Type 1 Improper Integrals

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Questions and Answers

Match the following improper integrals with their corresponding limit expressions:

$\int_{a}^{\infty} f(x) , dx$ = $\lim_{t \to \infty} \int_{a}^{t} f(x) , dx$ $\int_{-\infty}^{a} f(x) , dx$ = $\lim_{t \to -\infty} \int_{t}^{a} f(x) , dx$ $\int_{-\infty}^{\infty} f(x) , dx$ = $\int_{-\infty}^{c} f(x) , dx + \int_{c}^{\infty} f(x) , dx$ $\int_{a}^{b} f(x) , dx$ = Not an improper integral

Match each condition with the corresponding behavior of an improper integral:

The limit exists and is finite = The integral converges The limit is infinite or does not exist = The integral diverges The function has a vertical asymptote within the interval of integration = The integral is improper and requires special consideration The function is continuous on a finite interval = The integral is proper

Match the following concepts related to improper integrals with their descriptions:

Improper Integral = An integral over an unbounded interval or with a discontinuity Convergence = The limit of the integral exists and is finite Divergence = The limit of the integral does not exist or is infinite FTC = May not be directly applicable when evaluating an improper integral

Match the type of improper integral with its appropriate calculation method:

$\int_{a}^{\infty} f(x) , dx$ = Evaluate $\lim_{t \to \infty} \int_{a}^{t} f(x) , dx$ $\int_{-\infty}^{b} f(x) , dx$ = Evaluate $\lim_{t \to -\infty} \int_{t}^{b} f(x) , dx$ $\int_{a}^{b} f(x) , dx$ where f(x) has a vertical asymptote at c in (a,b) = Evaluate $\lim_{t \to c^-} \int_{a}^{t} f(x) , dx + \lim_{s \to c^+} \int_{s}^{b} f(x) , dx$ $\int_{a}^{b} f(x) , dx$ where f(x) is continuous on [a,b] = Apply the Fundamental Theorem of Calculus directly Signup and view all the answers

Match the application with the type of integral often used:

Net area under a curve over an infinite interval = Improper Integral Area under a curve over a finite interval = Definite Integral Finding a function given its rate of change = Indefinite Integral Discrete Summation = Series Signup and view all the answers

Match the expression with the concept it represents when evaluating improper integrals:

$\lim_{t \to \infty} A(t)$ = Net area under $f(x)$ from $a$ to infinity A(t) = Net area under $f(x)$ from $a$ to $t$ $\int_{a}^{\infty} f(x) , dx = L$ = Net area under $f(x)$ is bounded by $L$ A(t) is unbounded = Net area under $f(x)$ does not have an upper bound Signup and view all the answers

Match the properties to what needs to occur for Type I Improper Integrals.

$\int_{a}^{\infty} f(x) , dx$ = The existence of $\lim_{t \to \infty} \int_{a}^{t} f(x) , dx$ $\int_{-\infty}^{b} f(x) , dx$ = The existence of $\lim_{t \to -\infty} \int_{t}^{b} f(x) , dx$ $\int_{-\infty}^{\infty} f(x) , dx$ = Both $\int_{-\infty}^{c} f(x) , dx$ and $\int_{c}^{\infty} f(x) , dx$ must converge Improper Integral Diverges = The limit does not exist. Signup and view all the answers

Match each integral property with its corresponding value when solved:

$\int_{a}^{\infty} f(x) , dx = 5$ = The improper integral is said to converge. $\int_{a}^{\infty} f(x) , dx = \infty$ = The improper integral diverges to positive infinity $\int_{a}^{\infty} f(x) , dx = -\infty$ = The improper integral diverges to negative infinity The limit does not exist = The improper integral diverges Signup and view all the answers

Match each part of the area under $f(x)$ over $[a,\infty)$ to it being bounded by $a$.

A(t) = The area under $f(x)$ and the x-axis over [a,t] $\lim_{t \to \infty} A(t)$ = The entire area over its entire range $\int_{a}^{t} f(x) , dx$ = The upper bound of area under $f(x)$ The integral does not exist = A(t) grows infinitely when t > a Signup and view all the answers

When using an improper integral to integrate a function $f(x)$, match each value to whether the improper integral has a net area.

Converges = Function has a net area. Diverges = Function does not have a net area. Evaluate A(t) for any t > a. = Area bounded by $f(x)$ and the x-axis over [a,t] Evaluate the area A(t) as $t \to \infty$ = Area bounded by $f(x)$ and the x-axis over $[a,\infty)$ Signup and view all the answers

Flashcards

Improper Integration

Integrals over intervals of infinite length or with vertical asymptotes.

Type 1 Improper Integral

A type of improper integral where a function f(x) is integrated over an interval of infinite length, such as [a, ∞), (-∞, a], or (-∞, ∞).

Convergent Improper Integral

When the limit defining the improper integral exists.

Divergent Improper Integral

When the limit defining the improper integral does not exist.

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Study Notes

Definite integrals calculated so far have been over intervals of finite length [a, b].
Improper integration calculates integrals over intervals of infinite length, such as [a, ∞) or (-∞, a].
Improper integration also calculates integrals of functions where the function has a vertical asymptote.

Type 1 Improper Integrals

Type 1 Improper Integrals arise when integrating a function f(x) over an interval of infinite length.
These integrals have the form [a,∞), (-∞, a], or (-∞,∞), where a ∈ R is a constant.
Integrals have the form ∫ₐ^∞ f(x) dx, ∫₋∞^a f(x) dx, or ∫₋∞^∞ f(x) dx.
It is not possible to calculate such integrals using the Fundamental Theorem of Calculus (FTC) because functions cannot be evaluated at infinity.

Definition 5.9.1: Type 1 Improper Integrals

Type 1 improper integral is an expression of the form ∫ₐ^∞ f(x) dx = lim(t→∞) ∫ₐ^t f(x) dx or ∫₋∞^a f(x) dx = lim(t→-∞) ∫ₜ^a f(x) dx, where f(x) is continuous on [a,∞) or (-∞, a].
If the limit defining the improper integral exists, the improper integral is said to converge (is convergent).
If the limit defining the improper integral does not exist, the integral is said to diverge (is divergent).
Definite integral ∫ₐ^b f(x) dx is the net area bounded by the x-axis and f(x) over [a, b].
∫ₐ^∞ f(x) dx is the net area bounded by f(x) and the x-axis over [a,∞).
To calculate, evaluate ∫ₐ^t f(x) dx for values of t > a.
A(t) represents the area bounded by f(x) and the x-axis over [a,t].
Net area over [a, ∞) can be found by taking the limit of A(t) as t approaches infinity.