Area Between Curves: Functions of y - AP Calculus BC

1. Which integral represents area between $x=5-2y$ (right) and $x=y^2$ (left) on $-1\le y\le 1$?
   Answer: $\displaystyle \int_{-1}^{1}\big[(5-2y)-y^2\big]dy$ 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.
2. Which integral gives area between $x=3y+2$ (right) and $x=y^2$ (left) on $0\le y\le 2$?
   Answer: $\displaystyle \int_{0}^{2}\big[(3y+2)-y^2\big]dy$ 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 3y + 2 and x_left is y², so the integrand is (3y + 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.
3. Which integral represents area between $x=y^2-1$ (right) and $x=y-3$ (left) for $0 \le y \le 2$?
   Answer: $\displaystyle \int_{0}^{2}\big[(y^2-1)-(y-3)\big]dy$ 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 $y^2 - 1$ and x_left is $y - 3$, so the integrand is ($y^2 - 1$) - ($y - 3$). 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.
4. Set up the area between $x=-y+6$ (right) and $x=y+1$ (left) on $0 \le y \le 2$.
   Answer: $\displaystyle \int_{0}^{2}\big[(-y+6)-(y+1)\big]dy$ This problem assesses dy integration for areas. Subtract left from right in integrand. Right: $x = -y + 6$, left: $x = y + 1$, so $(-y + 6) - (y + 1)$. From 0 to 2. Choice B reverses, negative area. Verify with $y=1$ comparison.
5. Set up the area between $x=9- y$ (right) and $x= y^2+3$ (left) for $-1\le y\le 1$.
   Answer: $\displaystyle \int_{-1}^{1}\big[(9-y)-(y^2+3)\big]dy$ 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 9 - y and x_left is y² + 3, so the integrand is (9 - y) - (y² + 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.
6. Set up area between $x=2y^2+1$ (left) and $x=9$ (right) for $0 \le y \le 2$.
   Answer: $\displaystyle \int_{0}^{2}\big[9-(2y^2+1)\big]dy$ The skill involves setting up area with y. Integral of right - left dy. Right: x = 9, left: $x = 2y^2 + 1$, integrand $9 - (2y^2 + 1)$. Limits 0 to 2. Choice A subtracts wrong, negative. Compare at $y=1$.
7. Area between $x=2y+2$ (right) and $x=y^2-2$ (left) on $0\le y\le 2$ is which integral?
   Answer: $\displaystyle \int_{0}^{2}\big[(2y+2)-(y^2-2)\big]dy$ 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 2y + 2 and x_left is y² - 2, so the integrand is (2y + 2) - (y² - 2). The limits run from y = 0 to y = 2 to cover the region. A tempting distractor is choice B, 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.
8. Area between $x=4-y$ (right) and $x=y^2+1$ (left) on $-1\le y\le 1$ is which integral?
   Answer: $\displaystyle \int_{-1}^{1}\big[(4-y)-(y^2+1)\big]dy$ 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 y² + 1, so the integrand is (4 - y) - (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.
9. Area between $x=3$ (left) and $x=y^2+7$ (right) for $-1 \le y \le 1$ equals which integral?
   Answer: $\displaystyle \int_{-1}^{1}\big[(y^2+7)-3\big]dy$ 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 $y^2 + 7$ and x_left is 3, so the integrand is $(y^2 + 7) - 3$. 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.
10. Which integral represents the area between $x=1-y$ (left) and $x=3+y^2$ (right) on $-1\le y\le 1$?
    Answer: $\displaystyle \int_{-1}^{1}\big[(3+y^2)-(1-y)\big]\,dy$ This question evaluates setting up area integrals with respect to y for curves. Area is the integral of x_right minus x_left dy from lower to upper y. Here, x = 3 + y² is right and x = 1 - y is left, so use (3 + y²) - (1 - y). Limits are y = -1 to y = 1. Choice A reverses the order, producing a negative result instead of area. To orient correctly, compare x-values at y=0 to see which curve is farther right.
11. Set up the area between $x=8$ (right) and $x=2+y$ (left) on $-2 \le y \le 2$.
    Answer: $\displaystyle \int_{-2}^{2}\big[8-(2+y)\big]dy$ 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 $8$ and x_left is $2 + y$, so the integrand is $8 - (2 + y)$. 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.
12. Which integral gives the area between $x=1-y^2$ (right) and $x=-2$ (left) on $-1 \le y \le 1$?
    Answer: $\displaystyle \int_{-1}^{1}\big[(1-y^2)-(-2)\big]dy$ This tests area between curves in y-terms. $\int (x_{\text{right}} - x_{\text{left}}) \, dy$. Right: $x = 1 - y^2$, left: $x = -2$, so $(1 - y^2) - (-2)$. From -1 to 1. Choice A reverses, negative. Evaluate at y=0 to orient.
13. Set up the area between $x=4-y^2$ (right) and $x=y$ (left) for $-1\le y\le 1$.
    Answer: $\displaystyle \int_{-1}^{1}\big[(4-y^2)-y\big]\,dy$ The skill here involves calculating areas between curves by integrating with respect to y. The integral is formed by subtracting the left curve's x from the right curve's x over the y-range. For this, x = 4 - y² is right and x = y is left, making the integrand (4 - y²) - y. Integrate from y = -1 to y = 1. Choice A subtracts incorrectly as left minus right, leading to negative area. Remember to verify curve positions by plugging in a midpoint y-value.
14. Area between $x=9-y^2$ (right) and $x=3y$ (left) on $-1\le y\le 2$ is which integral?
    Answer: $\displaystyle \int_{-1}^{2}\big[(9-y^2)-3y\big]dy$ This question tests y-based area between curves. Integral of right minus left over y-range. Right: x = 9 - y², left: x = 3y, so (9 - y²) - 3y. From -1 to 2. Choice A reverses, negative. Compare x at y=0 for orientation.
15. Area between $x=y^2+4$ (right) and $x=2y$ (left) on $-2\le y\le 2$ is which integral?
    Answer: $\displaystyle \int_{-2}^{2}\big[(y^2+4)-2y\big]dy$ 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.
16. Area between $x=y^2+1$ (right) and $x=0$ (left) for $-2\le y\le 2$ is which integral?
    Answer: $\displaystyle \int_{-2}^{2}\big[(y^2+1)-0\big]dy$ 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.
17. Set up area between $x=2-y$ (left) and $x=y^2+2$ (right) for $0\le y\le 1$.
    Answer: $\displaystyle \int_{0}^{1}\big[(y^2+2)-(2-y)\big]dy$ 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.
18. Set up the area between $x=7-3y$ (right) and $x=1+y$ (left) on $-1 \le y \le 1$.
    Answer: $\displaystyle \int_{-1}^{1}\big[(7-3y)-(1+y)\big]dy$ This exercise assesses y-based area integrals. Subtract left from right. Right: $x = 7 - 3y$, left: $x = 1 + y$, integrand $(7 - 3y) - (1 + y)$. From $-1$ to $1$. Choice A reverses, negative. Evaluate midpoint.
19. Area between $x=6-2y$ (right) and $x=y^2-1$ (left) for $0 \le y \le 2$ equals which integral?
    Answer: $\displaystyle \int_{0}^{2}\big[(6-2y)-(y^2-1)\big]dy$ This exercise tests finding areas by integrating with respect to y. The area integral subtracts the left x from the right x over y-limits. Right is $x = 6 - 2y$, left is $x = y^2 - 1$, so $(6 - 2y) - (y^2 - 1)$. From $y = 0$ to $y = 2$. Choice A reverses subtraction, resulting in negative integrand. A transferable tip is to evaluate both functions at the interval's midpoint to identify the right curve.
20. Which integral represents area between $x=2-y^2$ (left) and $x=4-y$ (right) for $0\le y\le 2$?
    Answer: $\displaystyle \int_{0}^{2}\big[(4-y)-(2-y^2)\big]dy$ 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.

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