Integral Calculus and Areas Under Curves
This domain covers antiderivatives (immediate integrals, linearity, simple substitution), the definite integral and Barrow's rule, and the computation of areas: between a curve and the x-axis and between two curves. Exam items almost always ask for the area of a region defined by a parabola and a line, or by a curve and the axes.
Primitives you must recognise
- ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1; ∫x⁻¹ dx = ln|x| + C.
- ∫eˣ dx = eˣ + C, ∫k dx = kx + C.
- Linearity: ∫(a·f + b·g) = a∫f + b∫g. Expand polynomials before integrating rather than forcing substitutions.
- With a linear change of variable u = ax + b, du = a·dx, so the result is divided by a.
Definite integral and Barrow's rule
If F is any primitive of f, then ∫ from a to b of f(x)dx = F(b) − F(a). The constant C cancels, so it is not written in definite integrals. If you substitute, either change the limits to the new variable or undo the substitution before evaluating.
Areas
- Area between f and the x-axis on [a,b] is ∫|f|: locate the zeros of f, split the interval there, integrate each piece and add the absolute values.
- Area between two curves: solve f(x) = g(x) to get the limits a < b, then integrate |f − g|; on an interval where no crossing occurs, integrate the upper minus the lower.
Worked example
Area enclosed by y = x² and y = x + 2. Intersections: x² = x + 2 → x² − x − 2 = 0 → x = −1 and x = 2. On [−1,2] the line is above the parabola, so Area = ∫ from −1 to 2 of (x + 2 − x²)dx = [x²/2 + 2x − x³/3] from −1 to 2 = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 + 7/6 = 27/6 = 4.5 square units.
Traps that cost marks
- Integrating a region that lies below the axis and reporting a negative "area".
- Forgetting to split at the zeros of f, which gives a net value instead of a total area.
- Using the wrong limits because the intersections were not computed, or solving the intersection equation incorrectly.
- Subtracting the curves in the wrong order, or mishandling signs when a limit is negative.