Integration and Applications
Integration and Applications
Integration questions cover standard integrals, definite integrals and areas, the trapezium rule, and the techniques of substitution and integration by parts. Applications in this paper centre on area under a curve and between curves, with volume of revolution appearing as an extension.
Standard forms and techniques
∫xⁿ dx = x^(n+1)/(n + 1) + c for n ≠ −1; ∫x⁻¹ dx = ln|x| + c; ∫e^(kx) dx = e^(kx)/k + c; ∫sin kx dx = −(cos kx)/k + c; ∫cos kx dx = (sin kx)/k + c. For a product use parts: ∫u dv = uv − ∫v du, choosing u to simplify on differentiation (ln x, x, then trig or exponential). For a composite use substitution and change the limits as well as the variable.
Definite integrals and area
∫ from a to b of f(x) dx is the signed area. Where the curve crosses the axis, split the interval and add the magnitudes of the pieces, otherwise areas cancel. Area between two curves is ∫(upper − lower) dx with the intersection points as limits. Volume about the x-axis is π∫y² dx.
Trapezium rule
With n strips of width h = (b − a)/n, the estimate is (h/2)[y₀ + 2(y₁ + … + y_{n−1}) + yₙ]. It under-estimates a curve that is concave up over the interval and over-estimates one that is concave down.
Worked example
Evaluate ∫₁^e ln x dx. Use parts with u = ln x and dv = dx, so du = (1/x) dx and v = x. Then ∫ln x dx = x ln x − ∫x · (1/x) dx = x ln x − x. Applying the limits: (e ln e − e) − (1 ln 1 − 1) = (e − e) − (0 − 1) = 1. The value is positive and modest, consistent with ln x being small over most of [1, e].
Traps that cost marks
- Omitting the constant of integration for an indefinite integral.
- Forgetting to change limits when substituting in a definite integral.
- Subtracting areas across a root without splitting the interval.
- Sign errors from ∫sin kx dx, which is negative cosine.
- Mixing upper and lower functions in the bracket for area between curves.