Area Between Curves with Multiple Intersections - AP Calculus BC
Card 1 of 30
Find area between $f(x)=x^2$ and $g(x)=4-x^2$ from $x=-2$ to $x=2$.
Find area between $f(x)=x^2$ and $g(x)=4-x^2$ from $x=-2$ to $x=2$.
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$\text{Area} = \int_{-2}^{2} (4-x^2 - x^2) , dx$. Since $4-x^2 > x^2$ on $(-2,2)$, upper minus lower.
$\text{Area} = \int_{-2}^{2} (4-x^2 - x^2) , dx$. Since $4-x^2 > x^2$ on $(-2,2)$, upper minus lower.
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What is the area between $y=3x$ and $y=x^3$ on $[0, 1]$?
What is the area between $y=3x$ and $y=x^3$ on $[0, 1]$?
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$\text{Area} = \int_{0}^{1} (3x - x^3) , dx$. Since $3x > x^3$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (3x - x^3) , dx$. Since $3x > x^3$ on $(0,1)$, upper minus lower.
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How do you find the limits of integration for the area between curves?
How do you find the limits of integration for the area between curves?
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Use the intersection points as limits. Intersection points become integration boundaries.
Use the intersection points as limits. Intersection points become integration boundaries.
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What is the first step in finding the area between curves that intersect at multiple points?
What is the first step in finding the area between curves that intersect at multiple points?
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Identify and solve for intersection points. Essential to determine interval boundaries for integration.
Identify and solve for intersection points. Essential to determine interval boundaries for integration.
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What is the next step after finding intersection points for area between curves?
What is the next step after finding intersection points for area between curves?
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Determine the top and bottom functions in each interval. Determines correct integrand order for positive area.
Determine the top and bottom functions in each interval. Determines correct integrand order for positive area.
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Calculate the area between $y=2x$ and $y=x^2$ on $[0, 1]$.
Calculate the area between $y=2x$ and $y=x^2$ on $[0, 1]$.
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$\text{Area} = \int_{0}^{1} (2x - x^2) , dx$. Since $2x > x^2$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (2x - x^2) , dx$. Since $2x > x^2$ on $(0,1)$, upper minus lower.
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Find the area between $y=\frac{1}{x}$ and $y=x^2$ over $x=1$ to $x=2$.
Find the area between $y=\frac{1}{x}$ and $y=x^2$ over $x=1$ to $x=2$.
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$\text{Area} = \int_{1}^{2} (x^2 - \frac{1}{x}) , dx$. Since $x^2 > \frac{1}{x}$ on $(1,2)$, upper minus lower.
$\text{Area} = \int_{1}^{2} (x^2 - \frac{1}{x}) , dx$. Since $x^2 > \frac{1}{x}$ on $(1,2)$, upper minus lower.
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How do you handle situations where curves switch positions in the interval $[a, b]$?
How do you handle situations where curves switch positions in the interval $[a, b]$?
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Divide the integral at the intersection points. Split integral where curves change relative position.
Divide the integral at the intersection points. Split integral where curves change relative position.
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State the integral expression for the area between two curves $y=f(x)$ and $y=g(x)$.
State the integral expression for the area between two curves $y=f(x)$ and $y=g(x)$.
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$\text{Area} = \int_{a}^{b} (f(x) - g(x)) , dx$. Where $f(x) \geq g(x)$ on $[a,b]$ gives positive area.
$\text{Area} = \int_{a}^{b} (f(x) - g(x)) , dx$. Where $f(x) \geq g(x)$ on $[a,b]$ gives positive area.
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When integrating to find area, why is it important to know which curve is upper and which is lower?
When integrating to find area, why is it important to know which curve is upper and which is lower?
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To ensure the integrand is positive. Area requires non-negative integrand values.
To ensure the integrand is positive. Area requires non-negative integrand values.
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Find the area between $y=3x$ and $y=x^2$ from $x=0$ to $x=1$.
Find the area between $y=3x$ and $y=x^2$ from $x=0$ to $x=1$.
Tap to reveal answer
$\text{Area} = \int_{0}^{1} (3x - x^2) , dx$. Since $3x > x^2$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (3x - x^2) , dx$. Since $3x > x^2$ on $(0,1)$, upper minus lower.
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Solve for area: $y=x^2$ and $y=x^3$ intersect at $x=0$ and $x=1$.
Solve for area: $y=x^2$ and $y=x^3$ intersect at $x=0$ and $x=1$.
Tap to reveal answer
$\text{Area} = \int_{0}^{1} (x^2 - x^3) , dx$. Since $x^2 > x^3$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (x^2 - x^3) , dx$. Since $x^2 > x^3$ on $(0,1)$, upper minus lower.
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Find area between $y=x^2$ and $y=4-x^2$ from $x=0$ to $x=2$.
Find area between $y=x^2$ and $y=4-x^2$ from $x=0$ to $x=2$.
Tap to reveal answer
$\text{Area} = \int_{0}^{2} (4-x^2 - x^2) , dx$. Since $4-x^2 > x^2$ on $(0,2)$, upper minus lower.
$\text{Area} = \int_{0}^{2} (4-x^2 - x^2) , dx$. Since $4-x^2 > x^2$ on $(0,2)$, upper minus lower.
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What is the area between $y=x^2$ and $y=4$ on $x=0$ to $x=2$?
What is the area between $y=x^2$ and $y=4$ on $x=0$ to $x=2$?
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$\text{Area} = \int_{0}^{2} (4 - x^2) , dx$. Since $4 > x^2$ on $(0,2)$, upper minus lower.
$\text{Area} = \int_{0}^{2} (4 - x^2) , dx$. Since $4 > x^2$ on $(0,2)$, upper minus lower.
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What is the integral expression for the area between $y=\frac{1}{x}$ and $y=1$ from $x=1$ to $x=2$?
What is the integral expression for the area between $y=\frac{1}{x}$ and $y=1$ from $x=1$ to $x=2$?
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$\text{Area} = \int_{1}^{2} (1 - \frac{1}{x}) , dx$. Since $1 > \frac{1}{x}$ on $(1,2)$, upper minus lower.
$\text{Area} = \int_{1}^{2} (1 - \frac{1}{x}) , dx$. Since $1 > \frac{1}{x}$ on $(1,2)$, upper minus lower.
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Identify the error: $f(x)=x^3$ and $g(x)=x$ intersect at $x=0$ and $x=1$; $\text{Area} = \int_{0}^{1} (x^3 - x) , dx$.
Identify the error: $f(x)=x^3$ and $g(x)=x$ intersect at $x=0$ and $x=1$; $\text{Area} = \int_{0}^{1} (x^3 - x) , dx$.
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Correct to $\text{Area} = \int_{0}^{1} (x - x^3) , dx$. Since $x > x^3$ on $(0,1)$, should be $x - x^3$.
Correct to $\text{Area} = \int_{0}^{1} (x - x^3) , dx$. Since $x > x^3$ on $(0,1)$, should be $x - x^3$.
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What is the integral expression for area if $f(x) = x^2$ and $g(x) = x$?
What is the integral expression for area if $f(x) = x^2$ and $g(x) = x$?
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$\text{Area} = \int_{0}^{1} (x - x^2) , dx$. Since $x > x^2$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (x - x^2) , dx$. Since $x > x^2$ on $(0,1)$, upper minus lower.
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Identify the error in this: $\text{Area} = \int_{a}^{b} (f(x) - g(x)) , dx$ for $f(x) < g(x)$.
Identify the error in this: $\text{Area} = \int_{a}^{b} (f(x) - g(x)) , dx$ for $f(x) < g(x)$.
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Correct to $\text{Area} = \int_{a}^{b} (g(x) - f(x)) , dx$. Must have upper minus lower function for positive area.
Correct to $\text{Area} = \int_{a}^{b} (g(x) - f(x)) , dx$. Must have upper minus lower function for positive area.
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For $y=x^2$ and $y=2x$, find the area between $x=0$ and $x=2$.
For $y=x^2$ and $y=2x$, find the area between $x=0$ and $x=2$.
Tap to reveal answer
$\text{Area} = \int_{0}^{2} (2x - x^2) , dx$. Since $2x > x^2$ on $(0,2)$, upper minus lower.
$\text{Area} = \int_{0}^{2} (2x - x^2) , dx$. Since $2x > x^2$ on $(0,2)$, upper minus lower.
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Find the area between $y=x$ and $y=x^3$ over $x=0$ to $x=1$.
Find the area between $y=x$ and $y=x^3$ over $x=0$ to $x=1$.
Tap to reveal answer
$\text{Area} = \int_{0}^{1} (x - x^3) , dx$. Since $x > x^3$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (x - x^3) , dx$. Since $x > x^3$ on $(0,1)$, upper minus lower.
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Find the area between $y=\frac{1}{x}$ and $y=2$ from $x=1$ to $x=4$.
Find the area between $y=\frac{1}{x}$ and $y=2$ from $x=1$ to $x=4$.
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$\text{Area} = \int_{1}^{4} (2 - \frac{1}{x}) , dx$. Since $2 > \frac{1}{x}$ on $(1,4)$, upper minus lower.
$\text{Area} = \int_{1}^{4} (2 - \frac{1}{x}) , dx$. Since $2 > \frac{1}{x}$ on $(1,4)$, upper minus lower.
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What does a negative integrand indicate when calculating area between curves?
What does a negative integrand indicate when calculating area between curves?
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The order of the functions is incorrect. Means lower function was subtracted from upper.
The order of the functions is incorrect. Means lower function was subtracted from upper.
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How do you verify if the integrand is set up correctly for area between curves?
How do you verify if the integrand is set up correctly for area between curves?
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Ensure $f(x) - g(x)$ is non-negative over $[a, b]$. Check if upper function minus lower throughout interval.
Ensure $f(x) - g(x)$ is non-negative over $[a, b]$. Check if upper function minus lower throughout interval.
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For $f(x)=\frac{1}{x}$ and $g(x)=x$, what is the area between $x=1$ and $x=3$?
For $f(x)=\frac{1}{x}$ and $g(x)=x$, what is the area between $x=1$ and $x=3$?
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$\text{Area} = \int_{1}^{3} (x - \frac{1}{x}) , dx$. Since $x > \frac{1}{x}$ on $(1,3)$, upper minus lower.
$\text{Area} = \int_{1}^{3} (x - \frac{1}{x}) , dx$. Since $x > \frac{1}{x}$ on $(1,3)$, upper minus lower.
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What is the formula to find the intersection points of $y=h(x)$ and $y=k(x)$?
What is the formula to find the intersection points of $y=h(x)$ and $y=k(x)$?
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Set $h(x) = k(x)$ and solve for $x$. Standard method to find curve intersections.
Set $h(x) = k(x)$ and solve for $x$. Standard method to find curve intersections.
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What are the intersection points if $f(x)=2x$ and $g(x)=x^2$?
What are the intersection points if $f(x)=2x$ and $g(x)=x^2$?
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Intersection points are $x=0$ and $x=2$. Solve $2x = x^2$ gives $x(2-x) = 0$.
Intersection points are $x=0$ and $x=2$. Solve $2x = x^2$ gives $x(2-x) = 0$.
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What is the area between $y=x$ and $y=2x$ over $x=0$ to $x=1$?
What is the area between $y=x$ and $y=2x$ over $x=0$ to $x=1$?
Tap to reveal answer
$\text{Area} = \int_{0}^{1} (2x - x) , dx$. Since $2x > x$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (2x - x) , dx$. Since $2x > x$ on $(0,1)$, upper minus lower.
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Which method is used to find intersection points of $y=f(x)$ and $y=g(x)$?
Which method is used to find intersection points of $y=f(x)$ and $y=g(x)$?
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Set $f(x) = g(x)$ and solve for $x$. Equating functions finds where curves cross.
Set $f(x) = g(x)$ and solve for $x$. Equating functions finds where curves cross.
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Find area between $y=4-x^2$ and $y=3x$ from $x=0$ to $x=2$.
Find area between $y=4-x^2$ and $y=3x$ from $x=0$ to $x=2$.
Tap to reveal answer
$\text{Area} = \int_{0}^{2} (4-x^2 - 3x) , dx$. Since $4-x^2 > 3x$ on $(0,2)$, upper minus lower.
$\text{Area} = \int_{0}^{2} (4-x^2 - 3x) , dx$. Since $4-x^2 > 3x$ on $(0,2)$, upper minus lower.
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What is the area between $y=2x$ and $y=x^3$ over $x=0$ to $x=1$?
What is the area between $y=2x$ and $y=x^3$ over $x=0$ to $x=1$?
Tap to reveal answer
$\text{Area} = \int_{0}^{1} (2x - x^3) , dx$. Since $2x > x^3$ on $(0,1)$, upper minus lower.
$\text{Area} = \int_{0}^{1} (2x - x^3) , dx$. Since $2x > x^3$ on $(0,1)$, upper minus lower.
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