Area Between Curves: Functions of x - AP Calculus BC
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What is the integral setup for area: $f(x) = \frac{1}{x}$, $g(x) = 0$, $x \text{ from } 1 \text{ to } 2$?
What is the integral setup for area: $f(x) = \frac{1}{x}$, $g(x) = 0$, $x \text{ from } 1 \text{ to } 2$?
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$\text{integral from } 1 \text{ to } 2 \text{ of } \frac{1}{x} , dx$. Since $g(x) = 0$, the integrand is just $f(x)$.
$\text{integral from } 1 \text{ to } 2 \text{ of } \frac{1}{x} , dx$. Since $g(x) = 0$, the integrand is just $f(x)$.
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Find the area under $y=x^2$ above $y=0$ from $x=0$ to $x=2$.
Find the area under $y=x^2$ above $y=0$ from $x=0$ to $x=2$.
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$\frac{8}{3}$. Area = $\int_0^2 x^2 dx = [\frac{x^3}{3}]_0^2 = \frac{8}{3}$.
$\frac{8}{3}$. Area = $\int_0^2 x^2 dx = [\frac{x^3}{3}]_0^2 = \frac{8}{3}$.
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Find area between $y=x$ and $y=x^3$ from $x=0$ to $x=1$.
Find area between $y=x$ and $y=x^3$ from $x=0$ to $x=1$.
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$\frac{1}{2}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.
$\frac{1}{2}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.
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In which scenario is $g(x)$ subtracted from $f(x)$?
In which scenario is $g(x)$ subtracted from $f(x)$?
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When $f(x) \text{ is above } g(x)$. The upper curve minus the lower curve gives positive area.
When $f(x) \text{ is above } g(x)$. The upper curve minus the lower curve gives positive area.
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Find the area between $y=\frac{1}{x}$ and $y=0$ from $x=1$ to $x=3$.
Find the area between $y=\frac{1}{x}$ and $y=0$ from $x=1$ to $x=3$.
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$\text{ln}(3)$. Area = $\int_1^3 \frac{1}{x} dx = [\ln(x)]_1^3 = \ln(3)$.
$\text{ln}(3)$. Area = $\int_1^3 \frac{1}{x} dx = [\ln(x)]_1^3 = \ln(3)$.
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What is the integral setup for area: curves $y=\text{cos}(x)$, $y=\text{sin}(x)$, $x$ from $0$ to $\frac{\text{pi}}{4}$?
What is the integral setup for area: curves $y=\text{cos}(x)$, $y=\text{sin}(x)$, $x$ from $0$ to $\frac{\text{pi}}{4}$?
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$\text{integral from } 0 \text{ to } \frac{\text{pi}}{4} \text{ of } (\text{cos}(x) - \text{sin}(x)) , dx$. For $x \in [0, \frac{\pi}{4}]$, $\cos(x) \geq \sin(x)$.
$\text{integral from } 0 \text{ to } \frac{\text{pi}}{4} \text{ of } (\text{cos}(x) - \text{sin}(x)) , dx$. For $x \in [0, \frac{\pi}{4}]$, $\cos(x) \geq \sin(x)$.
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Find the area between $y=x^2$ and $y=x$ from $x=0$ to $x=1$.
Find the area between $y=x^2$ and $y=x$ from $x=0$ to $x=1$.
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$\frac{1}{6}$. Area = $\int_0^1 (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{6}$.
$\frac{1}{6}$. Area = $\int_0^1 (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{6}$.
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Find the area under $y=x^3$ and above $y=0$ from $x=0$ to $x=1$.
Find the area under $y=x^3$ and above $y=0$ from $x=0$ to $x=1$.
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$\frac{1}{4}$. Area = $\int_0^1 x^3 dx = [\frac{x^4}{4}]_0^1 = \frac{1}{4}$.
$\frac{1}{4}$. Area = $\int_0^1 x^3 dx = [\frac{x^4}{4}]_0^1 = \frac{1}{4}$.
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Which curve is on top in the area formula if $f(x) > g(x)$ on $[a, b]$?
Which curve is on top in the area formula if $f(x) > g(x)$ on $[a, b]$?
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$f(x)$ is on top. The curve with larger $y$-values is the upper curve.
$f(x)$ is on top. The curve with larger $y$-values is the upper curve.
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What is the area under $y=4x-x^2$ and above $y=0$ from $x=0$ to $x=4$?
What is the area under $y=4x-x^2$ and above $y=0$ from $x=0$ to $x=4$?
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$\frac{32}{3}$. Area = $\int_0^4 (4x - x^2) dx = [2x^2 - \frac{x^3}{3}]_0^4 = \frac{32}{3}$.
$\frac{32}{3}$. Area = $\int_0^4 (4x - x^2) dx = [2x^2 - \frac{x^3}{3}]_0^4 = \frac{32}{3}$.
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State the method to find points of intersection of $f(x)$ and $g(x)$.
State the method to find points of intersection of $f(x)$ and $g(x)$.
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Set $f(x) = g(x)$ and solve for $x$. Setting functions equal finds where curves cross.
Set $f(x) = g(x)$ and solve for $x$. Setting functions equal finds where curves cross.
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Determine the area between $y=\text{sin}(x)$ and $y=0$ from $x=0$ to $x=\frac{\text{pi}}{2}$.
Determine the area between $y=\text{sin}(x)$ and $y=0$ from $x=0$ to $x=\frac{\text{pi}}{2}$.
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- Area = $\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 1$.
- Area = $\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 1$.
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Identify the area formula if $f(x)$ is below $g(x)$ on $[a, b]$.
Identify the area formula if $f(x)$ is below $g(x)$ on $[a, b]$.
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$\text{integral from } a \text{ to } b \text{ of } (g(x) - f(x)) , dx$. When $g(x) > f(x)$, subtract $f(x)$ from $g(x)$ for positive area.
$\text{integral from } a \text{ to } b \text{ of } (g(x) - f(x)) , dx$. When $g(x) > f(x)$, subtract $f(x)$ from $g(x)$ for positive area.
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What is the area between $y=x^2$ and $y=3x$ from $x=0$ to $x=3$?
What is the area between $y=x^2$ and $y=3x$ from $x=0$ to $x=3$?
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$\frac{9}{2}$. Area = $\int_0^3 (3x - x^2) dx = [\frac{3x^2}{2} - \frac{x^3}{3}]_0^3 = \frac{9}{2}$.
$\frac{9}{2}$. Area = $\int_0^3 (3x - x^2) dx = [\frac{3x^2}{2} - \frac{x^3}{3}]_0^3 = \frac{9}{2}$.
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Find area between $y=x^3$ and $y=x$ from $x=0$ to $x=1$.
Find area between $y=x^3$ and $y=x$ from $x=0$ to $x=1$.
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$\frac{1}{4}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.
$\frac{1}{4}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.
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What is the difference $f(x) - g(x)$ if $f(x)=3x$ and $g(x)=x^2$?
What is the difference $f(x) - g(x)$ if $f(x)=3x$ and $g(x)=x^2$?
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$3x - x^2$. Subtracting the functions gives the integrand for area calculation.
$3x - x^2$. Subtracting the functions gives the integrand for area calculation.
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What is the integral for area: $f(x) = \text{ln}(x)$, $g(x) = 0$, on $[1, \text{e}]$?
What is the integral for area: $f(x) = \text{ln}(x)$, $g(x) = 0$, on $[1, \text{e}]$?
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$\text{integral from } 1 \text{ to } \text{e} \text{ of } \text{ln}(x) , dx$. Since $g(x) = 0$, the area is the integral of $\ln(x)$.
$\text{integral from } 1 \text{ to } \text{e} \text{ of } \text{ln}(x) , dx$. Since $g(x) = 0$, the area is the integral of $\ln(x)$.
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What is the integral for area: $f(x) = \text{sin}(x)$, $g(x) = 0$, on $[0, \frac{\text{pi}}{2}]$?
What is the integral for area: $f(x) = \text{sin}(x)$, $g(x) = 0$, on $[0, \frac{\text{pi}}{2}]$?
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$\text{integral from } 0 \text{ to } \frac{\text{pi}}{2} \text{ of } \text{sin}(x) , dx$. Since $g(x) = 0$, the area is the integral of $\sin(x)$.
$\text{integral from } 0 \text{ to } \frac{\text{pi}}{2} \text{ of } \text{sin}(x) , dx$. Since $g(x) = 0$, the area is the integral of $\sin(x)$.
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State the condition for $f(x)$ and $g(x)$ in $[a, b]$ to use area formula.
State the condition for $f(x)$ and $g(x)$ in $[a, b]$ to use area formula.
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$f(x) \text{ and } g(x)$ must be continuous on $[a, b]$. Ensures the area formula is valid throughout the interval.
$f(x) \text{ and } g(x)$ must be continuous on $[a, b]$. Ensures the area formula is valid throughout the interval.
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What is the integral form for area: $f(x) = x^2$, $g(x) = 0$, between $x=0$ and $x=3$?
What is the integral form for area: $f(x) = x^2$, $g(x) = 0$, between $x=0$ and $x=3$?
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$\int_0^3 x^2 , dx$. Since $g(x) = 0$, the area is just $\int f(x) dx$.
$\int_0^3 x^2 , dx$. Since $g(x) = 0$, the area is just $\int f(x) dx$.
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What is the general formula for the area between two curves $y=f(x)$ and $y=g(x)$?
What is the general formula for the area between two curves $y=f(x)$ and $y=g(x)$?
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$\text{Area} = \text{integral from } a \text{ to } b \text{ of } (f(x) - g(x)) , dx$. Integrates the difference of the upper and lower functions.
$\text{Area} = \text{integral from } a \text{ to } b \text{ of } (f(x) - g(x)) , dx$. Integrates the difference of the upper and lower functions.
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Identify the curve to be subtracted when finding area: $y=f(x)$ above $y=g(x)$.
Identify the curve to be subtracted when finding area: $y=f(x)$ above $y=g(x)$.
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Subtract $g(x)$ from $f(x)$. The lower curve is subtracted from the upper curve.
Subtract $g(x)$ from $f(x)$. The lower curve is subtracted from the upper curve.
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Determine the limits of integration for $y=x^2$ and $y=2x-x^2$.
Determine the limits of integration for $y=x^2$ and $y=2x-x^2$.
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$x=0$ to $x=2$. Found by solving $x^2 = 2x - x^2$, giving $x = 0$ and $x = 2$.
$x=0$ to $x=2$. Found by solving $x^2 = 2x - x^2$, giving $x = 0$ and $x = 2$.
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Which step is first: setting up limits or finding intersections?
Which step is first: setting up limits or finding intersections?
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Finding intersections. Intersection points determine the integration limits.
Finding intersections. Intersection points determine the integration limits.
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Which function represents the upper curve: $y=2x$ or $y=x^2$, on $[0, 1]$?
Which function represents the upper curve: $y=2x$ or $y=x^2$, on $[0, 1]$?
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$y=2x$. For $x \in [0,1]$, $2x \geq x^2$ since $2x - x^2 = x(2-x) \geq 0$.
$y=2x$. For $x \in [0,1]$, $2x \geq x^2$ since $2x - x^2 = x(2-x) \geq 0$.
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For $y=\text{e}^x$ and $y=1$, which function is on top for $x > 0$?
For $y=\text{e}^x$ and $y=1$, which function is on top for $x > 0$?
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$y=\text{e}^x$. For $x > 0$, $e^x > 1$ since the exponential function grows rapidly.
$y=\text{e}^x$. For $x > 0$, $e^x > 1$ since the exponential function grows rapidly.
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Calculate area between $y=4-x^2$ and $y=0$ from $x=0$ to $x=2$.
Calculate area between $y=4-x^2$ and $y=0$ from $x=0$ to $x=2$.
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$\frac{16}{3}$. Area = $\int_0^2 (4-x^2) dx = [4x - \frac{x^3}{3}]_0^2 = \frac{16}{3}$.
$\frac{16}{3}$. Area = $\int_0^2 (4-x^2) dx = [4x - \frac{x^3}{3}]_0^2 = \frac{16}{3}$.
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Find the area between $y=x$ and $y=0$ from $x=0$ to $x=1$.
Find the area between $y=x$ and $y=0$ from $x=0$ to $x=1$.
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$\frac{1}{2}$. Area = $\int_0^1 x dx = [\frac{x^2}{2}]_0^1 = \frac{1}{2}$.
$\frac{1}{2}$. Area = $\int_0^1 x dx = [\frac{x^2}{2}]_0^1 = \frac{1}{2}$.
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Identify the top curve: $y=\text{ln}(x)$ or $y=0$, for $x>1$.
Identify the top curve: $y=\text{ln}(x)$ or $y=0$, for $x>1$.
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$y=\text{ln}(x)$. For $x > 1$, $\ln(x) > 0$, so it's above the $x$-axis.
$y=\text{ln}(x)$. For $x > 1$, $\ln(x) > 0$, so it's above the $x$-axis.
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What is the integral setup for area: $f(x)=x^2$, $g(x)=x^4$, on $[0, 1]$?
What is the integral setup for area: $f(x)=x^2$, $g(x)=x^4$, on $[0, 1]$?
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$\text{integral from } 0 \text{ to } 1 \text{ of } (x^2 - x^4) , dx$. For $x \in [0,1]$, $x^2 \geq x^4$ since $x^2(1-x^2) \geq 0$.
$\text{integral from } 0 \text{ to } 1 \text{ of } (x^2 - x^4) , dx$. For $x \in [0,1]$, $x^2 \geq x^4$ since $x^2(1-x^2) \geq 0$.
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