An infinite integration endpoint is replaced by a finite bound and a limit. At a singular point inside the interval, split the integral and require both one-sided integrals to converge separately. Symmetric cancellation can define a Cauchy principal value but does not prove convergence of the ordinary improper integral. For 1/x across zero, the two sides diverge even though symmetric cutoffs cancel.
The integral of x^(−p) from 1 to infinity converges exactly when p>1. From zero to 1 it converges exactly when p<1. The same exponent behaves differently at the two boundaries. For positive integrands, comparison transfers convergence from a larger integrable function or divergence from a smaller nonintegrable one; keep the inequality direction correct. For an integral of a maximum or minimum, solve branch crossings and check which expression dominates each interval. For max(√(1−x²),x+1) on [−1,1], the interior crossing is zero; use the semicircle on [−1,0] and the line on [0,1]. The area is π/4+3/2, not the integral of either branch over the entire interval.
For rotation around the x-axis, disks or washers integrate π(R²−r²) dx. Cylindrical shells use 2π times radius times height and integrate in the matching variable. Select the method by the geometry and verify nonnegative radii. Area between curves integrates upper minus lower, splitting where their order changes. A signed integral is not automatically geometric area.
For a differentiable plane curve y=f(x), arc length is the integral of sqrt(1+(f′(x))²) dx. A surface formed by rotating a nonnegative f around the x-axis has area integral 2πf sqrt(1+(f′)²) dx. These are different quantities from volume. If an interval is unbounded, the geometric formula still needs a convergence test.
