AP Calculus AB/BC · Unit 6 · Topic 6.13 · BC Only
Evaluating Improper Integrals
Replace infinite bounds or unbounded integrands with limits and test convergence.
1. Topic Focus
Interpret definite integrals as accumulated change, connect sums to integrals, apply both Fundamental Theorems, and select antiderivative techniques.
This topic: Replace infinite bounds or unbounded integrands with limits and test convergence.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
The integral ∫₁^∞ 1/x² dx converges to 1, while ∫₁^∞ 1/x dx diverges.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. Improper integrals are limit problems
An ordinary definite integral assumes a finite interval and a bounded integrand. If either condition fails, the integral is improper. Its notation is shorthand for one or more limits of proper definite integrals. The integral converges only when every required limit exists as a finite real number.
2. Recognize the two sources of impropriety
- Infinite interval: at least one bound is \(\infty\) or \(-\infty\).
- Unbounded integrand: the function grows without bound at an endpoint or at a point inside the interval.
Inspect both the bounds and the domain of the integrand before using the Fundamental Theorem of Calculus.
3. Infinity is not an endpoint value
The symbol \(\infty\) describes unbounded behavior; it is not a number that can be substituted into an antiderivative. Introduce a finite variable first, evaluate the proper integral, and only then take a limit.
4. Infinite upper bound
If the limit is finite, that limit is the value of the improper integral. If it is infinite or does not exist, the integral diverges.
5. Infinite lower bound
The replacement variable approaches \(-\infty\) from within the interval. The same finite-limit requirement applies.
6. Both bounds infinite
Choose any convenient finite number \(c\), usually \(0\), and test the two sides independently:
The whole integral converges only if both improper integrals converge. A single symmetric limit from \(-b\) to \(b\) is not the definition of ordinary convergence.
7. Unbounded at an endpoint
If \(f\) is unbounded as \(x\to a^+\), approach the endpoint from inside the interval:
For a singularity at \(b\), use \(t\to b^-\). The direction on the limit is part of the definition.
8. An interior singularity requires a split
If \(f\) is unbounded at \(x=c\) with \(a<c<b\), create two one-sided integrals:
Do not integrate across the singularity in one step. Each side must produce a finite limit.
9. Every improper piece must converge
A finite result on one side cannot rescue divergence on another. Once any required component diverges, the original integral diverges and the remaining pieces need not be combined.
10. The tail \(p\)-integral benchmark
For \(p>1\), the tail decays fast enough to have finite accumulated area. For \(p\le 1\), it diverges.
11. The near-zero \(p\)-integral benchmark
The inequality reverses because the question is now how sharply the function blows up near zero, not how quickly it decays at infinity.
12. Why \(p=1\) is the boundary
Both \(\int_1^\infty 1/x\,dx\) and \(\int_0^1 1/x\,dx\) produce logarithms whose magnitudes grow without bound. Memorize the two \(p\)-rules, but connect their common boundary to logarithmic divergence.
13. A reliable evaluation workflow
- Locate every infinite bound and every point where the integrand is unbounded.
- Split the interval at each improper point.
- Write each piece as a correctly directed limit.
- Evaluate the finite definite integral before taking the limit.
- State converges to ... or diverges, with the limiting evidence.
14. Fast decay can produce finite accumulation
An interval may have infinite length while its integral is finite. Negative exponential functions and powers \(x^{-p}\) with \(p>1\) are standard examples because their values decrease rapidly enough.
15. A graph guides setup, not proof
A graph can reveal an infinite tail, a vertical asymptote, or a hidden domain break. It also helps check whether a positive-area answer should be finite or infinite. The convergence conclusion, however, must come from the required limits.
16. Symmetric cancellation is not ordinary convergence
For an odd function, \(\int_{-b}^{b}f(x)\,dx\) may equal zero for every \(b\). The improper integral from \(-\infty\) to \(\infty\) still diverges if either one-sided integral diverges. Such symmetric cancellation is a different idea called a principal value.
17. Keep signs and area interpretations separate
An improper integral is signed accumulation. If \(f\ge0\), convergence means the unbounded region has finite area. If \(f\) changes sign, a finite integral is a net value; ordinary convergence still requires all separately defined improper pieces to converge.
18. Final error check
Do not write \(F(\infty)\), ignore a vertical asymptote, combine \(\infty-\infty\), or stop after finding an antiderivative. The limits are the calculation, and a complete answer names convergence or divergence explicitly.
6. Detailed Worked Example and Error Check
Example 1: A convergent infinite tail.
The limit is finite, so the integral converges to \(1/8\).
Example 2: A logarithmically divergent tail.
The integral diverges. The curve approaches zero, but approaching zero alone does not guarantee finite accumulated area.
Example 3: A convergent endpoint singularity.
Although the integrand is unbounded at zero, the improper integral converges.
Example 4: A divergent endpoint singularity.
Here \(p=3/2>1\), so the near-zero \(p\)-integral rule also predicts divergence.
Example 5: An interior vertical asymptote.
For \(\int_0^3 1/(x-1)^2\,dx\), split at \(x=1\). On the left,
That one divergent piece is enough to conclude that the original integral diverges.
Example 6: Both bounds infinite.
Each value follows from a separate one-sided limit of \(\arctan x\); therefore the entire integral converges.
Example 7: Exponential decay on an infinite interval.
The positive area is finite because \(e^{-2b}\to0\).
7. AP Reasoning Routine
Identify the accumulating quantity and units, preserve bounds, choose a valid integration technique, and check answers by differentiation.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Evaluate each improper integral or state that it diverges. Show every required limit.
(a) \(\int_1^\infty x^{-4}\,dx\).
(b) \(\int_4^\infty x^{-1/2}\,dx\).
(c) \(\int_0^8 x^{-2/3}\,dx\).
(d) \(\int_0^1 x^{-5/4}\,dx\).
(e) \(\int_{-\infty}^{0}e^{3x}\,dx\).
(f) \(\int_{-\infty}^{\infty}e^{-|x|}\,dx\).
(g) \(\int_{-1}^{2}\frac{1}{(x-1)^2}\,dx\).
(h) \(\int_0^2\frac{1}{\sqrt{2-x}}\,dx\).
(i) Explain why \(\int_{-\infty}^{\infty}x\,dx\) is not zero even though \(\int_{-b}^{b}x\,dx=0\).
(j) Give the convergence conditions for \(\int_1^\infty x^{-p}\,dx\) and \(\int_0^1x^{-p}\,dx\), and explain the role of \(p=1\).
Check the solution
(a) \(\lim\limits_{b\to\infty}[-1/(3x^3)]_1^b=1/3\), so it converges to \(1/3\).
(b) \(\lim\limits_{b\to\infty}[2\sqrt{x}]_4^b=\lim\limits_{b\to\infty}(2\sqrt b-4)=\infty\); it diverges.
(c) \(\lim\limits_{a\to0^+}[3x^{1/3}]_a^8=6\), so it converges to \(6\).
(d) \(\lim\limits_{a\to0^+}[-4x^{-1/4}]_a^1=\lim\limits_{a\to0^+}(-4+4a^{-1/4})=\infty\); it diverges.
(e) \(\lim\limits_{a\to-\infty}[e^{3x}/3]_a^0=1/3\), so it converges to \(1/3\).
(f) Split at zero. Each half equals \(1\): on the left integrate \(e^x\), and on the right integrate \(e^{-x}\). Both converge, so the total is \(2\).
(g) The integrand is unbounded at the interior point \(x=1\). The left-hand integral already tends to \(\infty\), so the original integral diverges.
(h) \(\lim\limits_{b\to2^-}[-2\sqrt{2-x}]_0^b=2\sqrt2\), so it converges to \(2\sqrt2\).
(i) Ordinary convergence requires separate integrals on \(( -\infty,0]\) and \([0,\infty)\). They diverge to \(-\infty\) and \(\infty\), respectively, so their values cannot be canceled. The symmetric value zero is only a principal value.
(j) The tail integral converges for \(p>1\); the near-zero integral converges for \(p<1\). At \(p=1\), both reduce to logarithmic limits and diverge.