Integration reverses differentiation: \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + c\) for \(n \neq -1\). Definite integrals give signed areas - regions below the axis count negative - and the Fundamental Theorem of Calculus links \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\). The trapezium rule and its over/under-estimate are examined every year.
Contents: Indefinite and definite integrals (MM7.2, MM7.4)Areas and the trapezium rule (MM7.1, MM7.3, MM7.5)Must-knowTrapsReal questions
Indefinite and definite integrals (MM7.2, MM7.4)
- Rewrite before integrating: \(\dfrac{3x^2 - 5}{\sqrt x} = 3x^{3/2} - 5x^{-1/2}\).
- \(\int_a^b f(x)\,dx = F(b) - F(a)\); \(\int_b^a = -\int_a^b\); \(\int_a^b + \int_b^c = \int_a^c\); \(\int (f \pm g) = \int f \pm \int g\); constants come out.
- \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\) (FTC): if \(F(x) = \int_2^x (t^2 - 3t)\,dt\) then \(F'(x) = x^2 - 3x\).
- Odd functions integrate to \(0\) over symmetric intervals; even functions give twice the integral from \(0\).
- \(\frac{dy}{dx} = f(x)\) with a point on the curve: integrate, then find \(c\).
Areas and the trapezium rule (MM7.1, MM7.3, MM7.5)
- Area between a curve and the axis: split at the roots and add the absolute values of the pieces.
- Area between two curves: \(\int (\text{upper} - \text{lower})\,dx\) between the intersection points.
- Trapezium rule with strips of width \(h\): \(\frac h2\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right]\).
- Overestimate when the curve is convex (bends upwards, chords above the curve, e.g. \(x^2\), \(2^x\)); underestimate when concave (\(\sqrt x\), \(\log x\), \(-x^2\)).
- Symmetry shortcuts: \(y = x(x - 1)(x - 2)\) is odd about \((1, 0)\), so the two enclosed regions are equal.
Must-know (no formula booklet)
- \(\int x^n dx = \frac{x^{n+1}}{n+1} + c\)
- \(\int_a^b f = F(b) - F(a)\)
- \(\frac{d}{dx}\int_a^x f(t)dt = f(x)\)
- trapezium: \(\frac h2[y_0 + 2(\ldots) + y_n]\)
- convex \(\Rightarrow\) overestimate
Classic traps
- A definite integral can be negative or zero while the area is positive.
- \(\int (x + \frac1x)^2 dx\): expand first; you cannot integrate a bracket by raising the power.
- \(\int_{-1}^{1} x^{-2}dx\) is not \(-2\): the integrand is undefined at \(0\), inside the interval.
Worked example
Q. Area between \(y = x^2\) and \(y = 2x + 3\).
A. Meet at \(x = -1, 3\). \(\int_{-1}^{3}(2x + 3 - x^2)\,dx = \frac{32}{3}\).