AP Calculus BC Flashcards: Area Between Curves Functions Of X

What this deck covers

This deck focuses on Area Between Curves Functions Of X, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the integral setup for area: $f(x) = \frac{1}{x}$, $g(x) = 0$, $x \text{ from } 1 \text{ to } 2$?

Answer: $\text{integral from } 1 \text{ to } 2 \text{ of } \frac{1}{x} \, dx$. Since $g(x) = 0$, the integrand is just $f(x)$.

Flashcard 2: Find the area under $y=x^2$ above $y=0$ from $x=0$ to $x=2$.

Answer: $\frac{8}{3}$. Area = $\int_0^2 x^2 dx = [\frac{x^3}{3}]_0^2 = \frac{8}{3}$.

Flashcard 3: Find area between $y=x$ and $y=x^3$ from $x=0$ to $x=1$.

Answer: $\frac{1}{2}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.

Flashcard 4: In which scenario is $g(x)$ subtracted from $f(x)$?

Answer: When $f(x) \text{ is above } g(x)$. The upper curve minus the lower curve gives positive area.

Flashcard 5: Find the area between $y=\frac{1}{x}$ and $y=0$ from $x=1$ to $x=3$.

Answer: $\text{ln}(3)$. Area = $\int_1^3 \frac{1}{x} dx = [\ln(x)]_1^3 = \ln(3)$.

Flashcard 6: What is the integral setup for area: curves $y=\text{cos}(x)$, $y=\text{sin}(x)$, $x$ from $0$ to $\frac{\text{pi}}{4}$?

Answer: $\text{integral from } 0 \text{ to } \frac{\text{pi}}{4} \text{ of } (\text{cos}(x) - \text{sin}(x)) \, dx$. For $x \in [0, \frac{\pi}{4}]$, $\cos(x) \geq \sin(x)$.

Flashcard 7: Find the area between $y=x^2$ and $y=x$ from $x=0$ to $x=1$.

Answer: $\frac{1}{6}$. Area = $\int_0^1 (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_0^1 = \frac{1}{6}$.

Flashcard 8: Find the area under $y=x^3$ and above $y=0$ from $x=0$ to $x=1$.

Answer: $\frac{1}{4}$. Area = $\int_0^1 x^3 dx = [\frac{x^4}{4}]_0^1 = \frac{1}{4}$.

Flashcard 9: Which curve is on top in the area formula if $f(x) > g(x)$ on $[a, b]$?

Answer: $f(x)$ is on top. The curve with larger $y$-values is the upper curve.

Flashcard 10: What is the area under $y=4x-x^2$ and above $y=0$ from $x=0$ to $x=4$?

Answer: $\frac{32}{3}$. Area = $\int_0^4 (4x - x^2) dx = [2x^2 - \frac{x^3}{3}]_0^4 = \frac{32}{3}$.

Flashcard 11: State the method to find points of intersection of $f(x)$ and $g(x)$.

Answer: Set $f(x) = g(x)$ and solve for $x$. Setting functions equal finds where curves cross.

Flashcard 12: Determine the area between $y=\text{sin}(x)$ and $y=0$ from $x=0$ to $x=\frac{\text{pi}}{2}$.

Answer:

1. Area = $\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 1$.

Flashcard 13: Identify the area formula if $f(x)$ is below $g(x)$ on $[a, b]$.

Answer: $\text{integral from } a \text{ to } b \text{ of } (g(x) - f(x)) \, dx$. When $g(x) > f(x)$, subtract $f(x)$ from $g(x)$ for positive area.

Flashcard 14: What is the area between $y=x^2$ and $y=3x$ from $x=0$ to $x=3$?

Answer: $\frac{9}{2}$. Area = $\int_0^3 (3x - x^2) dx = [\frac{3x^2}{2} - \frac{x^3}{3}]_0^3 = \frac{9}{2}$.

Flashcard 15: Find area between $y=x^3$ and $y=x$ from $x=0$ to $x=1$.

Answer: $\frac{1}{4}$. Area = $\int_0^1 (x - x^3) dx = [\frac{x^2}{2} - \frac{x^4}{4}]_0^1 = \frac{1}{4}$.

Flashcard 16: What is the difference $f(x) - g(x)$ if $f(x)=3x$ and $g(x)=x^2$?

Answer: $3x - x^2$. Subtracting the functions gives the integrand for area calculation.

Flashcard 17: What is the integral for area: $f(x) = \text{ln}(x)$, $g(x) = 0$, on $[1, \text{e}]$?

Answer: $\text{integral from } 1 \text{ to } \text{e} \text{ of } \text{ln}(x) \, dx$. Since $g(x) = 0$, the area is the integral of $\ln(x)$.

Flashcard 18: What is the integral for area: $f(x) = \text{sin}(x)$, $g(x) = 0$, on $[0, \frac{\text{pi}}{2}]$?

Answer: $\text{integral from } 0 \text{ to } \frac{\text{pi}}{2} \text{ of } \text{sin}(x) \, dx$. Since $g(x) = 0$, the area is the integral of $\sin(x)$.

Flashcard 19: State the condition for $f(x)$ and $g(x)$ in $[a, b]$ to use area formula.

Answer: $f(x) \text{ and } g(x)$ must be continuous on $[a, b]$. Ensures the area formula is valid throughout the interval.

Flashcard 20: What is the integral form for area: $f(x) = x^2$, $g(x) = 0$, between $x=0$ and $x=3$?

Answer: $\int_0^3 x^2 \, dx$. Since $g(x) = 0$, the area is just $\int f(x) dx$.

Flashcard 21: What is the general formula for the area between two curves $y=f(x)$ and $y=g(x)$?

Answer: $\text{Area} = \text{integral from } a \text{ to } b \text{ of } (f(x) - g(x)) \, dx$. Integrates the difference of the upper and lower functions.

Flashcard 22: Identify the curve to be subtracted when finding area: $y=f(x)$ above $y=g(x)$.

Answer: Subtract $g(x)$ from $f(x)$. The lower curve is subtracted from the upper curve.

Flashcard 23: Determine the limits of integration for $y=x^2$ and $y=2x-x^2$.

Answer: $x=0$ to $x=2$. Found by solving $x^2 = 2x - x^2$, giving $x = 0$ and $x = 2$.

Flashcard 24: Which step is first: setting up limits or finding intersections?

Answer: Finding intersections. Intersection points determine the integration limits.

Flashcard 25: Which function represents the upper curve: $y=2x$ or $y=x^2$, on $[0, 1]$?

Answer: $y=2x$. For $x \in [0,1]$, $2x \geq x^2$ since $2x - x^2 = x(2-x) \geq 0$.

Flashcard 26: For $y=\text{e}^x$ and $y=1$, which function is on top for $x > 0$?

Answer: $y=\text{e}^x$. For $x > 0$, $e^x > 1$ since the exponential function grows rapidly.

Flashcard 27: Calculate area between $y=4-x^2$ and $y=0$ from $x=0$ to $x=2$.

Answer: $\frac{16}{3}$. Area = $\int_0^2 (4-x^2) dx = [4x - \frac{x^3}{3}]_0^2 = \frac{16}{3}$.

Flashcard 28: Find the area between $y=x$ and $y=0$ from $x=0$ to $x=1$.

Answer: $\frac{1}{2}$. Area = $\int_0^1 x dx = [\frac{x^2}{2}]_0^1 = \frac{1}{2}$.

Flashcard 29: Identify the top curve: $y=\text{ln}(x)$ or $y=0$, for $x>1$.

Answer: $y=\text{ln}(x)$. For $x > 1$, $\ln(x) > 0$, so it's above the $x$-axis.

Flashcard 30: What is the integral setup for area: $f(x)=x^2$, $g(x)=x^4$, on $[0, 1]$?

Answer: $\text{integral from } 0 \text{ to } 1 \text{ of } (x^2 - x^4) \, dx$. For $x \in [0,1]$, $x^2 \geq x^4$ since $x^2(1-x^2) \geq 0$.