Integral Calculus
What the exam tests
The paper tests standard indefinite integrals, integration by substitution and by parts, evaluation of definite integrals, and simple applications such as finding total cost from marginal cost.
Core rules
- ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1; ∫(1/x) dx = log|x| + C.
- ∫eˣ dx = eˣ + C; ∫(1/(a+x)) dx = log|a+x| + C.
- By parts: ∫u dv = uv − ∫v du, choosing u by the LIATE order (logarithmic, inverse, algebraic, trigonometric, exponential).
- Definite integral: ∫ₐᵇ f(x) dx = F(b) − F(a); the constant C cancels.
- For substitution, change the limits when the variable is changed.
Worked example
Evaluate ∫x eˣ dx. Take u = x (algebraic) and dv = eˣ dx, so du = dx and v = eˣ. ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.
Evaluate ∫₀¹ (3x² + 2x) dx. Antiderivative: x³ + x². Then F(1) − F(0) = (1 + 1) − 0 = 2.
Application: if marginal cost is MC = 6x + 5 and fixed cost is 100, total cost C(x) = ∫(6x + 5)dx = 3x² + 5x + 100.
Traps
- Dropping + C in indefinite integrals.
- Choosing u and dv badly in by parts, which makes the new integral harder.
- Forgetting to change limits after substitution in a definite integral.
- Sign errors when subtracting F(a) from F(b).
- Integrating a product term-by-term instead of using by parts.