Math · Calculus II · Worked example
Integrate across a vertical asymptote
Evaluate the integral of 1/√x from 0 to 1, and decide whether the integral of 1/x² from −1 to 1 converges.
Blow-up at an endpoint
1/√x is unbounded near 0, so integrate from t to 1 and let t → 0⁺. The integral from t to 1 is 2 − 2√t, which tends to 2.
Blow-up inside the interval
1/x² is unbounded at 0, inside [−1, 1], so split the integral there. The integral from t to 1 of dx/x² is 1/t − 1, which grows without bound as t → 0⁺.
Draw the conclusion
One piece diverges, so the whole integral from −1 to 1 diverges. The integrand is positive, so its area could never be negative anyway.
See the trap
Applying the Fundamental Theorem to −1/x from −1 to 1 gives −2, a negative answer for a positive integrand. The theorem needs f continuous on the whole interval, and 1/x² is not continuous at 0.
Result
The integral of 1/√x from 0 to 1 converges to 2; the integral of 1/x² from −1 to 1 diverges.
Your turn
Evaluate the integral of 1/∛x from 0 to 8.
Show the answer and explanation
6.
The integral from t to 8 of x^(−1/3) dx is (3/2)(8^(2/3) − t^(2/3)) = 6 − (3/2)t^(2/3), which tends to 6 as t → 0⁺.
Keep exploring
Limits & one-sided behavior checks 2 − 2√t as t → 0 from the right and verifies 2; change the function to 1/t − 1 to see it grow without bound.
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