Math: Improper Integrals and Convergence Tests
Evaluating improper integrals and deciding convergence
Math: Improper Integrals and Convergence Tests
Evaluating improper integrals and deciding convergence
Math - Grade 9-12
- 1
Evaluate the improper integral ∫ from 1 to infinity of 1/x^2 dx, or state that it diverges.
Use the antiderivative of x^-2.
The integral converges. Writing it as lim as b approaches infinity of ∫ from 1 to b x^-2 dx gives lim as b approaches infinity of [-1/x] from 1 to b, which is lim as b approaches infinity of 1 - 1/b = 1. - 2
Determine whether ∫ from 1 to infinity of 1/x dx converges or diverges. Explain your reasoning.
The antiderivative of 1/x is ln x.
The integral diverges. Writing it as lim as b approaches infinity of ∫ from 1 to b 1/x dx gives lim as b approaches infinity of ln b - ln 1, which grows without bound. - 3
Evaluate the improper integral ∫ from 0 to 1 of 1/sqrt(x) dx, or state that it diverges.
The problem is improper because the function is undefined at x = 0.
The integral converges. Writing it as lim as a approaches 0 from the right of ∫ from a to 1 x^-1/2 dx gives lim as a approaches 0 from the right of [2sqrt(x)] from a to 1, which equals 2. - 4
Determine whether ∫ from 0 to 1 of 1/x^2 dx converges or diverges. Explain your reasoning.
The integral diverges. Writing it as lim as a approaches 0 from the right of ∫ from a to 1 x^-2 dx gives lim as a approaches 0 from the right of [-1/x] from a to 1 = lim as a approaches 0 from the right of -1 + 1/a, which grows without bound. - 5
Use the p-integral test to determine whether ∫ from 1 to infinity of 1/x^p dx converges when p = 3/2. Do not compute the exact value unless needed.
For ∫ from 1 to infinity 1/x^p dx, compare p to 1.
The integral converges because it is a p-integral on [1, infinity) with p = 3/2, and p-integrals of the form ∫ from 1 to infinity 1/x^p dx converge when p > 1. - 6
Use the p-integral test to determine whether ∫ from 1 to infinity of 1/x^0.8 dx converges or diverges.
The integral diverges because it is a p-integral on [1, infinity) with p = 0.8, and p-integrals of the form ∫ from 1 to infinity 1/x^p dx diverge when p is less than or equal to 1. - 7
Determine whether ∫ from 2 to infinity of 5/(x^2 + 1) dx converges or diverges. Use comparison with a simpler function.
Compare the denominator x^2 + 1 to x^2.
The integral converges. For x greater than or equal to 2, x^2 + 1 is greater than x^2, so 5/(x^2 + 1) is less than 5/x^2. Since ∫ from 2 to infinity 5/x^2 dx converges, the given integral converges by direct comparison. - 8
Determine whether ∫ from 1 to infinity of x/(x^2 + 1) dx converges or diverges.
The integral diverges. An antiderivative is (1/2)ln(x^2 + 1), so the improper integral becomes lim as b approaches infinity of (1/2)ln(b^2 + 1) - (1/2)ln 2, which grows without bound. - 9
Use limit comparison to determine whether ∫ from 1 to infinity of (3x + 2)/(x^2 + 4) dx converges or diverges.
For large x, the expression behaves like 3x/x^2.
The integral diverges. Compare the integrand to 1/x. The limit as x approaches infinity of [(3x + 2)/(x^2 + 4)] divided by [1/x] is the limit of x(3x + 2)/(x^2 + 4), which equals 3. Since ∫ from 1 to infinity 1/x dx diverges and the limit is a positive finite number, the given integral diverges. - 10
Evaluate ∫ from negative infinity to 0 of e^x dx, or state that it diverges.
As x approaches negative infinity, e^x approaches 0.
The integral converges. Writing it as lim as a approaches negative infinity of ∫ from a to 0 e^x dx gives lim as a approaches negative infinity of [e^x] from a to 0, which is 1 - 0 = 1. - 11
Determine whether ∫ from 0 to infinity of e^-x dx converges, and find its value if it converges.
The integral converges. Writing it as lim as b approaches infinity of ∫ from 0 to b e^-x dx gives lim as b approaches infinity of [-e^-x] from 0 to b, which is lim as b approaches infinity of 1 - e^-b = 1. - 12
An improper integral is written as ∫ from 0 to 3 of 1/(x - 2)^2 dx. Explain why it must be split before testing convergence, then determine whether it converges or diverges.
An infinite discontinuity inside the interval creates two separate improper integrals.
The integral must be split because the integrand has a vertical asymptote at x = 2, which lies inside the interval from 0 to 3. It must be considered as ∫ from 0 to 2 of 1/(x - 2)^2 dx plus ∫ from 2 to 3 of 1/(x - 2)^2 dx. At least one side diverges because the function behaves like 1/u^2 near u = 0, so the original improper integral diverges.