Evaluate $\displaystyle\int_{-}$∞^∞ (1/(4+x²)) dx.
Infinite limits of integration, infinite discontinuities, and convergence/divergence.
53 questions
AP Learning Target(s): LT 13: Evaluating Improper Integrals (BC only)
Evaluate $\displaystyle\int_{-}$∞^∞ (1/(4+x²)) dx.
π/4
π
π/2
The integral diverges
For which of the following reasons is the integral $$\int_{0}^{1} \frac{1}{\sqrt{x}} dx$$ an improper integral?
Unbounded integrand at 0
Infinite upper limit
Infinite lower limit
Finite integrand
For which values of p does the improper integral $\displaystyle\int_{1}$^∞ 1/x^p dx converge?
p ≠ 1
p > 1
p < 1
p ≥ 1
The integral $$\displaystyle\int_{1}^{\infty} \frac{3}{\sqrt{x}} dx$$ is
$$3$$
$$\text{divergent}$$
$$\displaystyle\frac{3}{2}$$
$$6$$
The improper integral $$\displaystyle\int_{2}^{\infty} \frac{1}{x(\ln x)^{2}} dx$$ is
$$\displaystyle\frac{2}{\ln 2}$$
$$\ln 2$$
$$\displaystyle\frac{1}{\ln 2}$$
$$\text{divergent}$$
Which of the following is true about the improper integral $$\int_{2}^{\infty} \frac{1}{x} dx$$?
The integral converges to 2
The integral diverges
The integral converges to 0
The integral is finite
Evaluate the improper integral $$\int_{2}^{\infty} \frac{1}{x} dx$$.
Converges to 0
Converges to 1
Diverges
Converges to 2
Which statement accurately describes the improper integral $\displaystyle\int_{0}$^1 1/√(1-x²) dx?
Convergent with value 1
Divergent at both endpoints
Divergent at x = 1
Convergent with value π/2
Evaluate the improper integral $$\int_{1}^{\infty} \frac{1}{x^{3/2}} dx$$.
Converges to 1
Diverges
Converges to 2
Converges to 0
Which of the following statements about an improper integral with an infinite interval of integration is true?
The integral is always convergent.
The integral is always divergent.
The convergence or divergence depends on the function.
The convergence or divergence depends on the limit of integration.
Which of the following improper integrals converges?
$\displaystyle\int_{1}$^∞ x⁻¹ dx
$\displaystyle\int_{1}$^∞ x⁻¹(ln x)⁻¹ dx
$\displaystyle\int_{1}$^∞ x⁻³/² dx
$\displaystyle\int_{0}$^∞ xe^{-x^2} dx
Why is the integral $$\int_{1}^{\infty} \frac{1}{x^2} dx$$ considered an improper integral?
Infinite upper limit
Infinite lower limit
Unbounded integrand
Finite limits
For which of the following reasons is the integral $$\int_{-\infty}^{0} e^x dx$$ classified as an improper integral?
Finite integrand
Infinite lower limit
Finite limits
Unbounded integrand
Which of the following correctly describes the evaluation of the improper integral $$\int_{1}^{\infty} \frac{1}{x^2} dx$$?
Direct integration without limits
Limit as $$b \to \infty$$ of $$\int_{1}^{b} \frac{1}{x^2} dx$$
Substituting $$x = \tan(t)$$
Using partial fraction decomposition
An improper integral is described as divergent if:
the function is discontinuous.
the limit of integration is infinite.
the integrand approaches infinity.
the integrand approaches zero.
What is the value of $$\displaystyle\int_{0}^{2} \frac{1}{\sqrt{4-x^{2}}} dx$$?
$$\displaystyle\frac{\pi}{2}$$
$$\text{divergent}$$
$$2$$
$$\pi$$
The improper integral $$\displaystyle\int_{2}^{\infty} \frac{8}{(x-1)^{2}} dx$$ is
$$\text{divergent}$$
$$4$$
$$8$$
$$16$$
The integral $$\int_{1}^{\infty} \frac{6}{(x+3)^{3/2}} dx$$ is
$$\frac{3}{4}$$
$$\text{divergent}$$
$$3$$
$$6$$
Evaluate the integral $$\displaystyle\int_{0}^{1} \frac{\ln x}{x} dx$$.
$$-\displaystyle\frac{1}{2}$$
$$\displaystyle\frac{1}{2}$$
$$-1$$
$$\text{divergent}$$
The improper integral $$\int_{1}^{\infty} \frac{1}{x^2} dx$$
Converges to 2
Diverges
Converges to 0
Converges to 1
Which of the following is true for the improper integral $$\int_{0}^{1} \frac{1}{\sqrt{x}} dx$$?
Converges to 2
Diverges
Converges to 0
Converges to 1
The improper integral $$\displaystyle \int_{1}^{\infty} \frac{x^2}{(x^3 + 2)^2} dx$$ is
$$\displaystyle 1$$
$$\displaystyle -\frac{1}{9}$$
$$\displaystyle \frac{1}{3}$$
$$\displaystyle \frac{1}{9}$$
Which of the following describes the essential step for evaluating an improper integral with an infinite upper bound that distinguishes it from the evaluation of a standard definite integral?
It requires taking the limit as the upper bound approaches infinity
It necessitates the use of L'Hôpital's rule in all cases
It uses complex integration techniques not applicable to definite integrals
It always results in an infinite value
Let $$g$$ be a twice-differentiable function such that $$g(1) = 0.5$$ and $$\displaystyle \lim_{x \to \infty} g(x) = 4$$. The value of $$\displaystyle \int_{1}^{\infty} g'(x) dx$$ is
$$-3.5$$
$$3.5$$
nonexistent
$$45$$
Which of the following is an example of an improper integral?
$$\int_{1}^{2} \frac{1}{x} dx$$
$$\int_{0}^{\infty} e^{-x} dx$$
$$\int_{-1}^{1} x^3 dx$$
$$\int_{0}^{1} x^2 dx$$
The improper integral $\displaystyle\int_{0}$^π/2 tan x dx is defined as
lim[t→0⁺] $$\int$$ₜ^π/2 tan x dx
lim[t→(π/2)⁻] $\displaystyle\int_{0}$^t tan x dx
$\displaystyle\int_{0}$^π/2 sin x/cos x dx
π/4
Which of the following statements best describes a convergent improper integral?
The integral has an infinite value.
The integral oscillates between positive and negative values.
The integral approaches zero.
The integral has a finite value.
If the limit of the integrand of an improper integral is zero at the point of impropriety, then the integral:
diverges
is always zero.
depends on the function.
converges
If an improper integral has an infinite limit of integration and a positive, decreasing integrand, the integral:
converges.
oscillates between positive and negative values.
depends on the function.
diverges
The evaluation of an improper integral with an infinite upper bound differs from that of a standard definite integral because the improper integral:
It always results in an infinite value
It involves integrating the reciprocal of the function
It uses complex numbers instead of real numbers
It requires taking the limit as the upper bound approaches infinity
$$\displaystyle \int_{1}^{\infty} xe^{-x^2} dx =$$
$$\frac{1}{e}$$
$$-\frac{1}{e}$$
$$\frac{2}{e}$$
$$\frac{1}{2e}$$
The improper integral $$\displaystyle\int_{1}^{\infty} \frac{4}{x^{3}} dx$$ is
$$2$$
$$\text{divergent}$$
$$1$$
$$4$$
Which of the following describes the evaluation of the improper integral $$$\int$^∞_1 \frac{1}{x^2} dx$$ and its convergence?
Substitution method; oscillates indefinitely
Comparison test; conditionally converges
Limit of antiderivative as upper bound approaches infinity; converges to a finite value
Integration by parts; diverges to infinity
Evaluate $$\int_{1}^{3} \frac{2x^3 - 3x^2 + 4x - 7}{x^2 - 1} dx$$.
The integral diverges.
$$\ln 8 - 3\ln 4$$
$$\ln 8 - 5\ln 4 + 2$$
$$\ln 8 + 2\ln 2$$
The improper integral $\displaystyle\int_{0}$^π/2 tan(x) dx is
Divergent due to a vertical asymptote at x = 0
Divergent due to a vertical asymptote at x = π/2
Convergent with value ln(2)
Convergent with value π/4
What is the value of the improper integral $$\int_{0}^{\infty} e^{-x} dx$$?
Undefined
1
Infinity
0
What is the value of $$\int_{0}^{1} \frac{2x}{x^2 - 1} dx$$?
$$\frac{1}{2}$$
$$\ln(2)$$
$$\ln(3)$$
$$0$$
Evaluate the integral $\displaystyle\int_{1}^{2}$ (4x-3)/(x²-4x+3) dx.
4ln(2) - 3ln(3/2)
The integral diverges.
ln(3/2)
2ln(2) - ln(3)
The improper integral $$\displaystyle\int_{-\infty}^{\infty} \frac{2}{x^{2}+4} dx$$ is
$$\pi$$
$$\text{divergent}$$
$$\displaystyle\frac{\pi}{2}$$
$$2\pi$$
The improper integral $$\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}+1} dx$$ is
$$\pi$$
$$\text{divergent}$$
$$\displaystyle\frac{\pi}{4}$$
$$\displaystyle\frac{\pi}{2}$$
The integral $$\displaystyle\int_{0}^{1} \frac{5}{x^{1/3}} dx$$ is
$$\displaystyle\frac{5}{2}$$
$$\displaystyle\frac{15}{2}$$
$$15$$
$$\text{divergent}$$
The improper integral $$\displaystyle\int_{0}^{\infty} xe^{-2x} dx$$ is
$$1$$
$$\displaystyle\frac{1}{2}$$
$$\displaystyle\frac{1}{4}$$
$$\text{divergent}$$
Evaluate the integral $$\int_{0}^{1} \frac{3x-1}{x^2-x} dx$$.
ln(2) - 3
3ln(2) - 1
The integral is improper and diverges.
ln(2)
The value of the improper integral $\displaystyle\int_{0}$^∞ xe^(-x²) dx is
1/2
The integral diverges
√π/2
1
What is the value of the improper integral $\displaystyle\int_{0}$^∞ x²e^(-x³) dx?
Divergent
1
∞
1/3
Which of the following describes the improper integral $\displaystyle\int_{0}$^π tan(x) dx?
Equal to 0 by symmetry
Divergent due to a vertical asymptote at x = π/2
Divergent as x approaches π
Convergent with value ln(2)
Which statement is true regarding an improper integral that has an infinite discontinuity?
The convergence or divergence depends on the interval of integration.
The integral is always divergent.
The convergence or divergence depends on the function.
The integral is always convergent.
For what values of p is the improper integral $\displaystyle\int_{1}$^∞ 1/x^p dx convergent?
p < 1
p > 1
p ≠ 1
p ≥ 1
Which of the following best describes the purpose of using limits when evaluating improper integrals?
To apply the mean value theorem
To find the derivative
To simplify the integrand
To handle infinite or undefined bounds
The table shows values for right Riemann sum approximations of the integral $$\displaystyle \int_{0}^{1} \frac{1}{x} dx$$ using $$n$$ subintervals of equal width. Which of the following statements best describes the limit of the Riemann sums as $$n$$ approaches infinity?
| $$n$$ | $$\displaystyle \sum_{k=1}^{n} \left(\frac{1}{x_k}\right)\left(\frac{1}{n}\right)$$ |
|---|---|
| 100 | 5.19 |
| 200 | 5.88 |
| 300 | 6.28 |
| 400 | 6.57 |
| 500 | 6.79 |
The limit of the Riemann sums does not exist because $$\displaystyle \int_{0}^{1} \frac{1}{x} dx$$ does not exist.
The limit of the Riemann sums is a finite number less than 10.
The limit of the Riemann sums is a finite number greater than 10.
The limit of the Riemann sums does not exist because it is a sum of infinitely many positive numbers.
Which of the following is the most appropriate technique for determining whether the improper integral $\displaystyle\int_{0}$^∞ xe^(-x²) dx converges?
Comparison test with 1/x²
Substitution followed by limit evaluation
Integration by parts
Limit comparison test with e^(-x)
Which of the following identifies the primary theorem used for evaluating improper integrals and correctly describes its application in that context?
The Mean Value Theorem, applied to infinite intervals
The Intermediate Value Theorem, extended to unbounded functions
The Fundamental Theorem of Calculus, applied by taking limits of antiderivatives
L'Hôpital's Rule, used to evaluate discontinuities at endpoints
Which type of improper integral is evaluated using a limit as the upper bound of integration approaches infinity?
Both Type I and Type II improper integrals.
Type I improper integral.
Neither Type I nor Type II improper integrals.
Type II improper integral.