PAU Mathematics II (Galicia) study materials
PAU Mathematics II (Galicia) study resources

PAU Mathematics II (Galicia): Integral calculus and applications to area and volume

PAU Mathematics II (Galicia) · Work through your exam, one topic at a time.

All guide sections

Integral calculus and applications to area and volume

Integral calculus and applications to area and volume

Each option usually contains one integral question: a direct antiderivative (rational function, substitution, integration by parts) or an applied area or volume of revolution.

Core rules

  • Barrow's rule: the definite integral from a to b of f is F(b) - F(a) for any antiderivative F. The constant of integration is irrelevant in the definite case.
  • Techniques: substitution when the integrand contains g(x) times g'(x); integration by parts with the integral of u dv equal to uv minus the integral of v du, choosing u to become simpler when differentiated (ln x, inverse trigonometric functions, polynomials) and dv as the rest; partial fractions for rational functions, treating distinct linear, repeated linear and irreducible quadratic factors separately.
  • Area between curves: find the intersection points, then integrate (upper curve minus lower curve) over each subinterval. If the curves cross inside the interval, split it, since the integral of f - g can cancel. Integrating with respect to y, use right minus left.
  • Volume of revolution: about the x-axis, V = pi times the integral of f(x)^2; between two curves, pi times the integral of (f^2 - g^2); about the y-axis by shells, 2 pi times the integral of x f(x).
  • Improper integrals are defined as limits of proper integrals and converge only when that limit is finite.

Worked example

Find the area enclosed by y = x^2 and y = x + 2. The intersections satisfy x^2 = x + 2, so x = -1 and x = 2; on that interval the line lies above the parabola. The area is the integral from -1 to 2 of (x + 2 - x^2) dx, whose antiderivative is x^2/2 + 2x - x^3/3. Evaluating gives 10/3 at x = 2 and -7/6 at x = -1, so the area is 10/3 + 7/6 = 9/2.

Traps that cost marks

Not locating the intersection points; integrating f - g across a crossing point and reporting a near-zero area; losing signs when evaluating at a negative limit; omitting the factor pi in a volume of revolution; mixing the disk and washer formulas when a gap exists between the curve and the axis; and treating an improper integral as if the endpoint could simply be substituted.