y=1xy = \frac{1}{x}, y=xy = x, and x=2x=2?

  1. 32−ln⁡(2)\frac{3}{2} - \ln(2) (correct answer)
  2. 52+ln⁡(2)\frac{5}{2} + \ln(2)
  3. ln⁡(2)−12\ln(2) - \frac{1}{2}
  4. 32\frac{3}{2}
Explanation: The curves y=1/xy = 1/x and y=xy = x intersect when x=1/x⇒x2=1x = 1/x \Rightarrow x^2 = 1. In the first quadrant, this is x=1x=1. The region is bounded by x=1x=1 and x=2x=2. In the interval [1,2][1, 2], x≥1/xx \ge 1/x. The area is given by A=∫12(x−1x)dxA = \int_{1}^{2} (x - \frac{1}{x}) dx. The antiderivative is x22−ln⁡∣x∣\frac{x^2}{2} - \ln|x|. Evaluating from 1 to 2: [x22−ln⁡(x)]12=(222−ln⁡(2))−(122−ln⁡(1))=(2−ln⁡(2))−(12−0)=2−12−ln⁡(2)=32−ln⁡(2)[\frac{x^2}{2} - \ln(x)]_{1}^{2} = (\frac{2^2}{2} - \ln(2)) - (\frac{1^2}{2} - \ln(1)) = (2 - \ln(2)) - (\frac{1}{2} - 0) = 2 - \frac{1}{2} - \ln(2) = \frac{3}{2} - \ln(2).

Question 6

The area of the region enclosed by the graphs of y=x3−xy = x^3 - x and y=3xy = 3x is:

  1. 0
  2. 4
  3. 8 (correct answer)
  4. 16
Explanation: First, find the points of intersection by setting the functions equal: x3−x=3x⇒x3−4x=0⇒x(x−2)(x+2)=0x^3 - x = 3x \Rightarrow x^3 - 4x = 0 \Rightarrow x(x-2)(x+2) = 0. The intersection points are x=−2,0,2x = -2, 0, 2. The region consists of two parts. For x∈[−2,0]x \in [-2, 0], x3−x≥3xx^3 - x \ge 3x. For x∈[0,2]x \in [0, 2], 3x≥x3−x3x \ge x^3 - x. The total area is the sum of two integrals: A=∫−20((x3−x)−3x)dx+∫02(3x−(x3−x))dx=∫−20(x3−4x)dx+∫02(4x−x3)dxA = \int_{-2}^{0} ((x^3 - x) - 3x) dx + \int_{0}^{2} (3x - (x^3 - x)) dx = \int_{-2}^{0} (x^3 - 4x) dx + \int_{0}^{2} (4x - x^3) dx. Evaluating the first integral: [x44−2x2]−20=0−(164−2(4))=−(4−8)=4[\frac{x^4}{4} - 2x^2]_{-2}^{0} = 0 - (\frac{16}{4} - 2(4)) = -(4-8) = 4. Evaluating the second integral: [2x2−x44]02=(2(4)−164)−0=8−4=4[2x^2 - \frac{x^4}{4}]_{0}^{2} = (2(4) - \frac{16}{4}) - 0 = 8 - 4 = 4. The total area is 4+4=84 + 4 = 8.

Question 7

What is the area of the region bounded by the graphs of f(x)=∣x∣f(x)=|x| and g(x)=−x2+2g(x)=-x^2+2?

  1. 11
  2. 7/67/6
  3. 10/310/3
  4. 7/37/3 (correct answer)
Explanation: When finding the area between two curves, you need to identify where they intersect, determine which function is on top, and set up the appropriate integral. First, find the intersection points by solving ∣x∣=−x2+2|x| = -x^2 + 2. Since we're dealing with absolute value, consider two cases. For x≥0x \geq 0: x=−x2+2x = -x^2 + 2, so x2+x−2=0x^2 + x - 2 = 0, giving (x+2)(x−1)=0(x+2)(x-1) = 0. Since x≥0x \geq 0, we get x=1x = 1. For x<0x < 0: −x=−x2+2-x = -x^2 + 2, so x2−x−2=0x^2 - x - 2 = 0, giving (x−2)(x+1)=0(x-2)(x+1) = 0. Since x<0x < 0, we get x=−1x = -1. The curves intersect at (−1,1)(-1, 1) and (1,1)(1, 1). Between x=−1x = -1 and x=1x = 1, the parabola g(x)=−x2+2g(x) = -x^2 + 2 is above f(x)=∣x∣f(x) = |x|. The area is: ∫−11[(−x2+2)−∣x∣] dx\int_{-1}^{1} [(-x^2 + 2) - |x|] \, dx Since ∣x∣=−x|x| = -x for x∈[−1,0]x \in [-1, 0] and ∣x∣=x|x| = x for x∈[0,1]x \in [0, 1]: ∫−10[(−x2+2)−(−x)] dx+∫01[(−x2+2)−x] dx\int_{-1}^{0} [(-x^2 + 2) - (-x)] \, dx + \int_{0}^{1} [(-x^2 + 2) - x] \, dx =∫−10(−x2+x+2) dx+∫01(−x2−x+2) dx= \int_{-1}^{0} (-x^2 + x + 2) \, dx + \int_{0}^{1} (-x^2 - x + 2) \, dx =[−x33+x22+2x]−10+[−x33−x22+2x]01= \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^{0} + \left[-\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_{0}^{1} =76+76=73= \frac{7}{6} + \frac{7}{6} = \frac{7}{3} Answer (A) 1 likely comes from forgetting the absolute value. Answer (B) 76\frac{7}{6} is just half the correct answer. Answer (C) 103\frac{10}{3} might result from calculation errors with the absolute value. Always sketch both functions first to visualize which is on top in each interval—this prevents sign errors in your integral setup.

Question 8

The area of the region bounded by y=exy = e^x, y=1y=1 and x=2x=2 is:

  1. e2−1e^2 - 1
  2. e2−2e^2 - 2
  3. e2−3e^2 - 3 (correct answer)
  4. e2e^2
Explanation: The curve y=exy=e^x intersects y=1y=1 at x=0x=0. The region is bounded by x=0x=0 and x=2x=2. In this interval, ex≥1e^x \ge 1. The area is given by A=∫02(ex−1)dxA = \int_{0}^{2} (e^x - 1) dx. The antiderivative is ex−xe^x - x. Evaluating the definite integral: A=[ex−x]02=(e2−2)−(e0−0)=(e2−2)−(1)=e2−3A = [e^x - x]_{0}^{2} = (e^2 - 2) - (e^0 - 0) = (e^2 - 2) - (1) = e^2 - 3.

Question 9

Find the area of the region bounded by the graphs of y=sin⁡(x)y = \sin(x) and y=sin⁡(2x)y = \sin(2x) between x=0x=0 and x=π/3x=\pi/3.

  1. 1/21/2
  2. 1/41/4 (correct answer)
  3. 3/43/4
  4. 11
Explanation: When finding the area between two curves, you need to determine which function is on top, find their intersection points, and integrate the difference. This requires careful analysis of where the curves cross within your given interval. First, let's find where y=sin⁡(x)y = \sin(x) and y=sin⁡(2x)y = \sin(2x) intersect between x=0x = 0 and x=π/3x = \pi/3. Setting them equal: sin⁡(x)=sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(x) = \sin(2x) = 2\sin(x)\cos(x). This gives us sin⁡(x)(1−2cos⁡(x))=0\sin(x)(1 - 2\cos(x)) = 0, so x=0x = 0 or cos⁡(x)=1/2\cos(x) = 1/2, which means x=π/3x = \pi/3 in our interval. Next, determine which function is larger between these points. At x=π/6x = \pi/6: sin⁡(π/6)=1/2\sin(\pi/6) = 1/2 and sin⁡(π/3)=3/2≈0.866\sin(\pi/3) = \sqrt{3}/2 \approx 0.866. So sin⁡(2x)>sin⁡(x)\sin(2x) > \sin(x) in the interior. The area is: ∫0π/3[sin⁡(2x)−sin⁡(x)]dx=[−12cos⁡(2x)+cos⁡(x)]0π/3\int_0^{\pi/3} [\sin(2x) - \sin(x)] dx = \left[-\frac{1}{2}\cos(2x) + \cos(x)\right]_0^{\pi/3} Evaluating: (−12⋅12+12)−(−12⋅1+1)=14−12=−14\left(-\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2}\right) - \left(-\frac{1}{2} \cdot 1 + 1\right) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} Taking the absolute value gives 14\frac{1}{4}, which is answer B. Answer A (1/21/2) likely comes from forgetting the factor of 1/21/2 when integrating cos⁡(2x)\cos(2x). Answer C (3/43/4) and D (1) probably result from calculation errors or incorrectly identifying which function is on top. Always sketch the curves when possible—visualization helps you avoid sign errors and ensures you're integrating the correct difference.

Question 10

Find the area of the region bounded by the curves y=∣x∣y = |x| and y=2−x2y = 2 - x^2.

  1. 7/37/3 (correct answer)
  2. 10/310/3
  3. 13/313/3
  4. 14/314/3
Explanation: The region is symmetric with respect to the y-axis because both y=∣x∣y=|x| and y=2−x2y=2-x^2 are even functions. We can find the area for x≥0x \ge 0 and double it. First, find the intersection points. For x≥0x \ge 0, we solve x=2−x2x = 2 - x^2, which gives x2+x−2=0x^2 + x - 2 = 0, or (x+2)(x−1)=0(x+2)(x-1) = 0. Since x≥0x \ge 0, the intersection is at x=1x=1. By symmetry, the other intersection is at x=−1x=-1. For x∈[0,1]x \in [0, 1], 2−x2≥x2-x^2 \ge x. The area in the first quadrant is A1=∫01((2−x2)−x)dxA_1 = \int_{0}^{1} ((2 - x^2) - x) dx. Evaluating the integral: A1=[2x−x33−x22]01=(2−13−12)−0=12−2−36=76A_1 = [2x - \frac{x^3}{3} - \frac{x^2}{2}]_{0}^{1} = (2 - \frac{1}{3} - \frac{1}{2}) - 0 = \frac{12-2-3}{6} = \frac{7}{6}. The total area is twice this value due to symmetry: A=2⋅A1=2⋅76=73A = 2 \cdot A_1 = 2 \cdot \frac{7}{6} = \frac{7}{3}.

Question 11

Find the area of the region in the first quadrant bounded by the curves y=8/x2y = 8/x^2, y=xy = x, and y=8xy = 8x.

  1. 5/25/2
  2. 7/27/2
  3. 6 (correct answer)
  4. 14
Explanation: First, find the intersection points. y=xy=x and y=8xy=8x intersect at (0,0)(0,0). y=xy=x and y=8/x2y=8/x^2 intersect when x=8/x2⇒x3=8⇒x=2x = 8/x^2 \Rightarrow x^3=8 \Rightarrow x=2, at point (2,2)(2,2). y=8xy=8x and y=8/x2y=8/x^2 intersect when 8x=8/x2⇒x3=1⇒x=18x = 8/x^2 \Rightarrow x^3=1 \Rightarrow x=1, at point (1,8)(1,8). The region must be split into two parts at x=1x=1. From x=0x=0 to x=1x=1, the region is bounded above by y=8xy=8x and below by y=xy=x. From x=1x=1 to x=2x=2, the region is bounded above by y=8/x2y=8/x^2 and below by y=xy=x. Area = ∫01(8x−x)dx+∫12(8x−2−x)dx\int_{0}^{1} (8x - x) dx + \int_{1}^{2} (8x^{-2} - x) dx. First integral: ∫017xdx=[7x22]01=72\int_{0}^{1} 7x dx = [\frac{7x^2}{2}]_{0}^{1} = \frac{7}{2}. Second integral: [−8x−1−x22]12=(−82−42)−(−8−12)=(−4−2)−(−8.5)=−6+8.5=2.5=52[ -8x^{-1} - \frac{x^2}{2} ]_{1}^{2} = (-\frac{8}{2} - \frac{4}{2}) - (-8 - \frac{1}{2}) = (-4-2) - (-8.5) = -6 + 8.5 = 2.5 = \frac{5}{2}. Total area = 72+52=122=6\frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6.

Question 12

Let R be the region bounded by y=cos⁡(x)y = \cos(x), the x-axis, x=−π/2x = -\pi/2, and x=π/2x = \pi/2. A vertical line x=cx=c is drawn such that it divides R into two regions of equal area. What must be true about cc?

  1. c=0c=0 (correct answer)
  2. c=π/4c = \pi/4
  3. cos⁡(c)=1/2\cos(c) = 1/2
  4. ∫0ccos⁡(x)dx=1\int_{0}^{c} \cos(x) dx = 1
Explanation: The function y=cos⁡(x)y = \cos(x) is an even function, meaning its graph is symmetric with respect to the y-axis. The interval [−π/2,π/2][-\pi/2, \pi/2] is also symmetric about x=0x=0. Because of this symmetry, the total area of the region R is A=∫−π/2π/2cos⁡(x)dx=2∫0π/2cos⁡(x)dxA = \int_{-\pi/2}^{\pi/2} \cos(x) dx = 2 \int_{0}^{\pi/2} \cos(x) dx. The line that divides this symmetric region into two equal areas must be the line of symmetry itself, which is the y-axis, or x=0x=0. Therefore, c=0c=0.

Question 13

What is the total area of the finite region(s) bounded by the curve y=x1−x2y = x\sqrt{1-x^2} and the x-axis?

  1. 0
  2. 1/31/3
  3. 2/32/3 (correct answer)
  4. 4/34/3
Explanation: The domain of y=x1−x2y = x\sqrt{1-x^2} is [−1,1][-1, 1]. The curve intersects the x-axis at x=−1,0,1x=-1, 0, 1. The function is positive for x∈(0,1]x \in (0,1] and negative for x∈[−1,0)x \in [-1,0). Since this is an odd function (f(−x)=−f(x)f(-x) = -f(x)), the areas of the two regions are equal. We can calculate the area of the region for x∈[0,1]x \in [0,1] and double it. Area of one region: A1=∫01x1−x2dxA_1 = \int_{0}^{1} x\sqrt{1-x^2} dx. Use u-substitution: let u=1−x2u = 1-x^2, so du=−2xdxdu = -2x dx, or −du/2=xdx-du/2 = x dx. When x=0,u=1x=0, u=1. When x=1,u=0x=1, u=0. The integral becomes ∫10u(−12du)=12∫01u1/2du=12[23u3/2]01=12(23−0)=13\int_{1}^{0} \sqrt{u} (-\frac{1}{2} du) = \frac{1}{2} \int_{0}^{1} u^{1/2} du = \frac{1}{2} [\frac{2}{3}u^{3/2}]_{0}^{1} = \frac{1}{2} (\frac{2}{3} - 0) = \frac{1}{3}. The total area is the sum of the areas of the two regions, so Atotal=A1+A2=13+13=23A_{total} = A_1 + A_2 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}.

Question 14

The line x=cx=c divides the area of the region bounded by y=2x−x2y=2x-x^2 and the x-axis into two equal halves. Which equation must be satisfied by cc?

  1. ∫0c(2x−x2)dx=23\int_{0}^{c} (2x-x^2) dx = \frac{2}{3} (correct answer)
  2. ∫0c(2x−x2)dx=43\int_{0}^{c} (2x-x^2) dx = \frac{4}{3}
  3. 2c−c2=12c-c^2 = 1
  4. c2−c33=83c^2 - \frac{c^3}{3} = \frac{8}{3}
Explanation: First, find the total area of the region. The curve y=2x−x2=x(2−x)y=2x-x^2 = x(2-x) intersects the x-axis at x=0x=0 and x=2x=2. The total area is A=∫02(2x−x2)dx=[x2−x33]02=(4−83)−0=43A = \int_{0}^{2} (2x - x^2) dx = [x^2 - \frac{x^3}{3}]_{0}^{2} = (4 - \frac{8}{3}) - 0 = \frac{4}{3}. We want the line x=cx=c to divide this area in half, so the area of the region from x=0x=0 to x=cx=c must be A/2=(4/3)/2=2/3A/2 = (4/3)/2 = 2/3. This is represented by the equation ∫0c(2x−x2)dx=23\int_{0}^{c} (2x-x^2) dx = \frac{2}{3}.

Question 15

The base of a solid is the region in the xy-plane bounded by the parabolas y=x2y = x^2 and y=8−x2y = 8 - x^2. What is the area of the base?

  1. 32/332/3
  2. 64/364/3 (correct answer)
  3. 80/380/3
  4. 128/3128/3
Explanation: First, find the intersection points: x2=8−x2⇒2x2=8⇒x2=4⇒x=±2x^2 = 8 - x^2 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2. In the interval [−2,2][-2, 2], the parabola y=8−x2y = 8 - x^2 is the upper curve and y=x2y = x^2 is the lower curve. The area is given by A=∫−22((8−x2)−x2)dx=∫−22(8−2x2)dxA = \int_{-2}^{2} ((8 - x^2) - x^2) dx = \int_{-2}^{2} (8 - 2x^2) dx. Evaluating: [8x−2x33]−22=(16−163)−(−16+163)=16−163+16−163=32−323=96−323=643[8x - \frac{2x^3}{3}]_{-2}^{2} = (16 - \frac{16}{3}) - (-16 + \frac{16}{3}) = 16 - \frac{16}{3} + 16 - \frac{16}{3} = 32 - \frac{32}{3} = \frac{96-32}{3} = \frac{64}{3}.

Question 16

What is the area of the region in the first quadrant enclosed by the graph of y=exy = e^x, the line y=ey=e, and the y-axis?

  1. 1 (correct answer)
  2. e−1e-1
  3. e
  4. e+1e+1
Explanation: The region is in the first quadrant, so x≥0x \ge 0. The y-axis corresponds to the line x=0x=0. The curve y=exy=e^x intersects the line y=ey=e when ex=ee^x = e, which means x=1x=1. So, the region is bounded by x=0x=0 and x=1x=1. Over this interval, e≥exe \ge e^x. The area is given by the integral of the upper curve minus the lower curve: A=∫01(e−ex)dxA = \int_{0}^{1} (e - e^x) dx. The antiderivative is ex−exex - e^x. Evaluating the definite integral: A=[ex−ex]01=(e(1)−e1)−(e(0)−e0)=(e−e)−(0−1)=0−(−1)=1A = [ex - e^x]_{0}^{1} = (e(1) - e^1) - (e(0) - e^0) = (e - e) - (0 - 1) = 0 - (-1) = 1.

Question 17

The graphs of y=2xy = 2x and y=x2y = x^2 intersect at two points. What is the area of the region enclosed between them?

  1. 2/32/3
  2. 4/34/3 (correct answer)
  3. 8/38/3
  4. 4
Explanation: First, find the points of intersection by setting the equations equal: x2=2x⇒x2−2x=0⇒x(x−2)=0x^2 = 2x \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0. The intersection points are at x=0x=0 and x=2x=2. In the interval [0,2][0, 2], the line y=2xy=2x is above the parabola y=x2y=x^2 (e.g., at x=1x=1, 2>12 > 1). The area is given by the integral A=∫02(2x−x2)dxA = \int_{0}^{2} (2x - x^2) dx. The antiderivative is x2−x33x^2 - \frac{x^3}{3}. Evaluating the integral: A=[x2−x33]02=(22−233)−0=4−83=12−83=43A = [x^2 - \frac{x^3}{3}]_{0}^{2} = (2^2 - \frac{2^3}{3}) - 0 = 4 - \frac{8}{3} = \frac{12-8}{3} = \frac{4}{3}.

Question 18

What is the area of the region enclosed by the graphs of y=sin⁡(x)y = \sin(x) and y=cos⁡(x)y = \cos(x) for x∈[0,π]x \in [0, \pi]?

  1. 22
  2. 22−22\sqrt{2} - 2
  3. 222\sqrt{2} (correct answer)
  4. 00
Explanation: To find the area between two curves, we integrate the absolute difference of the functions. First, find the intersection points by setting sin⁡(x)=cos⁡(x)\sin(x) = \cos(x), which gives tan⁡(x)=1\tan(x) = 1. The solution in [0,π][0, \pi] is x=π/4x = \pi/4. This means the function that is greater ('on top') changes at this point. For x∈[0,π/4]x \in [0, \pi/4], cos⁡(x)≥sin⁡(x)\cos(x) \ge \sin(x). For x∈[π/4,π]x \in [\pi/4, \pi], sin⁡(x)≥cos⁡(x)\sin(x) \ge \cos(x). The total area requires two separate integrals: Area =∫0π/4(cos⁡(x)−sin⁡(x))dx+∫π/4π(sin⁡(x)−cos⁡(x))dx= \int_0^{\pi/4} (\cos(x) - \sin(x)) dx + \int_{\pi/4}^{\pi} (\sin(x) - \cos(x)) dx Calculating the first integral: [sin⁡(x)+cos⁡(x)]0π/4=(sin⁡(π/4)+cos⁡(π/4))−(sin⁡(0)+cos⁡(0))=(22+22)−(0+1)=2−1[\sin(x) + \cos(x)]_0^{\pi/4} = (\sin(\pi/4) + \cos(\pi/4)) - (\sin(0) + \cos(0)) = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1) = \sqrt{2} - 1. Calculating the second integral: [−cos⁡(x)−sin⁡(x)]π/4π=(−cos⁡(π)−sin⁡(π))−(−cos⁡(π/4)−sin⁡(π/4))=(−(−1)−0)−(−22−22)=1−(−2)=1+2[-\cos(x) - \sin(x)]_{\pi/4}^{\pi} = (-\cos(\pi) - \sin(\pi)) - (-\cos(\pi/4) - \sin(\pi/4)) = (-(-1) - 0) - (-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) = 1 - (-\sqrt{2}) = 1 + \sqrt{2}. Total area = (2−1)+(1+2)=22(\sqrt{2} - 1) + (1 + \sqrt{2}) = 2\sqrt{2}.

Question 19

For what positive value of cc is the area of the region bounded by the parabola y=x2y = x^2 and the horizontal line y=cy = c equal to 36?

  1. 333\sqrt{3}
  2. 66
  3. 99 (correct answer)
  4. 1818
Explanation: First, find the intersection points of y=x2y = x^2 and y=cy = c by setting them equal: x2=cx^2 = c, which gives x=±cx = \pm\sqrt{c}. These are the limits of integration. The line y=cy=c is the upper boundary and the parabola y=x2y=x^2 is the lower boundary. The area AA is given by the integral: A=∫−cc(c−x2)dxA = \int_{-\sqrt{c}}^{\sqrt{c}} (c - x^2) dx. Since the integrand (c−x2)(c - x^2) is an even function, we can simplify the calculation using symmetry: A=2∫0c(c−x2)dx=2[cx−x33]0cA = 2 \int_{0}^{\sqrt{c}} (c - x^2) dx = 2 [cx - \frac{x^3}{3}]_0^{\sqrt{c}}. Evaluating the integral: A=2(c(c)−(c)33)=2(c3/2−c3/23)=2(23c3/2)=43c3/2A = 2 (c(\sqrt{c}) - \frac{(\sqrt{c})^3}{3}) = 2 (c^{3/2} - \frac{c^{3/2}}{3}) = 2(\frac{2}{3}c^{3/2}) = \frac{4}{3}c^{3/2}. We are given that the area is 36. So, we set up the equation: 43c3/2=36\frac{4}{3}c^{3/2} = 36. Solving for cc: c3/2=36⋅34=27c^{3/2} = 36 \cdot \frac{3}{4} = 27. c=(27)2/3=(273)2=32=9c = (27)^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9.

Question 20

Find the area of the region bounded by the parabola y=x2+1y=x^2+1, the tangent line to this parabola at the point (1,2)(1,2), and the y-axis.

  1. 1/31/3 (correct answer)
  2. 2/32/3
  3. 11
  4. 4/34/3
Explanation: First, find the equation of the tangent line. The derivative of y=x2+1y=x^2+1 is y′=2xy' = 2x. At the point (1,2)(1,2), the slope of the tangent line is m=y′(1)=2(1)=2m = y'(1) = 2(1) = 2. Using the point-slope form, the equation of the tangent line is y−2=2(x−1)y - 2 = 2(x - 1), which simplifies to y=2xy = 2x. The region is bounded by y=x2+1y=x^2+1, y=2xy=2x, and the y-axis (x=0x=0). The curves intersect at x=1x=1. On the interval [0,1][0,1], we must determine the upper and lower functions. Testing a point, e.g., x=0.5x=0.5: y=(0.5)2+1=1.25y=(0.5)^2+1=1.25 and y=2(0.5)=1y=2(0.5)=1. Thus, y=x2+1y=x^2+1 is the upper function and y=2xy=2x is the lower function. The area is given by the integral of (upper function - lower function) from x=0x=0 to x=1x=1: A=∫01((x2+1)−2x)dx=∫01(x2−2x+1)dx=∫01(x−1)2dxA = \int_0^1 ((x^2+1) - 2x) dx = \int_0^1 (x^2 - 2x + 1) dx = \int_0^1 (x-1)^2 dx. Evaluating the integral: A=[(x−1)33]01=(1−1)33−(0−1)33=0−−13=13A = [\frac{(x-1)^3}{3}]_0^1 = \frac{(1-1)^3}{3} - \frac{(0-1)^3}{3} = 0 - \frac{-1}{3} = \frac{1}{3}.