Change of Variables

Stretch and squeeze the axis to rewrite an integral. Two different substitutions land on the same value.

Change the scale and the endpoints move along. Areas need correcting by the stretch factor. After the correction, a hard integral turns easy.

For example, measure with a 📏 ruler whose marks are stretched twofold and every reading doubles. Halving the reading corrects it back to the true length.

∫abf(φ(x))φ′(x) dx=∫φ(a)φ(b)f(u) du\int_a^b f(\varphi(x))\varphi'(x)\,dx = \int_{\varphi(a)}^{\varphi(b)} f(u)\,du
∫012xex2 dx=∫01eu du=e−1\int_0^1 2x e^{x^2}\,dx = \int_0^1 e^u\,du = e - 1

Examples & Exercises

Prove the substitution rule completely: if φ\varphi is C1C^1 and monotone on [a,b][a,b] and ff is continuous, then ∫abf(φ(x)) φ′(x) dx=∫φ(a)φ(b)f(u) du\int_a^b f(\varphi(x))\,\varphi'(x)\,dx = \int_{\varphi(a)}^{\varphi(b)} f(u)\,du.

Take an antiderivative FF with F′=fF' = f and apply the chain rule plus the fundamental theorem to F∘φF \circ \varphi.

∫abf(φ(x)) φ′(x) dx=∫φ(a)φ(b)f(u) du\int_a^b f(\varphi(x))\,\varphi'(x)\,dx = \int_{\varphi(a)}^{\varphi(b)} f(u)\,du
ddx[F(φ(x))]=F′(φ(x)) φ′(x)=f(φ(x)) φ′(x)\frac{d}{dx}\bigl[F(\varphi(x))\bigr] = F'(\varphi(x))\,\varphi'(x) = f(\varphi(x))\,\varphi'(x)
u=x2u = x^2 (N=50N = 50): original ≈1.7182820013\approx 1.7182820013, substituted ≈1.7182818300\approx 1.7182818300 — both are e−1e - 1. The proof below is what guarantees this match.
Show proof

  1. Since ff is continuous on an interval containing φ([a,b])\varphi([a,b]), an antiderivative FF with F′=fF' = f exists (a continuous function has one via integration). The composition H(x)=F(φ(x))H(x) = F(\varphi(x)) is C1C^1, and the chain rule gives H′(x)=F′(φ(x)) φ′(x)=f(φ(x)) φ′(x)H'(x) = F'(\varphi(x))\,\varphi'(x) = f(\varphi(x))\,\varphi'(x) — exactly the pre-substitution integrand.

  2. H′H' is continuous, hence integrable on [a,b][a,b], and the FTC gives ∫abf(φ(x)) φ′(x) dx=∫abH′(x) dx=H(b)−H(a)=F(φ(b))−F(φ(a))\int_a^b f(\varphi(x))\,\varphi'(x)\,dx = \int_a^b H'(x)\,dx = H(b) - H(a) = F(\varphi(b)) - F(\varphi(a)).

  3. Meanwhile the same FTC applied directly to ff gives ∫φ(a)φ(b)f(u) du=F(φ(b))−F(φ(a))\int_{\varphi(a)}^{\varphi(b)} f(u)\,du = F(\varphi(b)) - F(\varphi(a)) (monotonicity of φ\varphi guarantees φ(a)≤φ(b)\varphi(a) \le \varphi(b) or the reverse, so the interval is well-defined).

    Both sides agree, proving the substitution rule.

    Complete. (Tao 11.10.5–11.10.7.)

Compute ∫012x ex2 dx=e−1\int_0^1 2x\,e^{x^2}\,dx = e - 1 completely via the substitution u=x2u = x^2: determine dudu, transform the limits, and write out the antiderivative — no skipped steps.

∫012x ex2 dx=∫01eu du=[eu]01=e−1\int_0^1 2x\,e^{x^2}\,dx = \int_0^1 e^u\,du = \bigl[e^u\bigr]_0^1 = e - 1
N=50N = 50: Simpson's original integral ≈1.7182820013\approx 1.7182820013 matches e−1≈1.7182818285e - 1 \approx 1.7182818285 within 1.8×10−71.8 \times 10^{-7}.
Show proof

  1. φ(x)=x2\varphi(x) = x^2 is C1C^1 and increasing on [0,1][0,1], with φ′(x)=2x\varphi'(x) = 2x. The integrand 2x ex2=eφ(x) φ′(x)2x\,e^{x^2} = e^{\varphi(x)}\,\varphi'(x) is exactly in substitution form (with f(u)=euf(u) = e^u).

  2. With u=x2u = x^2 we get du=2x dxdu = 2x\,dx, and the limits transform as x=0↦u=0x = 0 \mapsto u = 0, x=1↦u=1x = 1 \mapsto u = 1. Hence ∫012x ex2 dx=∫01eu du\int_0^1 2x\,e^{x^2}\,dx = \int_0^1 e^u\,du.

  3. Since eue^u is its own antiderivative, ∫01eu du=[eu]01=e−1≈1.7182818285\int_0^1 e^u\,du = [e^u]_0^1 = e - 1 \approx 1.7182818285. Numerical check: with N=50N = 50 Simpson panels the original integral is 1.71828200131.7182820013 and the substituted one is 1.71828183001.7182818300 — both converge to e−1e - 1.

Verify explicitly how a decreasing φ\varphi flips the limits: apply φ(x)=−x\varphi(x) = -x on [−1,0][-1, 0] and check by direct computation that ∫−10e−x⋅(−1) dx=∫10eu du=−∫01eu du=1−e\int_{-1}^{0} e^{-x}\cdot(-1)\,dx = \int_{1}^{0} e^u\,du = -\int_0^1 e^u\,du = 1 - e.

∫−10e−x⋅(−1) dx=∫10eu du=−∫01eu du=1−e\int_{-1}^{0} e^{-x}\cdot(-1)\,dx = \int_{1}^{0} e^u\,du = -\int_0^1 e^u\,du = 1 - e
[e−x]−10=e0−e1=1−e\bigl[e^{-x}\bigr]_{-1}^{0} = e^0 - e^1 = 1 - e
The increasing substitution u=sin⁡xu = \sin x (N=50N = 50): [0,π/2]↦[0,1][0, \pi/2] \mapsto [0, 1] in order. The problem below is the decreasing case, where the order flips.
Show proof

  1. φ(x)=−x\varphi(x) = -x is C1C^1 and decreasing on [−1,0][-1, 0] with φ′(x)=−1\varphi'(x) = -1.

    Check the substitution rule's other hypothesis as well: f(u)=euf(u) = e^u is continuous on every interval.

    The substitution rule gives ∫−10f(φ(x)) φ′(x) dx=∫φ(−1)φ(0)f(u) du\int_{-1}^{0} f(\varphi(x))\,\varphi'(x)\,dx = \int_{\varphi(-1)}^{\varphi(0)} f(u)\,du; with φ(−1)=1\varphi(-1) = 1 and φ(0)=0\varphi(0) = 0 the right side is ∫10eu du=−∫01eu du=−(e−1)=1−e\int_{1}^{0} e^u\,du = -\int_0^1 e^u\,du = -(e - 1) = 1 - e — the orientation flipped.

  2. Verify by direct computation: in ∫−10e−x⋅(−1) dx\int_{-1}^{0} e^{-x}\cdot(-1)\,dx, −e−x-e^{-x} has antiderivative e−xe^{-x} (since ddxe−x=−e−x\frac{d}{dx}e^{-x} = -e^{-x}), so the integral is [e−x]−10=e0−e1=1−e[e^{-x}]_{-1}^{0} = e^0 - e^1 = 1 - e. Exactly what the substitution rule predicted.

11. Integration