U-substitution — technique, change of limits for definite integrals

WHAT it is

WHY the formula is true (derivation from scratch)

We do not memorize the rule — we build it.

Start from the chain rule: $\frac{d}{dx}\,F(g(x)) = F'(g(x))\cdot g'(x).$

Integrate both sides with respect to $x$ . The left side is the antiderivative of a derivative, so it returns the original function: $\int \frac{d}{dx}\,F(g(x))\,dx = F(g(x)) + C.$

Therefore: $\int F'(g(x))\,g'(x)\,dx = F(g(x)) + C.$

Now rename: let $u = g(x)$ and let $f = F'$ (so $F$ is an antiderivative of $f$ ). The right side is $F(u)+C$ , which is exactly $\int f(u)\,du$ . Hence: $\boxed{\;\int f(g(x))\,g'(x)\,dx = \int f(u)\,du\;}$

HOW to do it — the recipe

Choose $u = g(x)$ : usually the inner function, the thing inside a power/root/exp/log/trig, OR the function whose derivative also appears.
Differentiate: compute $du = g'(x)\,dx$ .
Replace every $x$ -piece (including $dx$ ) so the integral is entirely in $u$ .
Integrate in $u$ .
Indefinite: substitute back $u=g(x)$ . Definite: change the limits (see below) and never substitute back.

Figure — U-substitution — technique, change of limits for definite integrals

Worked examples

Worked example Example 3 — DEFINITE integral, change the limits

Evaluate $\displaystyle \int_{0}^{2} x\sqrt{x^2+1}\,dx$ .

Choose $u=x^2+1$ , $du=2x\,dx\Rightarrow x\,dx=\tfrac12 du$ .
Change limits (Why? variable is now $u$ ):
- $x=0 \Rightarrow u=0^2+1=1$
- $x=2 \Rightarrow u=2^2+1=5$
Replace: $\displaystyle \int_{1}^{5}\sqrt{u}\cdot\tfrac12\,du = \tfrac12\int_1^5 u^{1/2}\,du$ .
Integrate: $\tfrac12\cdot\frac{u^{3/2}}{3/2}\Big|_1^5=\frac{1}{3}\,u^{3/2}\Big|_1^5$ .
Evaluate: $\frac13\big(5^{3/2}-1\big)=\boxed{\dfrac{5\sqrt5-1}{3}}\approx 3.39$ .

Why no back-substitution? Because we already translated the endpoints; the answer is a pure number.

Steel-manned mistakes

Flashcards

U-substitution reverses which differentiation rule?

The chain rule.

u=g(x)

, what is

du

du = g'(x)\,dx

General u-sub identity?

\int f(g(x))g'(x)\,dx=\int f(u)\,du

In a DEFINITE integral, what happens to the limits after substituting?

They change to

u

-values:

\int_{g(a)}^{g(b)} f(u)\,du

Why can't you do back-substitution AND change limits together?

They are two equivalent finishing methods; doing both double-translates the endpoints and gives a wrong answer.

You have

x\,dx

but

du=2x\,dx

— what do you write?

x\,dx=\tfrac12\,du

(fix the constant only).

Can you pull a stray variable like

1/x

outside the integral?

No — only constants can leave; solve

u=g(x)

for

x

instead.

For

\int_0^2 x\sqrt{x^2+1}\,dx

with

u=x^2+1

, the new limits are?

u:1\to5

First step of the u-sub recipe?

Choose

u=g(x)

— usually the inner function whose derivative is present.

Recall Feynman: explain to a 12-year-old

Imagine a long messy word that's actually a nickname for something simple. U-substitution is giving the messy inside part a short nickname "u". Suddenly the whole problem is short and easy. The rule " $du =$ (slope of u) $\times dx$ " is how we make sure the amount of stuff stays the same when we rename. And if the problem said "from point 0 to point 2 on the old name," we have to figure out what those two points are called in the new name — that's why the limits change.

Connections

Chain Rule — u-sub is its exact inverse.
Antiderivatives & Indefinite Integrals
Definite Integral & Fundamental Theorem of Calculus — justifies limit-changing.
Integration by Parts — the product-rule inverse (contrast with chain-rule inverse).
Trigonometric Substitution — a specialized u-sub.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, U-substitution basically chain rule ka ulta hai. Jab aap $F(g(x))$ ko differentiate karte ho to chain rule se andar wala function aur uska derivative dono aate hain. To jab integral me aisa pattern dikhe — ek function aur uske paas hi uska derivative — tab hum andar wale part ko ek chhota naam "u" de dete hain. Isse pura integral simple ho jaata hai. Rule yaad rakho: $u=g(x)$ choose karo, phir $du=g'(x)\,dx$ nikaalo, aur saare $x$ wale parts ko (including $dx$ ) $u$ me badal do.

Sabse important cheez definite integral me hoti hai — limits change karna. Jab variable $x$ se $u$ ban gaya, to purane numbers (jaise 0 se 2) ab matlab nahi rakhte, kyunki ab to variable $u$ hai. Isliye limits ko bhi translate karo: $x=0$ dalo $u=g(0)$ , $x=2$ dalo $u=g(2)$ . Example me $u=x^2+1$ tha to limits 0,2 se badal ke 1,5 ho gaye. Aur yaad rakho — agar limits change kar di to wapas $x$ me back-substitute karne ki zaroorat nahi.

Ek common galti: agar ek extra constant missing ho (jaise $du=2x\,dx$ par aapke paas sirf $x\,dx$ hai), to $x\,dx=\tfrac12 du$ likh do — constant adjust karna allowed hai. Lekin variable ko bahar nikalna (jaise $1/x$ ) bilkul galat hai, kyunki sirf constants hi integral ke bahar ja sakte hain. Yeh technique 80/20 wala hai — calculus ke aadhe se zyada integrals isi ek trick se ho jaate hain, to ise solid bana lo.

U-substitution — technique, change of limits for definite integrals

WHAT it is

WHY the formula is true (derivation from scratch)

HOW to do it — the recipe

Worked examples

Steel-manned mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections