Calculus 2 Quiz: Area Between Curves X Functions
3 questions · exam conditions
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Area Between Curves X FunctionsQuestion 1 of 3

Find the area of the finite region bounded by y=x3−xy = x^3 - x and y=3xy = 3x.

88
66
44
1010
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Calculus 2 Quiz

Calculus 2 Quiz: Area Between Curves X Functions

Practice Area Between Curves X Functions in Calculus 2 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 2.

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Question 1

Find the area of the finite region bounded by y=x3−xy = x^3 - x and y=3xy = 3x.

  1. 88 (correct answer)
  2. 66
  3. 44
  4. 1010
Explanation: To find intersections, solve x3−x=3xx^3 - x = 3x, which gives x3−4x=0x^3 - 4x = 0, so x(x2−4)=0x(x^2 - 4) = 0, thus x(x−2)(x+2)=0x(x-2)(x+2) = 0. The curves intersect at x=−2,0,2x = -2, 0, 2. To determine which function is larger: For x∈(−2,0)x \in (-2, 0), test x=−1x = -1: y1=(−1)3−(−1)=−1+1=0y_1 = (-1)^3 - (-1) = -1 + 1 = 0 and y2=3(−1)=−3y_2 = 3(-1) = -3, so x3−x>3xx^3 - x > 3x. For x∈(0,2)x \in (0, 2), test x=1x = 1: y1=13−1=0y_1 = 1^3 - 1 = 0 and y2=3(1)=3y_2 = 3(1) = 3, so 3x>x3−x3x > x^3 - x. Area = ∫−20(x3−x−3x)dx+∫02(3x−(x3−x))dx=∫−20(x3−4x)dx+∫02(−x3+4x)dx=[x44−2x2]−20+[−x44+2x2]02=(0−(4−8))+((−4+8)−0)=4+4=8\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 (-x^3 + 4x) dx = \left[\frac{x^4}{4} - 2x^2\right]_{-2}^0 + \left[-\frac{x^4}{4} + 2x^2\right]_0^2 = (0 - (4 - 8)) + ((-4 + 8) - 0) = 4 + 4 = 8.

Question 2

The area between y=1x2y = \frac{1}{x^2} and y=4x2y = \frac{4}{x^2} from x=1x = 1 to x=2x = 2 is:

  1. 32\frac{3}{2} (correct answer)
  2. 52\frac{5}{2}
  3. 33
  4. 72\frac{7}{2}
Explanation: Since 4x2>1x2\frac{4}{x^2} > \frac{1}{x^2} for all x>0x > 0, the area is ∫12(4x2−1x2)dx=∫123x2dx=∫123x−2dx=3∫12x−2dx=3[−x−1]12=3[−1x]12=3(−12−(−11))=3(−12+1)=3⋅12=32\int_1^2 \left(\frac{4}{x^2} - \frac{1}{x^2}\right) dx = \int_1^2 \frac{3}{x^2} dx = \int_1^2 3x^{-2} dx = 3 \int_1^2 x^{-2} dx = 3[-x^{-1}]_1^2 = 3\left[-\frac{1}{x}\right]_1^2 = 3\left(-\frac{1}{2} - \left(-\frac{1}{1}\right)\right) = 3\left(-\frac{1}{2} + 1\right) = 3 \cdot \frac{1}{2} = \frac{3}{2}.

Question 3

The curves y=xy = \sqrt{x} and y=x2y = x^2 intersect at two points. The area between these curves is:

  1. 13\frac{1}{3} (correct answer)
  2. 23\frac{2}{3}
  3. 16\frac{1}{6}
  4. 512\frac{5}{12}
Explanation: To find intersections, solve x=x2\sqrt{x} = x^2. Squaring both sides: x=x4x = x^4, so x4−x=0x^4 - x = 0, giving x(x3−1)=0x(x^3 - 1) = 0, thus x(x−1)(x2+x+1)=0x(x-1)(x^2+x+1) = 0. Since we need x≥0x \geq 0 for x\sqrt{x} to be defined, and x2+x+1>0x^2 + x + 1 > 0 for all real xx, the intersections are at x=0x = 0 and x=1x = 1. To determine which function is larger in (0,1)(0,1), test x=14x = \frac{1}{4}: 14=12\sqrt{\frac{1}{4}} = \frac{1}{2} and (14)2=116(\frac{1}{4})^2 = \frac{1}{16}. Since 12>116\frac{1}{2} > \frac{1}{16}, we have x>x2\sqrt{x} > x^2 on (0,1)(0,1). Area = ∫01(x−x2)dx=∫01(x1/2−x2)dx=[x3/23/2−x33]01=[2x3/23−x33]01=23−13−0=13\int_0^1 (\sqrt{x} - x^2) dx = \int_0^1 (x^{1/2} - x^2) dx = \left[\frac{x^{3/2}}{3/2} - \frac{x^3}{3}\right]_0^1 = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} - 0 = \frac{1}{3}.