Area Between Curves: Functions of y
Help Questions
AP Calculus BC › Area Between Curves: Functions of y
Which integral represents area between $x=y^2-1$ (right) and $x=y-3$ (left) for $0 \le y \le 2$?
$\displaystyle \int_{0}^{2}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{2}^{0}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{-2}^{0}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y-3)-(y^2-1)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y^2-1)+(y-3)\big]dy$
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 $y^2 - 1$ 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.
Set up the area between $x=7-3y$ (right) and $x=1+y$ (left) on $-1 \le y \le 1$.
$\displaystyle \int_{-1}^{1}\big[(7-3y)+(1+y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(1+y)-(7-3y)\big]dy$
$\displaystyle \int_{0}^{1}\big[(7-3y)-(1+y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(7-3y)-(1+y)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(7-3y)-(1+y)\big]dy$
Explanation
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.
Area between $x=y^2+3$ (right) and $x=2y+1$ (left) for $-1 \le y \le 1$ is which integral?
$\displaystyle \int_{-1}^{1}\big[(2y+1)-(y^2+3)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(y^2+3)-(2y+1)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2+3)+(2y+1)\big]dy$
$\displaystyle \int_{0}^{1}\big[(y^2+3)-(2y+1)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2+3)-(2y+1)\big]dy$
Explanation
The task evaluates area with respect to y. Use right - left in integral. Right: x = $y^2 + 3$, left: x = $2y + 1$, integrand $(y^2 + 3) - (2y + 1)$. Limits $-1$ to $1$. Choice A reverses, negative. Verify at $y=0$.
Area between $x=4y$ (left) and $x=8-y^2$ (right) on $-1\le y\le 1$ equals which integral?
$\displaystyle \int_{-1}^{1}\big[4y-(8-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(8-y^2)+4y\big]dy$
$\displaystyle \int_{-1}^{1}\big[(8-y^2)-4y\big]dy$
$\displaystyle \int_{1}^{-1}\big[(8-y^2)-4y\big]dy$
$\displaystyle \int_{0}^{1}\big[(8-y^2)-4y\big]dy$
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.
Which integral represents area between $x=y^2-4$ (left) and $x=2$ (right) for $-1\le y\le 1$?
$\displaystyle \int_{-1}^{1}\big[2-(y^2-4)\big]dy$
$\displaystyle \int_{-1}^{1}\big[2+(y^2-4)\big]dy$
$\displaystyle \int_{1}^{-1}\big[2-(y^2-4)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2-4)-2\big]dy$
$\displaystyle \int_{0}^{1}\big[2-(y^2-4)\big]dy$
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.
Which integral gives area between $x=5+y$ (right) and $x=1+2y^2$ (left) for $-1\le y\le 1$?
$\displaystyle \int_{-1}^{1}\big[(5+y)+(1+2y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(1+2y^2)-(5+y)\big]dy$
$\displaystyle \int_{0}^{1}\big[(5+y)-(1+2y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5+y)-(1+2y^2)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(5+y)-(1+2y^2)\big]dy$
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.
Which integral represents area between $x=5-2y$ (right) and $x=y^2$ (left) on $-1\le y\le 1$?
$\displaystyle \int_{-1}^{1}\big[(5-2y)-y^2\big]dy$
$\displaystyle \int_{0}^{1}\big[(5-2y)-y^2\big]dy$
$\displaystyle \int_{-1}^{1}\big[y^2-(5-2y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5-2y)+y^2\big]dy$
$\displaystyle \int_{1}^{-1}\big[(5-2y)-y^2\big]dy$
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.
Area between $x=y^2+4$ (right) and $x=2y$ (left) on $-2\le y\le 2$ is which integral?
$\displaystyle \int_{-2}^{2}\big[(y^2+4)-2y\big]dy$
$\displaystyle \int_{-2}^{2}\big[2y-(y^2+4)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y^2+4)-2y\big]dy$
$\displaystyle \int_{2}^{-2}\big[(y^2+4)-2y\big]dy$
$\displaystyle \int_{-2}^{2}\big[(y^2+4)+2y\big]dy$
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.
Which integral represents area between $x=2$ (right) and $x=-y^2$ (left) for $-2\le y\le 2$?
$\displaystyle \int_{2}^{-2}\big[2-(-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[2-(-y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(-y^2)-2\big]dy$
$\displaystyle \int_{-2}^{2}\big[2-(-y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[2+(-y^2)\big]dy$
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.
Set up area between $x=1$ (left) and $x=5-y^2$ (right) for $-1\le y\le 1$.
$\displaystyle \int_{0}^{1}\big[(5-y^2)-1\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5-y^2)-1\big]dy$
$\displaystyle \int_{-1}^{1}\big[1-(5-y^2)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(5-y^2)-1\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5-y^2)+1\big]dy$
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.