Evaluating Improper Integrals - AP Calculus BC
Card 1 of 30
Evaluate: $\int_{1}^{\infty} e^{-x} , dx$.
Evaluate: $\int_{1}^{\infty} e^{-x} , dx$.
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Converges to $\frac{1}{e}$. $\lim_{b \to \infty} [-e^{-x}]_1^b = 0 - (-e^{-1}) = \frac{1}{e}$
Converges to $\frac{1}{e}$. $\lim_{b \to \infty} [-e^{-x}]_1^b = 0 - (-e^{-1}) = \frac{1}{e}$
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Find the limit: $\textstyle \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} dx$.
Find the limit: $\textstyle \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^3} dx$.
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Converges to $\frac{1}{2}$. $\lim_{b \to \infty} [-\frac{1}{2x^2}]_1^b = 0 - (-\frac{1}{2}) = \frac{1}{2}$.
Converges to $\frac{1}{2}$. $\lim_{b \to \infty} [-\frac{1}{2x^2}]_1^b = 0 - (-\frac{1}{2}) = \frac{1}{2}$.
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Evaluate limit: $\textstyle \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} dx$.
Evaluate limit: $\textstyle \lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} dx$.
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Diverges. $\lim_{a \to 0^+} [\ln x]_a^1 = 0 - (-\infty) = \infty$.
Diverges. $\lim_{a \to 0^+} [\ln x]_a^1 = 0 - (-\infty) = \infty$.
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Identify singularity: $\int_{1}^{\infty} \frac{1}{(x-1)^2} dx$.
Identify singularity: $\int_{1}^{\infty} \frac{1}{(x-1)^2} dx$.
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Unbounded integrand at $x=1$. Function becomes infinite at the lower limit $x = 1$.
Unbounded integrand at $x=1$. Function becomes infinite at the lower limit $x = 1$.
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Evaluate: $\textstyle \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} , dx$.
Evaluate: $\textstyle \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} , dx$.
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Converges to $\frac{\pi}{2}$. Arcsine function evaluation: $\arcsin(1) - \arcsin(0) = \frac{\pi}{2}$.
Converges to $\frac{\pi}{2}$. Arcsine function evaluation: $\arcsin(1) - \arcsin(0) = \frac{\pi}{2}$.
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Identify the singularity: $\textstyle \int_{0}^{1} \frac{1}{x} , dx$.
Identify the singularity: $\textstyle \int_{0}^{1} \frac{1}{x} , dx$.
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Unbounded integrand at $x = 0$. Integrand has vertical asymptote at the lower limit.
Unbounded integrand at $x = 0$. Integrand has vertical asymptote at the lower limit.
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Determine convergence: $\textstyle \int_{0}^{\infty} e^{-x^2} \text{dx}$
Determine convergence: $\textstyle \int_{0}^{\infty} e^{-x^2} \text{dx}$
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Converges. Gaussian integral converges due to exponential decay.
Converges. Gaussian integral converges due to exponential decay.
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Evaluate: $\textstyle \int_{1}^{\infty} \frac{1}{x (\ln x)^2} , dx$.
Evaluate: $\textstyle \int_{1}^{\infty} \frac{1}{x (\ln x)^2} , dx$.
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Converges. Higher power in denominator ensures convergence.
Converges. Higher power in denominator ensures convergence.
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Identify the type of improper integral: $\int_{0}^{1} \frac{1}{\sqrt{x}} \text{dx}$.
Identify the type of improper integral: $\int_{0}^{1} \frac{1}{\sqrt{x}} \text{dx}$.
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Unbounded integrand at $x = 0$. $\frac{1}{\sqrt{x}}$ becomes infinite as $x \to 0^+.$
Unbounded integrand at $x = 0$. $\frac{1}{\sqrt{x}}$ becomes infinite as $x \to 0^+.$
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Identify convergence: $\int_{1}^{\infty} \frac{1}{\sqrt{x}} , dx$.
Identify convergence: $\int_{1}^{\infty} \frac{1}{\sqrt{x}} , dx$.
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Diverges. Since $p = \frac{1}{2} \leq 1$, the integral diverges.
Diverges. Since $p = \frac{1}{2} \leq 1$, the integral diverges.
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Convert to limit: $\int_{0}^{1} \frac{1}{x} , \mathrm{d}x$.
Convert to limit: $\int_{0}^{1} \frac{1}{x} , \mathrm{d}x$.
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$\lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , \mathrm{d}x$. Integrand $\frac{1}{x}$ is unbounded at lower limit $x = 0$.
$\lim_{a \to 0^+} \int_{a}^{1} \frac{1}{x} , \mathrm{d}x$. Integrand $\frac{1}{x}$ is unbounded at lower limit $x = 0$.
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Determine convergence: $\textstyle \int_{1}^{\infty} \frac{1}{x} \text{dx}$.
Determine convergence: $\textstyle \int_{1}^{\infty} \frac{1}{x} \text{dx}$.
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Diverges. $\ln(x)$ grows without bound, so the integral diverges.
Diverges. $\ln(x)$ grows without bound, so the integral diverges.
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Determine behavior: $\int_{0}^{1} \frac{1}{\sqrt{x}} \mathrm{d}x$.
Determine behavior: $\int_{0}^{1} \frac{1}{\sqrt{x}} \mathrm{d}x$.
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Converges. Power $-\frac{1}{2} > -1$ ensures convergence at $x = 0$.
Converges. Power $-\frac{1}{2} > -1$ ensures convergence at $x = 0$.
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Convert to limit: $\textstyle \int_{0}^{\infty} x e^{-x} , dx$.
Convert to limit: $\textstyle \int_{0}^{\infty} x e^{-x} , dx$.
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$\textstyle \lim_{b \to \infty} \int_{0}^{b} x e^{-x} , dx$. Infinite upper limit requires limit evaluation.
$\textstyle \lim_{b \to \infty} \int_{0}^{b} x e^{-x} , dx$. Infinite upper limit requires limit evaluation.
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State the condition for the convergence of $\textstyle \int_{0}^{a} (a-x)^{-1/2} , dx$.
State the condition for the convergence of $\textstyle \int_{0}^{a} (a-x)^{-1/2} , dx$.
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Converges if $a > 0$. For $a > 0$, the singularity at $x = a$ is integrable.
Converges if $a > 0$. For $a > 0$, the singularity at $x = a$ is integrable.
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State the behavior of $\int_0^1 x^{-1} \text{dx}$
State the behavior of $\int_0^1 x^{-1} \text{dx}$
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Diverges. Same as $\int_0^1 \frac{1}{x} dx$ which diverges.
Diverges. Same as $\int_0^1 \frac{1}{x} dx$ which diverges.
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Evaluate: $\textstyle \text{∫}_{0}^{1} \frac{1}{\text{√}x} \text{dx}$.
Evaluate: $\textstyle \text{∫}_{0}^{1} \frac{1}{\text{√}x} \text{dx}$.
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Converges to $2$. $\lim_{a \to 0^+} [2\sqrt{x}]_a^1 = 2\sqrt{1} - 0 = 2$.
Converges to $2$. $\lim_{a \to 0^+} [2\sqrt{x}]_a^1 = 2\sqrt{1} - 0 = 2$.
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What does it mean if an improper integral converges?
What does it mean if an improper integral converges?
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The limit defining the integral exists and is finite. The limit exists and equals a finite real number.
The limit defining the integral exists and is finite. The limit exists and equals a finite real number.
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Explain divergence in the context of improper integrals.
Explain divergence in the context of improper integrals.
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The integral's limit is infinite or does not exist. The limit is infinite or undefined.
The integral's limit is infinite or does not exist. The limit is infinite or undefined.
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When does an integral $\int_{a}^{b} f(x) , dx$ become improper?
When does an integral $\int_{a}^{b} f(x) , dx$ become improper?
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If $a$, $b$, or $f(x)$ are infinite or unbounded. Any infinite bound or unbounded integrand makes it improper.
If $a$, $b$, or $f(x)$ are infinite or unbounded. Any infinite bound or unbounded integrand makes it improper.
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What is the behavior of $\int_{-\infty}^{\infty} e^{-x^2} , dx$?
What is the behavior of $\int_{-\infty}^{\infty} e^{-x^2} , dx$?
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Converges to $\sqrt{\pi}$. Famous Gaussian integral equals $\sqrt{\pi}$.
Converges to $\sqrt{\pi}$. Famous Gaussian integral equals $\sqrt{\pi}$.
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What is an improper integral?
What is an improper integral?
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An integral with infinite limits or an unbounded integrand. Defined when limits of integration or integrand are infinite.
An integral with infinite limits or an unbounded integrand. Defined when limits of integration or integrand are infinite.
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Convert to limit form: $\int_{-\infty}^{0} e^{x} , dx$.
Convert to limit form: $\int_{-\infty}^{0} e^{x} , dx$.
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$\lim_{a \to -\infty} \int_{a}^{0} e^{x} , dx$. Negative infinite lower limit requires limit form.
$\lim_{a \to -\infty} \int_{a}^{0} e^{x} , dx$. Negative infinite lower limit requires limit form.
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What is the behavior of $\textstyle \text{∫}_{0}^{1} x^{-1/3} \text{dx}$?
What is the behavior of $\textstyle \text{∫}_{0}^{1} x^{-1/3} \text{dx}$?
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Converges. Power $-\frac{1}{3} > -1$ ensures convergence.
Converges. Power $-\frac{1}{3} > -1$ ensures convergence.
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Evaluate: $\textstyle \text{∫}_{0}^{\text{∞}} e^{-x} \text{dx}$.
Evaluate: $\textstyle \text{∫}_{0}^{\text{∞}} e^{-x} \text{dx}$.
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Converges to $1$. $\lim_{b \to \infty} [-e^{-x}]_0^b = 0 - (-1) = 1$.
Converges to $1$. $\lim_{b \to \infty} [-e^{-x}]_0^b = 0 - (-1) = 1$.
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What is convergence for $\int_1^\infty \frac{1}{x^4} , dx$?
What is convergence for $\int_1^\infty \frac{1}{x^4} , dx$?
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Converges to $\frac{1}{3}$. $\lim_{b \to \infty} [-\frac{1}{3x^3}]_1^b = 0 - (-\frac{1}{3}) = \frac{1}{3}$.
Converges to $\frac{1}{3}$. $\lim_{b \to \infty} [-\frac{1}{3x^3}]_1^b = 0 - (-\frac{1}{3}) = \frac{1}{3}$.
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What is the behavior of $\textstyle \text{∫}_{-\text{∞}}^{0} e^x \text{dx}$?
What is the behavior of $\textstyle \text{∫}_{-\text{∞}}^{0} e^x \text{dx}$?
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Converges to $1$. $\lim_{a \to -\infty} [e^x]_a^0 = 1 - 0 = 1$.
Converges to $1$. $\lim_{a \to -\infty} [e^x]_a^0 = 1 - 0 = 1$.
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Evaluate: $\textstyle \text{∫}_{1}^{\text{∞}} \frac{1}{x^2} \text{dx}$.
Evaluate: $\textstyle \text{∫}_{1}^{\text{∞}} \frac{1}{x^2} \text{dx}$.
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Converges to $1$. $\lim_{b \to \infty} [-\frac{1}{x}]1^b = \lim{b \to \infty} (0 + 1) = 1$.
Converges to $1$. $\lim_{b \to \infty} [-\frac{1}{x}]1^b = \lim{b \to \infty} (0 + 1) = 1$.
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Determine behavior: $\textstyle \int_{0}^{1} \frac{1}{x^p} dx$ for $p<1$.
Determine behavior: $\textstyle \int_{0}^{1} \frac{1}{x^p} dx$ for $p<1$.
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Converges. When $p < 1$, the singularity at $x = 0$ is integrable.
Converges. When $p < 1$, the singularity at $x = 0$ is integrable.
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Identify convergence: $\int_{1}^{\infty} \frac{1}{\sqrt{x}} dx$.
Identify convergence: $\int_{1}^{\infty} \frac{1}{\sqrt{x}} dx$.
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Diverges. Since $p = \frac{1}{2} \leq 1$, the integral diverges.
Diverges. Since $p = \frac{1}{2} \leq 1$, the integral diverges.
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