Area Between Curves: Functions of y
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AP Calculus BC › Area Between Curves: Functions of y
Which integral gives area between $x=y-4$ (left) and $x=y^2$ (right) for $-1\le y\le 1$?
$\displaystyle \int_{-1}^{1}\big[y^2+(y-4)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y-4)-y^2\big]dy$
$\displaystyle \int_{-1}^{1}\big[y^2-(y-4)\big]dy$
$\displaystyle \int_{0}^{1}\big[y^2-(y-4)\big]dy$
$\displaystyle \int_{1}^{-1}\big[y^2-(y-4)\big]dy$
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.
Find the correct area expression for $x=y^2$ (left) and $x=2y+4$ (right) over $-1\le y\le 3$.
$\displaystyle \int_{-3}^{1}\big[(2y+4)-y^2\big]dy$
$\displaystyle \int_{-1}^{3}\big[(2y+4)-y^2\big]dy$
$\displaystyle \int_{3}^{-1}\big[(2y+4)-y^2\big]dy$
$\displaystyle \int_{-1}^{3}\big[y^2-(2y+4)\big]dy$
$\displaystyle \int_{-1}^{3}\big[(2y+4)+y^2\big]dy$
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.
Set up area between $x=2y^2+1$ (left) and $x=9$ (right) for $0 \le y \le 2$.
$\displaystyle \int_{0}^{2}\big[9+(2y^2+1)\big]dy$
$\displaystyle \int_{2}^{0}\big[9-(2y^2+1)\big]dy$
$\displaystyle \int_{0}^{2}\big[(2y^2+1)-9\big]dy$
$\displaystyle \int_{0}^{2}\big[9-(2y^2+1)\big]dy$
$\displaystyle \int_{-2}^{0}\big[9-(2y^2+1)\big]dy$
Explanation
The skill involves setting up area with y. Integral of right - left dy. Right: x = 9, left: $x = 2y^2 + 1$, integrand $9 - (2y^2 + 1)$. Limits 0 to 2. Choice A subtracts wrong, negative. Compare at $y=1$.
Area between $x=2y-3$ (left) and $x=1-y^2$ (right) for $-1\le y\le 1$ equals which integral?
$\displaystyle \int_{-1}^{1}\big[(1-y^2)-(2y-3)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(2y-3)-(1-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(1-y^2)+(2y-3)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(1-y^2)-(2y-3)\big]dy$
$\displaystyle \int_{0}^{1}\big[(1-y^2)-(2y-3)\big]dy$
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.
Which integral represents area between $x=y^2-1$ (right) and $x=y-3$ (left) for $0 \le y \le 2$?
$\displaystyle \int_{2}^{0}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{-2}^{0}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y^2-1)+(y-3)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y^2-1)-(y-3)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y-3)-(y^2-1)\big]dy$
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 $y^2 - 1$ 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.
Area between $x=y^2+3$ (right) and $x=2y+1$ (left) for $-1 \le y \le 1$ is which integral?
$\displaystyle \int_{1}^{-1}\big[(y^2+3)-(2y+1)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2+3)-(2y+1)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2+3)+(2y+1)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(2y+1)-(y^2+3)\big]dy$
$\displaystyle \int_{0}^{1}\big[(y^2+3)-(2y+1)\big]dy$
Explanation
The task evaluates area with respect to y. Use right - left in integral. Right: x = $y^2 + 3$, left: x = $2y + 1$, integrand $(y^2 + 3) - (2y + 1)$. Limits $-1$ to $1$. Choice A reverses, negative. Verify at $y=0$.
Set up the area between $x=7-3y$ (right) and $x=1+y$ (left) on $-1 \le y \le 1$.
$\displaystyle \int_{1}^{-1}\big[(7-3y)-(1+y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(7-3y)-(1+y)\big]dy$
$\displaystyle \int_{0}^{1}\big[(7-3y)-(1+y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(7-3y)+(1+y)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(1+y)-(7-3y)\big]dy$
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.
Which integral represents area between $x=y^2-4$ (left) and $x=2$ (right) for $-1\le y\le 1$?
$\displaystyle \int_{0}^{1}\big[2-(y^2-4)\big]dy$
$\displaystyle \int_{1}^{-1}\big[2-(y^2-4)\big]dy$
$\displaystyle \int_{-1}^{1}\big[2+(y^2-4)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(y^2-4)-2\big]dy$
$\displaystyle \int_{-1}^{1}\big[2-(y^2-4)\big]dy$
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.
Area between $x=4y$ (left) and $x=8-y^2$ (right) on $-1\le y\le 1$ equals which integral?
$\displaystyle \int_{-1}^{1}\big[4y-(8-y^2)\big]dy$
$\displaystyle \int_{0}^{1}\big[(8-y^2)-4y\big]dy$
$\displaystyle \int_{-1}^{1}\big[(8-y^2)+4y\big]dy$
$\displaystyle \int_{1}^{-1}\big[(8-y^2)-4y\big]dy$
$\displaystyle \int_{-1}^{1}\big[(8-y^2)-4y\big]dy$
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 gives area between $x=5+y$ (right) and $x=1+2y^2$ (left) for $-1\le y\le 1$?
$\displaystyle \int_{-1}^{1}\big[(1+2y^2)-(5+y)\big]dy$
$\displaystyle \int_{0}^{1}\big[(5+y)-(1+2y^2)\big]dy$
$\displaystyle \int_{1}^{-1}\big[(5+y)-(1+2y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5+y)-(1+2y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(5+y)+(1+2y^2)\big]dy$
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.