AP Calculus BC Quiz: Area Between Curves Functions Of Y
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Area Between Curves Functions Of YQuestion 1 of 20

Which integral gives area between x=y−4x=y-4 (left) and x=y2x=y^2 (right) for −1≤y≤1-1\le y\le 1?

∫−11[y2−(y−4)]dy\displaystyle \int_{-1}^{1}\big[y^2-(y-4)\big]dy
∫−11[(y−4)−y2]dy\displaystyle \int_{-1}^{1}\big[(y-4)-y^2\big]dy
∫1−1[y2−(y−4)]dy\displaystyle \int_{1}^{-1}\big[y^2-(y-4)\big]dy
∫−11[y2+(y−4)]dy\displaystyle \int_{-1}^{1}\big[y^2+(y-4)\big]dy
∫01[y2−(y−4)]dy\displaystyle \int_{0}^{1}\big[y^2-(y-4)\big]dy
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AP Calculus BC Quiz

AP Calculus BC Quiz: Area Between Curves Functions Of Y

Practice Area Between Curves Functions Of Y in AP Calculus BC 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 Area Between Curves Functions Of Y, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

All questions

Question 1

Which integral gives area between x=y−4x=y-4 (left) and x=y2x=y^2 (right) for −1≤y≤1-1\le y\le 1?

  1. ∫−11[y2−(y−4)]dy\displaystyle \int_{-1}^{1}\big[y^2-(y-4)\big]dy (correct answer)
  2. ∫−11[(y−4)−y2]dy\displaystyle \int_{-1}^{1}\big[(y-4)-y^2\big]dy
  3. ∫1−1[y2−(y−4)]dy\displaystyle \int_{1}^{-1}\big[y^2-(y-4)\big]dy
  4. ∫−11[y2+(y−4)]dy\displaystyle \int_{-1}^{1}\big[y^2+(y-4)\big]dy
  5. ∫01[y2−(y−4)]dy\displaystyle \int_{0}^{1}\big[y^2-(y-4)\big]dy
Explanation: The skill is area between curves via dy. Integral: right minus left. Right: x = y², left: x = y - 4, so y² - (y - 4). Limits -1 to 1. Choice B reverses, negative area. Compare at y=0.

Question 2

Find the correct area expression for x=y2x=y^2 (left) and x=2y+4x=2y+4 (right) over −1≤y≤3-1\le y\le 3.

  1. ∫−13[(2y+4)−y2]dy\displaystyle \int_{-1}^{3}\big[(2y+4)-y^2\big]dy (correct answer)
  2. ∫3−1[(2y+4)−y2]dy\displaystyle \int_{3}^{-1}\big[(2y+4)-y^2\big]dy
  3. ∫−13[y2−(2y+4)]dy\displaystyle \int_{-1}^{3}\big[y^2-(2y+4)\big]dy
  4. ∫−13[(2y+4)+y2]dy\displaystyle \int_{-1}^{3}\big[(2y+4)+y^2\big]dy
  5. ∫−31[(2y+4)−y2]dy\displaystyle \int_{-3}^{1}\big[(2y+4)-y^2\big]dy
Explanation: The problem focuses on the skill of area between curves via integration with respect to y. Compute area by integrating (right - left) dy over the interval. x = 2y + 4 is right, x = y² is left, so integrand is (2y + 4) - y². Limits from y = -1 to y = 3. Choice C subtracts left from right incorrectly, giving negative area. Always test with a y-value in the range to confirm right and left assignments.

Question 3

Set up area between x=2y2+1x=2y^2+1 (left) and x=9x=9 (right) for 0≤y≤20 \le y \le 2.

  1. ∫02[(2y2+1)−9]dy\displaystyle \int_{0}^{2}\big[(2y^2+1)-9\big]dy
  2. ∫20[9−(2y2+1)]dy\displaystyle \int_{2}^{0}\big[9-(2y^2+1)\big]dy
  3. ∫02[9−(2y2+1)]dy\displaystyle \int_{0}^{2}\big[9-(2y^2+1)\big]dy (correct answer)
  4. ∫02[9+(2y2+1)]dy\displaystyle \int_{0}^{2}\big[9+(2y^2+1)\big]dy
  5. ∫−20[9−(2y2+1)]dy\displaystyle \int_{-2}^{0}\big[9-(2y^2+1)\big]dy
Explanation: The skill involves setting up area with y. Integral of right - left dy. Right: x = 9, left: x=2y2+1x = 2y^2 + 1, integrand 9−(2y2+1)9 - (2y^2 + 1). Limits 0 to 2. Choice A subtracts wrong, negative. Compare at y=1y=1.

Question 4

Area between x=2y−3x=2y-3 (left) and x=1−y2x=1-y^2 (right) for −1≤y≤1-1\le y\le 1 equals which integral?

  1. ∫−11[(2y−3)−(1−y2)]dy\displaystyle \int_{-1}^{1}\big[(2y-3)-(1-y^2)\big]dy
  2. ∫−11[(1−y2)−(2y−3)]dy\displaystyle \int_{-1}^{1}\big[(1-y^2)-(2y-3)\big]dy (correct answer)
  3. ∫1−1[(1−y2)−(2y−3)]dy\displaystyle \int_{1}^{-1}\big[(1-y^2)-(2y-3)\big]dy
  4. ∫−11[(1−y2)+(2y−3)]dy\displaystyle \int_{-1}^{1}\big[(1-y^2)+(2y-3)\big]dy
  5. ∫01[(1−y2)−(2y−3)]dy\displaystyle \int_{0}^{1}\big[(1-y^2)-(2y-3)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is 1 - y² and x_left is 2y - 3, so the integrand is (1 - y²) - (2y - 3). The limits run from y = -1 to y = 1 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 5

Which integral represents area between x=y2−1x=y^2-1 (right) and x=y−3x=y-3 (left) for 0≤y≤20 \le y \le 2?

  1. ∫02[(y−3)−(y2−1)]dy\displaystyle \int_{0}^{2}\big[(y-3)-(y^2-1)\big]dy
  2. ∫02[(y2−1)−(y−3)]dy\displaystyle \int_{0}^{2}\big[(y^2-1)-(y-3)\big]dy (correct answer)
  3. ∫20[(y2−1)−(y−3)]dy\displaystyle \int_{2}^{0}\big[(y^2-1)-(y-3)\big]dy
  4. ∫02[(y2−1)+(y−3)]dy\displaystyle \int_{0}^{2}\big[(y^2-1)+(y-3)\big]dy
  5. ∫−20[(y2−1)−(y−3)]dy\displaystyle \int_{-2}^{0}\big[(y^2-1)-(y-3)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is y2−1y^2 - 1 and x_left is y−3y - 3, so the integrand is (y2−1y^2 - 1) - (y−3y - 3). The limits run from y = 0 to y = 2 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 6

Area between x=y2+3x=y^2+3 (right) and x=2y+1x=2y+1 (left) for −1≤y≤1-1 \le y \le 1 is which integral?

  1. ∫−11[(2y+1)−(y2+3)]dy\displaystyle \int_{-1}^{1}\big[(2y+1)-(y^2+3)\big]dy
  2. ∫−11[(y2+3)−(2y+1)]dy\displaystyle \int_{-1}^{1}\big[(y^2+3)-(2y+1)\big]dy (correct answer)
  3. ∫1−1[(y2+3)−(2y+1)]dy\displaystyle \int_{1}^{-1}\big[(y^2+3)-(2y+1)\big]dy
  4. ∫−11[(y2+3)+(2y+1)]dy\displaystyle \int_{-1}^{1}\big[(y^2+3)+(2y+1)\big]dy
  5. ∫01[(y2+3)−(2y+1)]dy\displaystyle \int_{0}^{1}\big[(y^2+3)-(2y+1)\big]dy
Explanation: The task evaluates area with respect to y. Use right - left in integral. Right: x = y2+3y^2 + 3, left: x = 2y+12y + 1, integrand (y2+3)−(2y+1)(y^2 + 3) - (2y + 1). Limits −1-1 to 11. Choice A reverses, negative. Verify at y=0y=0.

Question 7

Set up the area between x=7−3yx=7-3y (right) and x=1+yx=1+y (left) on −1≤y≤1-1 \le y \le 1.

  1. ∫−11[(1+y)−(7−3y)]dy\displaystyle \int_{-1}^{1}\big[(1+y)-(7-3y)\big]dy
  2. ∫−11[(7−3y)−(1+y)]dy\displaystyle \int_{-1}^{1}\big[(7-3y)-(1+y)\big]dy (correct answer)
  3. ∫1−1[(7−3y)−(1+y)]dy\displaystyle \int_{1}^{-1}\big[(7-3y)-(1+y)\big]dy
  4. ∫−11[(7−3y)+(1+y)]dy\displaystyle \int_{-1}^{1}\big[(7-3y)+(1+y)\big]dy
  5. ∫01[(7−3y)−(1+y)]dy\displaystyle \int_{0}^{1}\big[(7-3y)-(1+y)\big]dy
Explanation: This exercise assesses y-based area integrals. Subtract left from right. Right: x=7−3yx = 7 - 3y, left: x=1+yx = 1 + y, integrand (7−3y)−(1+y)(7 - 3y) - (1 + y). From −1-1 to 11. Choice A reverses, negative. Evaluate midpoint.

Question 8

Which integral represents area between x=y2−4x=y^2-4 (left) and x=2x=2 (right) for −1≤y≤1-1\le y\le 1?

  1. ∫−11[(y2−4)−2]dy\displaystyle \int_{-1}^{1}\big[(y^2-4)-2\big]dy
  2. ∫−11[2−(y2−4)]dy\displaystyle \int_{-1}^{1}\big[2-(y^2-4)\big]dy (correct answer)
  3. ∫1−1[2−(y2−4)]dy\displaystyle \int_{1}^{-1}\big[2-(y^2-4)\big]dy
  4. ∫−11[2+(y2−4)]dy\displaystyle \int_{-1}^{1}\big[2+(y^2-4)\big]dy
  5. ∫01[2−(y2−4)]dy\displaystyle \int_{0}^{1}\big[2-(y^2-4)\big]dy
Explanation: The skill involves areas via dy integrals. Use ∫ (right - left) dy. Right: x = 2, left: x = y² - 4, integrand 2 - (y² - 4). Limits -1 to 1. Choice A subtracts left minus right, negative. Test y=0 to confirm right curve.

Question 9

Area between x=4yx=4y (left) and x=8−y2x=8-y^2 (right) on −1≤y≤1-1\le y\le 1 equals which integral?

  1. ∫−11[4y−(8−y2)]dy\displaystyle \int_{-1}^{1}\big[4y-(8-y^2)\big]dy
  2. ∫−11[(8−y2)−4y]dy\displaystyle \int_{-1}^{1}\big[(8-y^2)-4y\big]dy (correct answer)
  3. ∫1−1[(8−y2)−4y]dy\displaystyle \int_{1}^{-1}\big[(8-y^2)-4y\big]dy
  4. ∫−11[(8−y2)+4y]dy\displaystyle \int_{-1}^{1}\big[(8-y^2)+4y\big]dy
  5. ∫01[(8−y2)−4y]dy\displaystyle \int_{0}^{1}\big[(8-y^2)-4y\big]dy
Explanation: This question evaluates area between curves in y-terms. Subtract left from right. Right: x = 8 - y², left: x = 4y, integrand (8 - y²) - 4y. From -1 to 1. Choice A subtracts wrong, negative. Test at y=0.

Question 10

Which integral gives area between x=5+yx=5+y (right) and x=1+2y2x=1+2y^2 (left) for −1≤y≤1-1\le y\le 1?

  1. ∫−11[(1+2y2)−(5+y)]dy\displaystyle \int_{-1}^{1}\big[(1+2y^2)-(5+y)\big]dy
  2. ∫−11[(5+y)−(1+2y2)]dy\displaystyle \int_{-1}^{1}\big[(5+y)-(1+2y^2)\big]dy (correct answer)
  3. ∫1−1[(5+y)−(1+2y2)]dy\displaystyle \int_{1}^{-1}\big[(5+y)-(1+2y^2)\big]dy
  4. ∫−11[(5+y)+(1+2y2)]dy\displaystyle \int_{-1}^{1}\big[(5+y)+(1+2y^2)\big]dy
  5. ∫01[(5+y)−(1+2y2)]dy\displaystyle \int_{0}^{1}\big[(5+y)-(1+2y^2)\big]dy
Explanation: This problem practices area calculations with respect to y. Area = ∫ (x_right - x_left) dy. Right: x = 5 + y, left: x = 1 + 2y², so (5 + y) - (1 + 2y²). From y = -1 to 1. Choice A reverses, giving negative. Test at y=0 to confirm which is right.

Question 11

Which integral represents area between x=5−2yx=5-2y (right) and x=y2x=y^2 (left) on −1≤y≤1-1\le y\le 1?

  1. ∫−11[y2−(5−2y)]dy\displaystyle \int_{-1}^{1}\big[y^2-(5-2y)\big]dy
  2. ∫−11[(5−2y)−y2]dy\displaystyle \int_{-1}^{1}\big[(5-2y)-y^2\big]dy (correct answer)
  3. ∫1−1[(5−2y)−y2]dy\displaystyle \int_{1}^{-1}\big[(5-2y)-y^2\big]dy
  4. ∫−11[(5−2y)+y2]dy\displaystyle \int_{-1}^{1}\big[(5-2y)+y^2\big]dy
  5. ∫01[(5−2y)−y2]dy\displaystyle \int_{0}^{1}\big[(5-2y)-y^2\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is 5 - 2y and x_left is y², so the integrand is (5 - 2y) - y². The limits run from y = -1 to y = 1 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 12

Area between x=y2+4x=y^2+4 (right) and x=2yx=2y (left) on −2≤y≤2-2\le y\le 2 is which integral?

  1. ∫−22[2y−(y2+4)]dy\displaystyle \int_{-2}^{2}\big[2y-(y^2+4)\big]dy
  2. ∫−22[(y2+4)−2y]dy\displaystyle \int_{-2}^{2}\big[(y^2+4)-2y\big]dy (correct answer)
  3. ∫2−2[(y2+4)−2y]dy\displaystyle \int_{2}^{-2}\big[(y^2+4)-2y\big]dy
  4. ∫−22[(y2+4)+2y]dy\displaystyle \int_{-2}^{2}\big[(y^2+4)+2y\big]dy
  5. ∫02[(y2+4)−2y]dy\displaystyle \int_{0}^{2}\big[(y^2+4)-2y\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is y² + 4 and x_left is 2y, so the integrand is (y² + 4) - 2y. The limits run from y = -2 to y = 2 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 13

Which integral represents area between x=2x=2 (right) and x=−y2x=-y^2 (left) for −2≤y≤2-2\le y\le 2?

  1. ∫−22[(−y2)−2]dy\displaystyle \int_{-2}^{2}\big[(-y^2)-2\big]dy
  2. ∫2−2[2−(−y2)]dy\displaystyle \int_{2}^{-2}\big[2-(-y^2)\big]dy
  3. ∫−22[2−(−y2)]dy\displaystyle \int_{-2}^{2}\big[2-(-y^2)\big]dy (correct answer)
  4. ∫−22[2+(−y2)]dy\displaystyle \int_{-2}^{2}\big[2+(-y^2)\big]dy
  5. ∫−11[2−(−y2)]dy\displaystyle \int_{-1}^{1}\big[2-(-y^2)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is 2 and x_left is -y², so the integrand is 2 - (-y²). The limits run from y = -2 to y = 2 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 14

Set up area between x=2−yx=2-y (left) and x=y2+2x=y^2+2 (right) for 0≤y≤10\le y\le 1.

  1. ∫01[(2−y)−(y2+2)]dy\displaystyle \int_{0}^{1}\big[(2-y)-(y^2+2)\big]dy
  2. ∫10[(y2+2)−(2−y)]dy\displaystyle \int_{1}^{0}\big[(y^2+2)-(2-y)\big]dy
  3. ∫01[(y2+2)−(2−y)]dy\displaystyle \int_{0}^{1}\big[(y^2+2)-(2-y)\big]dy (correct answer)
  4. ∫01[(y2+2)+(2−y)]dy\displaystyle \int_{0}^{1}\big[(y^2+2)+(2-y)\big]dy
  5. ∫−10[(y2+2)−(2−y)]dy\displaystyle \int_{-1}^{0}\big[(y^2+2)-(2-y)\big]dy
Explanation: The task is to set up area integrals in terms of y. Subtract left from right and integrate dy. Right: x = y² + 2, left: x = 2 - y, integrand (y² + 2) - (2 - y). Limits 0 to 1. Choice A subtracts wrong way, negative area. Evaluate at midpoint y to verify positions.

Question 15

Set up area between x=1x=1 (left) and x=5−y2x=5-y^2 (right) for −1≤y≤1-1\le y\le 1.

  1. ∫−11[1−(5−y2)]dy\displaystyle \int_{-1}^{1}\big[1-(5-y^2)\big]dy
  2. ∫−11[(5−y2)−1]dy\displaystyle \int_{-1}^{1}\big[(5-y^2)-1\big]dy (correct answer)
  3. ∫1−1[(5−y2)−1]dy\displaystyle \int_{1}^{-1}\big[(5-y^2)-1\big]dy
  4. ∫−11[(5−y2)+1]dy\displaystyle \int_{-1}^{1}\big[(5-y^2)+1\big]dy
  5. ∫01[(5−y2)−1]dy\displaystyle \int_{0}^{1}\big[(5-y^2)-1\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is 5 - y² and x_left is 1, so the integrand is (5 - y²) - 1. The limits run from y = -1 to y = 1 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 16

Set up the area between x=6+yx=6+y (right) and x=2y+1x=2y+1 (left) on −1≤y≤1-1 \le y \le 1.

  1. ∫−11[(2y+1)−(6+y)]dy\displaystyle \int_{-1}^{1}\big[(2y+1)-(6+y)\big]dy
  2. ∫1−1[(6+y)−(2y+1)]dy\displaystyle \int_{1}^{-1}\big[(6+y)-(2y+1)\big]dy
  3. ∫−11[(6+y)−(2y+1)]dy\displaystyle \int_{-1}^{1}\big[(6+y)-(2y+1)\big]dy (correct answer)
  4. ∫−11[(6+y)+(2y+1)]dy\displaystyle \int_{-1}^{1}\big[(6+y)+(2y+1)\big]dy
  5. ∫01[(6+y)−(2y+1)]dy\displaystyle \int_{0}^{1}\big[(6+y)-(2y+1)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference xright−xleftx_{\text{right}} - x_{\text{left}} over the y-interval. Here, x_right is 6+y6 + y and x_left is 2y+12y + 1, so the integrand is (6+y)−(2y+1)(6 + y) - (2y + 1). The limits run from y = −1-1 to y = 11 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 17

Which integral gives the area between x=y2−2yx=y^2-2y (right) and x=−1x=-1 (left) for −1≤y≤1-1 \leq y \leq 1?

  1. ∫−11[(−1)−(y2−2y)]dy\displaystyle \int_{-1}^{1}\big[(-1)-(y^2-2y)\big]dy
  2. ∫−11[(y2−2y)−(−1)]dy\displaystyle \int_{-1}^{1}\big[(y^2-2y)-(-1)\big]dy (correct answer)
  3. ∫1−1[(y2−2y)−(−1)]dy\displaystyle \int_{1}^{-1}\big[(y^2-2y)-(-1)\big]dy
  4. ∫−11[(y2−2y)+(−1)]dy\displaystyle \int_{-1}^{1}\big[(y^2-2y)+(-1)\big]dy
  5. ∫−22[(y2−2y)−(−1)]dy\displaystyle \int_{-2}^{2}\big[(y^2-2y)-(-1)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is y2−2yy^2 - 2y and x_left is −1-1, so the integrand is (y2−2y)−(−1)(y^2 - 2y) - (-1). The limits run from y = −1-1 to y = 11 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 18

Which integral represents area between x=2−y2x=2-y^2 (left) and x=4−yx=4-y (right) for 0≤y≤20\le y\le 2?

  1. ∫02[(2−y2)−(4−y)]dy\displaystyle \int_{0}^{2}\big[(2-y^2)-(4-y)\big]dy
  2. ∫20[(4−y)−(2−y2)]dy\displaystyle \int_{2}^{0}\big[(4-y)-(2-y^2)\big]dy
  3. ∫02[(4−y)−(2−y2)]dy\displaystyle \int_{0}^{2}\big[(4-y)-(2-y^2)\big]dy (correct answer)
  4. ∫02[(4−y)+(2−y2)]dy\displaystyle \int_{0}^{2}\big[(4-y)+(2-y^2)\big]dy
  5. ∫−20[(4−y)−(2−y2)]dy\displaystyle \int_{-2}^{0}\big[(4-y)-(2-y^2)\big]dy
Explanation: This problem assesses your skill in computing areas between curves by integrating with respect to y, which is particularly effective when the boundaries are given as x in terms of y. The area is found by integrating the difference x_right - x_left over the y-interval. Here, x_right is 4 - y and x_left is 2 - y², so the integrand is (4 - y) - (2 - y²). The limits run from y = 0 to y = 2 to cover the region. A tempting distractor is choice A, which reverses the subtraction to left minus right, resulting in a negative value that does not represent area. Always verify the order by checking which curve has larger x-values at a point inside the interval.

Question 19

Area between x=y2+1x=y^2+1 (right) and x=0x=0 (left) for −2≤y≤2-2\le y\le 2 is which integral?

  1. ∫−22[0−(y2+1)]dy\displaystyle \int_{-2}^{2}\big[0-(y^2+1)\big]dy
  2. ∫2−2[(y2+1)−0]dy\displaystyle \int_{2}^{-2}\big[(y^2+1)-0\big]dy
  3. ∫−22[(y2+1)−0]dy\displaystyle \int_{-2}^{2}\big[(y^2+1)-0\big]dy (correct answer)
  4. ∫−22[(y2+1)+0]dy\displaystyle \int_{-2}^{2}\big[(y^2+1)+0\big]dy
  5. ∫−11[(y2+1)−0]dy\displaystyle \int_{-1}^{1}\big[(y^2+1)-0\big]dy
Explanation: The problem focuses on areas with respect to y. ∫ (right - left) dy. Right: x = y² + 1, left: x = 0, so (y² + 1) - 0. Limits -2 to 2. Choice A subtracts wrong, negative. Test y=0.

Question 20

Set up area between x=3y−1x=3y-1 (left) and x=y2+1x=y^2+1 (right) for 1≤y≤21 \le y \le 2.

  1. ∫12[(3y−1)−(y2+1)]dy\displaystyle \int_{1}^{2}\big[(3y-1)-(y^2+1)\big]dy
  2. ∫21[(y2+1)−(3y−1)]dy\displaystyle \int_{2}^{1}\big[(y^2+1)-(3y-1)\big]dy
  3. ∫12[(y2+1)−(3y−1)]dy\displaystyle \int_{1}^{2}\big[(y^2+1)-(3y-1)\big]dy (correct answer)
  4. ∫12[(y2+1)+(3y−1)]dy\displaystyle \int_{1}^{2}\big[(y^2+1)+(3y-1)\big]dy
  5. ∫01[(y2+1)−(3y−1)]dy\displaystyle \int_{0}^{1}\big[(y^2+1)-(3y-1)\big]dy
Explanation: The skill is y-integration for curve areas. Subtract left from right. Right: x = y2+1y^2 + 1, left: x = 3y−13y - 1, integrand (y2+1y^2 + 1) - (3y−13y - 1). Limits 1 to 2. Choice A reverses, negative area. Test midpoint y for confirmation.