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u-Substitution (Definite Integrals)

MATH161

Reference: Stewart 4.5 • Chapter: 4 • Section: 5

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u-Substitution (Definite Integrals)

The Key Difference: Changing the Limits

When applying substitution to a definite integral $\int_a^b f(g(x))g'(x)\,dx$, you have two choices:

Method 1: Evaluate the indefinite integral first, then apply the limits
Method 2: Change the limits of integration when you substitute (usually better!)

Method 2 is typically more efficient because you never need to back-substitute—you evaluate directly in terms of $u$.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart 4.5
Course	MATH162
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Substitution Rule for Definite Integrals

Theorem: If $g'$ is continuous on $[a, b]$ and $f$ is continuous on the range of $u = g(x)$, then

$$\boxed{\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du}$$

Why Change the Limits?

When $u = g(x)$:

At the lower limit: when $x = a$, we have $u = g(a)$
At the upper limit: when $x = b$, we have $u = g(b)$

The limits transform along with the integrand. You're essentially moving from the "$x$-world" to the "$u$-world" completely.

Method 1 vs Method 2

Method 1: Evaluate Indefinite First

∫₀⁴ √(2x+1) dx → Find ∫√(2x+1) dx = ⅓(2x+1)^(3/2) + C
                → Evaluate: [⅓(2x+1)^(3/2)]₀⁴
                → = ⅓(9)^(3/2) - ⅓(1)^(3/2) = 9 - ⅓ = 26/3

Method 2: Change Limits (Preferred)

∫₀⁴ √(2x+1) dx, let u = 2x+1, du = 2dx
When x=0: u=1    When x=4: u=9
→ ∫₁⁹ √u · ½ du = ½ · ⅔u^(3/2)|₁⁹
→ = ⅓(27 - 1) = 26/3  ✓ No back-substitution needed!

The Procedure (Method 2)

Step 1: Choose $u = g(x)$ and find $du$.

Step 2: Change the limits:

Lower limit: $u_{\text{lower}} = g(a)$
Upper limit: $u_{\text{upper}} = g(b)$

Step 3: Substitute everything (integrand AND limits).

Step 4: Evaluate the integral in $u$. Do NOT back-substitute.

Critical Warning: Don't Mix Limits

After changing to $u$, the integral is entirely in the "$u$-world":

$$\int_1^9 \sqrt{u}\,du$$

The limits 1 and 9 are $u$-values, not $x$-values. You evaluate directly: $$\left[\frac{2}{3}u^{3/2}\right]_1^9 = \frac{2}{3}(27) - \frac{2}{3}(1) = \frac{52}{3}$$

Do not substitute $u = 2x + 1$ back in at this stage.

Worked Examples

Example 1: Basic Limit Change

Evaluate $\int_0^4 \sqrt{2x + 1}\,dx$.

Step 1: Let $u = 2x + 1$, so $du = 2\,dx$, meaning $dx = \frac{1}{2}du$.

Step 2: Change limits:

When $x = 0$: $u = 2(0) + 1 = 1$
When $x = 4$: $u = 2(4) + 1 = 9$

Step 3-4: Substitute and evaluate: $$\int_0^4 \sqrt{2x + 1}\,dx = \int_1^9 \sqrt{u} \cdot \frac{1}{2}\,du = \frac{1}{2}\int_1^9 u^{1/2}\,du$$

$$= \frac{1}{2} \cdot \frac{2}{3}\left[u^{3/2}\right]_1^9 = \frac{1}{3}(9^{3/2} - 1^{3/2}) = \frac{1}{3}(27 - 1) = \frac{26}{3}$$

Example 2: Limits May Reverse

Evaluate $\int_1^2 \frac{dx}{(3 - 5x)^2}$.

Step 1: Let $u = 3 - 5x$, so $du = -5\,dx$, meaning $dx = -\frac{1}{5}du$.

Step 2: Change limits:

When $x = 1$: $u = 3 - 5(1) = -2$
When $x = 2$: $u = 3 - 5(2) = -7$

Step 3-4: Substitute and evaluate: $$\int_1^2 \frac{dx}{(3 - 5x)^2} = \int_{-2}^{-7} \frac{1}{u^2} \cdot \left(-\frac{1}{5}\right)\,du = -\frac{1}{5}\int_{-2}^{-7} u^{-2}\,du$$

$$= -\frac{1}{5}\left[-u^{-1}\right]_{-2}^{-7} = -\frac{1}{5}\left(-\frac{1}{-7} - \left(-\frac{1}{-2}\right)\right)$$

$$= -\frac{1}{5}\left(\frac{1}{7} - \frac{1}{2}\right) = -\frac{1}{5} \cdot \left(-\frac{5}{14}\right) = \frac{1}{14}$$

Note: The limits went from $-2$ to $-7$ (decreasing), which is fine! The negative from the $du$ adjustment corrects the direction.

Example 3: Why Method 2 is Better

Evaluate $\int_0^{\pi/2} \sin^3 x \cos x\,dx$.

Using Method 2:

Let $u = \sin x$, so $du = \cos x\,dx$.

Change limits:

When $x = 0$: $u = \sin 0 = 0$
When $x = \pi/2$: $u = \sin(\pi/2) = 1$

$$\int_0^{\pi/2} \sin^3 x \cos x\,dx = \int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$$

Clean and simple! No back-substitution, no messy expressions.

Practice Problems

Level 1 Basic Limit Changes

Evaluate each integral by changing the limits:

$\displaystyle\int_0^1 (2x+1)^3\,dx$
$\displaystyle\int_0^{\pi} \cos(x/2)\,dx$

Thought Process

(a) Let $u = 2x + 1$. When $x = 0$, $u = 1$. When $x = 1$, $u = 3$.

(b) Let $u = x/2$. When $x = 0$, $u = 0$. When $x = \pi$, $u = \pi/2$.

Show Answer

(a) Let $u = 2x + 1$, $du = 2\,dx$, $dx = \frac{1}{2}du$.

$x = 0 \Rightarrow u = 1$
$x = 1 \Rightarrow u = 3$

$$\int_0^1 (2x+1)^3\,dx = \frac{1}{2}\int_1^3 u^3\,du = \frac{1}{2}\left[\frac{u^4}{4}\right]_1^3 = \frac{1}{8}(81 - 1) = \boxed{10}$$

(b) Let $u = x/2$, $du = \frac{1}{2}dx$, $dx = 2\,du$.

$x = 0 \Rightarrow u = 0$
$x = \pi \Rightarrow u = \pi/2$

$$\int_0^{\pi} \cos(x/2)\,dx = 2\int_0^{\pi/2} \cos u\,du = 2[\sin u]_0^{\pi/2} = 2(1 - 0) = \boxed{2}$$

Level 2 Derivative Present in Integrand

Evaluate $\displaystyle\int_0^1 x^2(1 + 2x^3)^5\,dx$.

Thought Process

Notice that $\frac{d}{dx}[1 + 2x^3] = 6x^2$. The $x^2$ in the integrand is (up to a constant) the derivative of the inside.

Let $u = 1 + 2x^3$.

Show Answer

Let $u = 1 + 2x^3$, $du = 6x^2\,dx$, so $x^2\,dx = \frac{1}{6}du$.

Change limits:

$x = 0 \Rightarrow u = 1 + 0 = 1$
$x = 1 \Rightarrow u = 1 + 2 = 3$

$$\int_0^1 x^2(1 + 2x^3)^5\,dx = \frac{1}{6}\int_1^3 u^5\,du = \frac{1}{6}\left[\frac{u^6}{6}\right]_1^3$$

$$= \frac{1}{36}(3^6 - 1^6) = \frac{1}{36}(729 - 1) = \frac{728}{36} = \boxed{\frac{182}{9}}$$

Level 3 Trigonometric Substitution

Evaluate $\displaystyle\int_0^{\pi/4} \tan^3\theta \sec^2\theta\,d\theta$.

Thought Process

Notice $\frac{d}{d\theta}[\tan\theta] = \sec^2\theta$.

Let $u = \tan\theta$. The $\sec^2\theta\,d\theta$ becomes $du$.

Show Answer

Let $u = \tan\theta$, $du = \sec^2\theta\,d\theta$.

Change limits:

$\theta = 0 \Rightarrow u = \tan 0 = 0$
$\theta = \pi/4 \Rightarrow u = \tan(\pi/4) = 1$

$$\int_0^{\pi/4} \tan^3\theta \sec^2\theta\,d\theta = \int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4} - 0 = \boxed{\frac{1}{4}}$$

Level 4 Reversing Limits

Evaluate $\displaystyle\int_1^4 \frac{dx}{\sqrt{x}(1 + \sqrt{x})^2}$.

Thought Process

Let $u = 1 + \sqrt{x}$. Then $du = \frac{1}{2\sqrt{x}}dx$, so $\frac{dx}{\sqrt{x}} = 2\,du$.

Show Answer

Let $u = 1 + \sqrt{x}$, so $du = \frac{1}{2\sqrt{x}}dx$, meaning $\frac{dx}{\sqrt{x}} = 2\,du$.

Change limits:

$x = 1 \Rightarrow u = 1 + 1 = 2$
$x = 4 \Rightarrow u = 1 + 2 = 3$

$$\int_1^4 \frac{dx}{\sqrt{x}(1 + \sqrt{x})^2} = \int_2^3 \frac{2\,du}{u^2} = 2\int_2^3 u^{-2}\,du$$

$$= 2\left[-u^{-1}\right]_2^3 = 2\left(-\frac{1}{3} + \frac{1}{2}\right) = 2 \cdot \frac{1}{6} = \boxed{\frac{1}{3}}$$

Level 5 Compare Both Methods

Evaluate $\displaystyle\int_0^2 x\sqrt{4 - x^2}\,dx$ using:

Method 1: Find the indefinite integral first, then apply limits
Method 2: Change the limits of integration

Which method was easier?

Thought Process

Let $u = 4 - x^2$, $du = -2x\,dx$.

For Method 1: Evaluate $\int x\sqrt{4-x^2}\,dx$ first, then plug in limits.

For Method 2: When $x = 0$, $u = 4$. When $x = 2$, $u = 0$. Note the limits reverse!

Show Answer

(a) Method 1: Indefinite First

Let $u = 4 - x^2$, $du = -2x\,dx$, so $x\,dx = -\frac{1}{2}du$.

$$\int x\sqrt{4 - x^2}\,dx = -\frac{1}{2}\int \sqrt{u}\,du = -\frac{1}{2} \cdot \frac{2}{3}u^{3/2} + C = -\frac{1}{3}(4-x^2)^{3/2} + C$$

Now apply limits: $$\left[-\frac{1}{3}(4-x^2)^{3/2}\right]_0^2 = -\frac{1}{3}(4-4)^{3/2} - \left(-\frac{1}{3}(4-0)^{3/2}\right)$$ $$= 0 + \frac{1}{3}(8) = \frac{8}{3}$$

(b) Method 2: Change Limits

Let $u = 4 - x^2$, $du = -2x\,dx$.

$x = 0 \Rightarrow u = 4$
$x = 2 \Rightarrow u = 0$

$$\int_0^2 x\sqrt{4 - x^2}\,dx = \int_4^0 \sqrt{u} \cdot \left(-\frac{1}{2}\right)\,du = \frac{1}{2}\int_0^4 \sqrt{u}\,du$$

(We flipped the limits to absorb the negative sign.)

$$= \frac{1}{2} \cdot \frac{2}{3}\left[u^{3/2}\right]_0^4 = \frac{1}{3}(8 - 0) = \frac{8}{3}$$

Both give: $\boxed{\frac{8}{3}}$

Method 2 was easier — no back-substitution needed, and the calculation is more direct.

Conceptual Questions (CCI-Style)

Level 2 Understanding Limit Changes

When evaluating $\int_0^1 2x e^{x^2}\,dx$ with $u = x^2$, a student writes:

$$\int_0^1 2x e^{x^2}\,dx = \int_0^1 e^u\,du$$

What is wrong with this setup?

Thought Process

The limits 0 and 1 are $x$-values. When we change to $u = x^2$, what are the corresponding $u$-values?

Show Answer

The error: The student kept the old limits (0 and 1), which are $x$-values.

When $u = x^2$:

At $x = 0$: $u = 0^2 = 0$ ✓ (happens to be the same)
At $x = 1$: $u = 1^2 = 1$ ✓ (also the same by coincidence!)

Actually, in this particular case, the limits happen to be the same. But the student got lucky! In general:

Correct procedure: $$u = x^2 \Rightarrow \text{when } x = 0, u = 0; \text{ when } x = 1, u = 1$$

The answer $\int_0^1 e^u\,du = e - 1$ is correct, but the reasoning should explicitly include the limit transformation.

Level 3 Why the Rule Works

The Substitution Rule for Definite Integrals says: $$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$$

Use the Fundamental Theorem of Calculus to prove this identity.

Hint: If $F' = f$, what is $\frac{d}{dx}[F(g(x))]$?

Thought Process

By the chain rule, $\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x) = f(g(x)) g'(x)$.

So $F(g(x))$ is an antiderivative of $f(g(x))g'(x)$.

Apply FTC to both sides.

Show Answer

Proof:

Let $F$ be an antiderivative of $f$ (so $F' = f$).

By the chain rule: $$\frac{d}{dx}[F(g(x))] = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$$

So $F(g(x))$ is an antiderivative of $f(g(x))g'(x)$.

By FTC Part 2 (evaluation): $$\int_a^b f(g(x))g'(x)\,dx = F(g(x))\Big\vert _a^b = F(g(b)) - F(g(a))$$

Also by FTC Part 2: $$\int_{g(a)}^{g(b)} f(u)\,du = F(u)\Big\vert _{g(a)}^{g(b)} = F(g(b)) - F(g(a))$$

Both sides equal $F(g(b)) - F(g(a))$. $\square$

Mastery Checklist

[ ] I can correctly change limits of integration when substituting
[ ] I know to compute new limits using $u = g(x)$ at the original limits
[ ] I do NOT back-substitute after changing limits
[ ] I can handle cases where the limits reverse direction
[ ] I can choose between Method 1 and Method 2 based on the problem
[ ] I can verify my answer using both methods

Mental Model

The GPS Coordinate Change Analogy:

Imagine driving from City A to City B on a road. You can measure your position in miles from A, or in kilometers. If you switch to kilometers mid-trip, you need to convert both your start AND end positions.

Similarly, when switching from "$x$-coordinates" to "$u$-coordinates," you must convert both limits of integration. The integral represents the same "journey" — just measured in different units.

Connections

Looking back:

u-Substitution (Indefinite) provides the core substitution technique
Fundamental Theorem of Calculus justifies evaluating at limits

Looking ahead:

Symmetric Function Integrals uses this technique for even/odd functions
Area Between Curves applies definite integrals geometrically

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u-Substitution (Indefinite)	Skills Index	Symmetric Function Integrals

Last updated: 2026-01-22