Change of Variables

Stretch and squeeze the axis with u=φ(x)u = \varphi(x) and the limits follow — ∫abf(φ)φ′=∫φ(a)φ(b)f\int_a^b f(\varphi)\varphi' = \int_{\varphi(a)}^{\varphi(b)} f. Watch two substitutions collapse onto the same integral.

∫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. 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.

10. Integration