Evaluation by Substitution: Explained with Examples

Substitution works for definite integrals too, but with one extra twist: the limits of integration must change along with the variable. When you set u = g(x), the interval [a, b] in x becomes [g(a), g(b)] in u. Handling the limits correctly — either by changing them or by back-substituting before evaluating — is the whole skill of this topic.

This matters because definite integrals are where calculus meets measurement: areas, volumes, work. A substitution that simplifies the integrand but leaves the limits in the wrong variable produces a wrong number, not just a wrong expression, so examiners test this step deliberately.

Key Concepts

• Change of limits: if u = g(x), then ∫ₐᵇ f(g(x))g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du.
• Two valid routes: (i) change the limits and never return to x, or (ii) keep x-limits and back-substitute before evaluating. Never mix them.
• Order of new limits: keep g(a) as the lower and g(b) as the upper limit, even if g(a) > g(b) — the sign works out automatically.
• dx must be replaced: write du = g′(x) dx and substitute for the whole dx factor.
• Common pairings: √(a² − x²) suggests x = a sin θ; x² + a² suggests x = a tan θ; √(x² − a²) suggests x = a sec θ.

Formulas

• ∫ₐᵇ f(g(x)) · g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du, u = g(x)
• ∫₀^{π/2} sinⁿx dx type integrals often simplify with x → π/2 − x (property route)
• ∫₀¹ x(x² + 1)³ dx = 15/8

Worked Example

Evaluate ∫₀¹ x(x² + 1)³ dx. Step 1: set u = x² + 1, so du = 2x dx, i.e. x dx = du/2. Step 2: change the limits: when x = 0, u = 1; when x = 1, u = 2. Step 3: rewrite: ∫₁² u³ · (du/2) = ½∫₁² u³ du. Step 4: integrate: ½·[u⁴/4]₁² = (1/8)(16 − 1) = 15/8. Checking by back-substitution instead: ½·[(x²+1)⁴/4]₀¹ gives the same 15/8, confirming the limit change was correct.

Common Misconceptions

• Changing variables but not limits: evaluating a u-antiderivative at x-limits is the single most common error in this topic.
• Changing limits AND back-substituting: do one or the other, never both — that double-counts the transformation.
• Reordering limits to look neat: if g(a) > g(b), leave them; swapping introduces a sign error unless you add the minus sign.

Real-World Applications

Definite substitution computes the work done by position-dependent forces, the arc length of parametric curves, and probabilities over transformed random variables. In engineering, it evaluates the energy in signals after a change of time scale. Any measurement that is easier in a new coordinate system uses exactly this move.

For definite integrals, substitution has three steps: new variable, new integrand, new limits. Change all three together, and the Fundamental Theorem finishes the job in the new variable without ever looking back.