Area Between Curves: Functions of y - AP Calculus BC
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State the integral setup for $x = y^3$ and $x = 4y - 1$ from $y = -1$ to $y = 1$.
State the integral setup for $x = y^3$ and $x = 4y - 1$ from $y = -1$ to $y = 1$.
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$\int_{-1}^{1} [(4y - 1) - y^3] , dy$. $(4y - 1)$ is rightmost for this interval, so subtract $y^3$ from it.
$\int_{-1}^{1} [(4y - 1) - y^3] , dy$. $(4y - 1)$ is rightmost for this interval, so subtract $y^3$ from it.
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Identify the integral setup for $x = 2y$ and $x = y^3$ from $y = -2$ to $y = 2$.
Identify the integral setup for $x = 2y$ and $x = y^3$ from $y = -2$ to $y = 2$.
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$\text{integral from } -2 \text{ to } 2 \text{ of } [(2y) - (y^3)] \text{ dy}$. $2y$ is rightmost for most of the interval, so subtract $y^3$ from it.
$\text{integral from } -2 \text{ to } 2 \text{ of } [(2y) - (y^3)] \text{ dy}$. $2y$ is rightmost for most of the interval, so subtract $y^3$ from it.
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What happens if $f(y)$ and $g(y)$ switch roles over the interval?
What happens if $f(y)$ and $g(y)$ switch roles over the interval?
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Split the integral at the intersection point. When curves cross, split the integral at intersection to maintain proper order.
Split the integral at the intersection point. When curves cross, split the integral at intersection to maintain proper order.
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What do you need to verify before integrating to find the area between curves?
What do you need to verify before integrating to find the area between curves?
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The top and bottom functions for each interval. Must identify which function is rightmost to ensure positive area calculation.
The top and bottom functions for each interval. Must identify which function is rightmost to ensure positive area calculation.
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What is the integral setup for $x = y^2 - 4$ and $x = y + 2$ from $y = -3$ to $y = 3$?
What is the integral setup for $x = y^2 - 4$ and $x = y + 2$ from $y = -3$ to $y = 3$?
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$\text{integral from } -3 \text{ to } 3 \text{ of } [(y + 2) - (y^2 - 4)] \text{ dy}$. $(y + 2)$ is rightmost, so subtract $(y^2 - 4)$ over the specified interval.
$\text{integral from } -3 \text{ to } 3 \text{ of } [(y + 2) - (y^2 - 4)] \text{ dy}$. $(y + 2)$ is rightmost, so subtract $(y^2 - 4)$ over the specified interval.
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Determine the area between $x = 4 - y^2$ and $x = y^2 - 4$ for $y \text{ in } [-2, 2]$.
Determine the area between $x = 4 - y^2$ and $x = y^2 - 4$ for $y \text{ in } [-2, 2]$.
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$\text{Area} = \text{integral from } -2 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2 - 4)] \text{ dy}$. $(4 - y^2)$ is always rightmost since it's positive while $(y^2 - 4)$ is negative.
$\text{Area} = \text{integral from } -2 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2 - 4)] \text{ dy}$. $(4 - y^2)$ is always rightmost since it's positive while $(y^2 - 4)$ is negative.
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State the condition for using horizontal slices in finding area between curves.
State the condition for using horizontal slices in finding area between curves.
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Functions must be expressed as $x = f(y)$. Horizontal slices integrate with respect to $y$, requiring $x$ as a function of $y$.
Functions must be expressed as $x = f(y)$. Horizontal slices integrate with respect to $y$, requiring $x$ as a function of $y$.
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State the integral setup for curves $x = y^2 + 1$ and $x = 4y - 2$ from $y = -1$ to $y = 3$.
State the integral setup for curves $x = y^2 + 1$ and $x = 4y - 2$ from $y = -1$ to $y = 3$.
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$\text{integral from } -1 \text{ to } 3 \text{ of } [(4y - 2) - (y^2 + 1)] \text{ dy}$. $(4y - 2)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.
$\text{integral from } -1 \text{ to } 3 \text{ of } [(4y - 2) - (y^2 + 1)] \text{ dy}$. $(4y - 2)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.
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Calculate the area between $x = y^2 + 2$ and $x = 3y + 1$ from $y = 0$ to $y = 2$.
Calculate the area between $x = y^2 + 2$ and $x = 3y + 1$ from $y = 0$ to $y = 2$.
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$$\int_0^2 [(3y + 1) - (y^2 + 2)] , dy$$. $(3y + 1)$ is rightmost, so subtract $(y^2 + 2)$ over the given bounds.
$$\int_0^2 [(3y + 1) - (y^2 + 2)] , dy$$. $(3y + 1)$ is rightmost, so subtract $(y^2 + 2)$ over the given bounds.
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What is the integral setup for area between $x = 2y^2$ and $x = y + 3$ from $y = -1$ to $y = 2$?
What is the integral setup for area between $x = 2y^2$ and $x = y + 3$ from $y = -1$ to $y = 2$?
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$\text{integral from } -1 \text{ to } 2 \text{ of } [(y + 3) - (2y^2)] \text{ dy}$. $(y + 3)$ is rightmost, so subtract $2y^2$ over the specified interval.
$\text{integral from } -1 \text{ to } 2 \text{ of } [(y + 3) - (2y^2)] \text{ dy}$. $(y + 3)$ is rightmost, so subtract $2y^2$ over the specified interval.
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Which curve should be on top when finding the area between curves expressed as functions of $y$?
Which curve should be on top when finding the area between curves expressed as functions of $y$?
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The curve with the greater $y$-value for each $y$ in the interval. The rightmost curve (larger $x$-value) is considered 'on top' for horizontal integration.
The curve with the greater $y$-value for each $y$ in the interval. The rightmost curve (larger $x$-value) is considered 'on top' for horizontal integration.
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How do you determine which function is the top function?
How do you determine which function is the top function?
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Evaluate $f(y)$ and $g(y)$ at points in the interval. Test points in the interval to see which function gives larger $x$-values.
Evaluate $f(y)$ and $g(y)$ at points in the interval. Test points in the interval to see which function gives larger $x$-values.
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Identify the setup to find the area between $x = y^2$ and $x = 4 - y^2$ from $y = -2$ to $y = 2$.
Identify the setup to find the area between $x = y^2$ and $x = 4 - y^2$ from $y = -2$ to $y = 2$.
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$\text{Area} = 2 \times \text{integral from } 0 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2)] \text{ dy}$. Uses symmetry since the region is symmetric about the $x$-axis.
$\text{Area} = 2 \times \text{integral from } 0 \text{ to } 2 \text{ of } [(4 - y^2) - (y^2)] \text{ dy}$. Uses symmetry since the region is symmetric about the $x$-axis.
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Identify the area calculation between $x = y + 2$ and $x = 3y + 1$ for $y$ in $[0, 2]$.
Identify the area calculation between $x = y + 2$ and $x = 3y + 1$ for $y$ in $[0, 2]$.
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$\text{integral from } 0 \text{ to } 2 \text{ of } [(3y + 1) - (y + 2)] \text{ dy}$. $(3y + 1)$ is rightmost, so subtract $(y + 2)$ over the given bounds.
$\text{integral from } 0 \text{ to } 2 \text{ of } [(3y + 1) - (y + 2)] \text{ dy}$. $(3y + 1)$ is rightmost, so subtract $(y + 2)$ over the given bounds.
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What is the integral setup for $x = 3y^2$ and $x = y + 4$ from $y = 0$ to $y = 2$?
What is the integral setup for $x = 3y^2$ and $x = y + 4$ from $y = 0$ to $y = 2$?
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$\text{integral from } 0 \text{ to } 2 \text{ of } [(y + 4) - (3y^2)] \text{ dy}$. $(y + 4)$ is rightmost, so subtract $3y^2$ over the specified interval.
$\text{integral from } 0 \text{ to } 2 \text{ of } [(y + 4) - (3y^2)] \text{ dy}$. $(y + 4)$ is rightmost, so subtract $3y^2$ over the specified interval.
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Which function is top: $x = y^2 - 2$ or $x = 2y$ for $y$ in $[-1, 2]$?
Which function is top: $x = y^2 - 2$ or $x = 2y$ for $y$ in $[-1, 2]$?
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$x = 2y$ is the top function. Compare values: at $y = 0$, $2y = 0$ and $y^2 - 2 = -2$, so $2y$ is rightmost.
$x = 2y$ is the top function. Compare values: at $y = 0$, $2y = 0$ and $y^2 - 2 = -2$, so $2y$ is rightmost.
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How do you find intersection points algebraically for $x = f(y)$ and $x = g(y)$?
How do you find intersection points algebraically for $x = f(y)$ and $x = g(y)$?
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Set $f(y) = g(y)$ and solve for $y$. Standard method: equate the functions and solve the resulting equation.
Set $f(y) = g(y)$ and solve for $y$. Standard method: equate the functions and solve the resulting equation.
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Determine the integral setup for $x = y^2$ and $x = 2 - y$ from $y = -1$ to $y = 1$.
Determine the integral setup for $x = y^2$ and $x = 2 - y$ from $y = -1$ to $y = 1$.
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$\text{integral from } -1 \text{ to } 1 \text{ of } [(2 - y) - (y^2)] \text{ dy}$. $(2 - y)$ is rightmost for this interval, so subtract $y^2$ from it.
$\text{integral from } -1 \text{ to } 1 \text{ of } [(2 - y) - (y^2)] \text{ dy}$. $(2 - y)$ is rightmost for this interval, so subtract $y^2$ from it.
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Calculate the area between $x = y + 2$ and $x = 2y + 1$ for $y$ in $[1, 3]$.
Calculate the area between $x = y + 2$ and $x = 2y + 1$ for $y$ in $[1, 3]$.
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$\text{integral from } 1 \text{ to } 3 \text{ of } [(2y + 1) - (y + 2)] \text{ dy}$. $(2y + 1)$ is rightmost, so subtract $(y + 2)$ over the given interval.
$\text{integral from } 1 \text{ to } 3 \text{ of } [(2y + 1) - (y + 2)] \text{ dy}$. $(2y + 1)$ is rightmost, so subtract $(y + 2)$ over the given interval.
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What must be true for $f(y)$ and $g(y)$ to use the area formula between curves?
What must be true for $f(y)$ and $g(y)$ to use the area formula between curves?
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Both must be continuous on the interval. Continuity ensures the integral exists and the area formula is valid.
Both must be continuous on the interval. Continuity ensures the integral exists and the area formula is valid.
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What is the significance of intersection points in finding the area between curves?
What is the significance of intersection points in finding the area between curves?
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They are the limits of integration. Intersection points define where curves meet and serve as integration boundaries.
They are the limits of integration. Intersection points define where curves meet and serve as integration boundaries.
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What are the limits of integration for $x = y^2$ and $x = 2y + 3$ from $y = 0$ to $y = 2$?
What are the limits of integration for $x = y^2$ and $x = 2y + 3$ from $y = 0$ to $y = 2$?
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$y = 0$ to $y = 2$. The integration bounds are simply the given interval endpoints.
$y = 0$ to $y = 2$. The integration bounds are simply the given interval endpoints.
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What should you check if the integral yields a negative area?
What should you check if the integral yields a negative area?
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Check the order of the functions in the integral. Negative area indicates the order of functions in the integrand was reversed.
Check the order of the functions in the integral. Negative area indicates the order of functions in the integrand was reversed.
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What is the general formula for the area between two curves expressed as functions of $y$?
What is the general formula for the area between two curves expressed as functions of $y$?
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$A = \text{integral from } c \text{ to } d \text{ of } [f(y) - g(y)] \text{ dy}$. Standard area formula with horizontal slices where $f(y)$ is the rightmost curve.
$A = \text{integral from } c \text{ to } d \text{ of } [f(y) - g(y)] \text{ dy}$. Standard area formula with horizontal slices where $f(y)$ is the rightmost curve.
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State the condition for using vertical slices in finding area between curves.
State the condition for using vertical slices in finding area between curves.
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Functions must be expressed as $y = f(x)$. Vertical slices integrate with respect to $x$, requiring $y$ as a function of $x$.
Functions must be expressed as $y = f(x)$. Vertical slices integrate with respect to $x$, requiring $y$ as a function of $x$.
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Find the top function between $x = 2 - y^2$ and $x = y + 1$ for $y$ in $[-1, 2]$.
Find the top function between $x = 2 - y^2$ and $x = y + 1$ for $y$ in $[-1, 2]$.
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$x = y + 1$ is the top function. For any $y$ in the interval, $y + 1$ gives larger $x$-values than $2 - y^2$.
$x = y + 1$ is the top function. For any $y$ in the interval, $y + 1$ gives larger $x$-values than $2 - y^2$.
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Identify the integral setup for curves $x = y^2 + 1$ and $x = 2y + 3$ between $y = 0$ and $y = 2$.
Identify the integral setup for curves $x = y^2 + 1$ and $x = 2y + 3$ between $y = 0$ and $y = 2$.
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$\text{integral from } 0 \text{ to } 2 \text{ of } [(2y + 3) - (y^2 + 1)] \text{ dy}$. Rightmost function $(2y + 3)$ minus leftmost function $(y^2 + 1)$ over given bounds.
$\text{integral from } 0 \text{ to } 2 \text{ of } [(2y + 3) - (y^2 + 1)] \text{ dy}$. Rightmost function $(2y + 3)$ minus leftmost function $(y^2 + 1)$ over given bounds.
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What is the integral setup for the area between $x = y^2 + 4$ and $x = 5y$ from $y = 0$ to $y = 3$?
What is the integral setup for the area between $x = y^2 + 4$ and $x = 5y$ from $y = 0$ to $y = 3$?
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$\int_0^3 [(5y) - (y^2 + 4)] , \mathrm{d}y$. $(5y)$ is rightmost, so subtract $(y^2 + 4)$ over the specified interval.
$\int_0^3 [(5y) - (y^2 + 4)] , \mathrm{d}y$. $(5y)$ is rightmost, so subtract $(y^2 + 4)$ over the specified interval.
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Determine the area between $x = y^2 + 1$ and $x = 5y - 3$ from $y = 0$ to $y = 3$.
Determine the area between $x = y^2 + 1$ and $x = 5y - 3$ from $y = 0$ to $y = 3$.
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$$\int_0^3 [(5y - 3) - (y^2 + 1)] , dy$$. $(5y - 3)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.
$$\int_0^3 [(5y - 3) - (y^2 + 1)] , dy$$. $(5y - 3)$ is rightmost, so subtract $(y^2 + 1)$ over the given bounds.
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What is the first step in finding the area between curves $x = f(y)$ and $x = g(y)$?
What is the first step in finding the area between curves $x = f(y)$ and $x = g(y)$?
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Determine points of intersection. Intersection points determine the integration bounds for the area calculation.
Determine points of intersection. Intersection points determine the integration bounds for the area calculation.
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