Calculus 2 Quiz: Area Between Multiple Intersections
3 questions · exam conditions
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Area Between Multiple IntersectionsQuestion 1 of 3

The curves y=x3−3xy = x^3 - 3x and y=xy = x intersect at three points. If the area between the curves in the middle region (where the cubic is below the line) is AA, and the total area of the two outer regions is BB, what is the ratio BA\frac{B}{A}?

11
22
12\frac{1}{2}
32\frac{3}{2}
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Calculus 2 Quiz

Calculus 2 Quiz: Area Between Multiple Intersections

Practice Area Between Multiple Intersections 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 Multiple Intersections, 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

The curves y=x3−3xy = x^3 - 3x and y=xy = x intersect at three points. If the area between the curves in the middle region (where the cubic is below the line) is AA, and the total area of the two outer regions is BB, what is the ratio BA\frac{B}{A}?

  1. 11 (correct answer)
  2. 22
  3. 12\frac{1}{2}
  4. 32\frac{3}{2}
Explanation: Setting x3−3x=xx^3 - 3x = x, we get x3−4x=0x^3 - 4x = 0, so x(x2−4)=0x(x^2 - 4) = 0, giving x=0,±2x = 0, \pm 2. The difference g(x)=x3−4xg(x) = x^3 - 4x satisfies g(−1)=3>0g(-1) = 3 > 0, g(1)=−3<0g(1) = -3 < 0. So the cubic is above the line on (−2,0)(-2,0) and below on (0,2)(0,2). We have A=∫02∣x3−4x∣dx=−∫02(x3−4x)dx=4A = \int_0^2 |x^3 - 4x|dx = -\int_0^2 (x^3 - 4x)dx = 4 and B=∫−20∣x3−4x∣dx=∫−20(x3−4x)dx=4B = \int_{-2}^0 |x^3 - 4x|dx = \int_{-2}^0 (x^3 - 4x)dx = 4 by symmetry of g(x)g(x) (odd function). Thus BA=1\frac{B}{A} = 1. Choice B assumes B=2AB = 2A, choice C reverses the ratio, choice D uses incorrect region identification.

Question 2

The function f(x)=x4−5x2+4f(x) = x^4 - 5x^2 + 4 intersects the xx-axis at four points. What is the total area between f(x)f(x) and the xx-axis?

  1. 2215\frac{22}{15}
  2. 4415\frac{44}{15} (correct answer)
  3. 3215\frac{32}{15}
  4. 6415\frac{64}{15}
Explanation: Setting x4−5x2+4=0x^4 - 5x^2 + 4 = 0 and substituting u=x2u = x^2, we get u2−5u+4=0u^2 - 5u + 4 = 0, so (u−1)(u−4)=0(u-1)(u-4) = 0, giving u=1,4u = 1, 4. Thus x=±1,±2x = \pm 1, \pm 2. Since f(0)=4>0f(0) = 4 > 0 and f(1.5)=(1.5)4−5(1.5)2+4=5.0625−11.25+4=−2.1875<0f(1.5) = (1.5)^4 - 5(1.5)^2 + 4 = 5.0625 - 11.25 + 4 = -2.1875 < 0, we have f(x)>0f(x) > 0 on (−2,−1)∪(1,2)(-2,-1) \cup (1,2) and f(x)<0f(x) < 0 on (−1,1)(-1,1). The area is 2∫12f(x)dx−∫−11f(x)dx=2[x55−5x33+4x]12−[x55−5x33+4x]−11=2⋅2215−0=44152\int_1^2 f(x)dx - \int_{-1}^1 f(x)dx = 2\left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_1^2 - \left[\frac{x^5}{5} - \frac{5x^3}{3} + 4x\right]_{-1}^1 = 2 \cdot \frac{22}{15} - 0 = \frac{44}{15}. Choice A gives area of one region only, C and D use computational errors.

Question 3

The curves y=x3−6x2+9xy = x^3 - 6x^2 + 9x and y=xy = x intersect at three points. What is the total area of the regions enclosed between these curves?

  1. 88 (correct answer)
  2. 323\frac{32}{3}
  3. 1616
  4. 643\frac{64}{3}
Explanation: First find intersection points by solving x3−6x2+9x=xx^3 - 6x^2 + 9x = x, which gives x3−6x2+8x=0x^3 - 6x^2 + 8x = 0, so x(x2−6x+8)=0x(x^2 - 6x + 8) = 0, yielding x=0,2,4x = 0, 2, 4. The difference function is f(x)=x3−6x2+8xf(x) = x^3 - 6x^2 + 8x. Since f(1)=3>0f(1) = 3 > 0 and f(3)=−9<0f(3) = -9 < 0, the cubic is positive on [0,2][0,2] and negative on [2,4][2,4]. The total area is ∫02∣f(x)∣dx+∫24∣f(x)∣dx=∫02f(x)dx−∫24f(x)dx=[x44−2x3+4x2]02−[x44−2x3+4x2]24=163−(−163)=8\int_0^2 |f(x)|dx + \int_2^4 |f(x)|dx = \int_0^2 f(x)dx - \int_2^4 f(x)dx = \left[\frac{x^4}{4} - 2x^3 + 4x^2\right]_0^2 - \left[\frac{x^4}{4} - 2x^3 + 4x^2\right]_2^4 = \frac{16}{3} - \left(-\frac{16}{3}\right) = 8. Choice B gives the area of just one region, C doubles the total incorrectly, and D assumes no sign change.