← Integration methods

9 Definite integrals: you MUST change the limits

This is the single biggest trap in the exam. If you substitute u, the limits are no longer x-values. ✕

The trap

The limits on ∫₀² …dx mean x = 0 to x = 2. Once you switch to u, those numbers are meaningless. You must convert them into u-values using your substitution, or convert back to x before evaluating.

Two valid methods (pick one and be consistent)

  • Method A. Change the limits (recommended). Substitute each x-limit into u = g(x) to get new u-limits. Then evaluate entirely in u. No back-substitution needed.
  • Method B. Back-substitute first. Ignore limits, find the antiderivative in x, then apply the original x-limits.
  • Method A is faster and less error-prone. Method B is safer if you find limit-swapping confusing.
Worked: Method A ∫₀² 2x(x² + 1)³ dx Evaluate ∫₀² 2x(x² + 1)³ dx, changing the limits. Let u = x² + 1 , du = 2x dx . Change the limits: when x = 0, u = 0² + 1 = 1. When x = 2, u = 2² + 1 = 5. The integral becomes ∫₁⁵ u³ du . Note the limits are now 1 and 5, not 0 and 2. Integrate: [u⁴/4]₁⁵ . Evaluate: 5⁴/4 − 1⁴/4 = 625/4 − 1/4 = 624/4 . ✓ 156 Worked: same problem, Method B Same integral, back-substituting instead Evaluate ∫₀² 2x(x² + 1)³ dx without changing limits. Find the indefinite integral first: ∫2x(x²+1)³dx = (x²+1)⁴/4 + c . Now apply the ORIGINAL x-limits: [(x²+1)⁴/4]₀² . = (5)⁴/4 − (1)⁴/4 = 625/4 − 1/4 . ✓ 156: same answer, as it must be. Both methods are valid; never mix them. Level 4: definite with a constant fix ∫₀^(√π) x·sin(x²) dx Evaluate ∫₀^(√π) x sin(x²) dx. Let u = x² , du = 2x dx ⟹ x dx = ½du . Change limits: x = 0 → u = 0. x = √π → u = (√π)² = π. = ½ ∫₀^π sin u du = ½ [−cos u]₀^π . = ½ (−cos π − (−cos 0)) = ½ (−(−1) + 1) = ½(2) . ✓ 1

Practise just this