Substitution: Definite Integrals
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 1.5: “Substitution” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/1-5-substitution |
| Textbook used in class | Stewart, Calculus, Section 4.5: “The Substitution Rule” |
Opening Scenario
For indefinite integrals, after substituting $u = g(x)$, you back-substitute at the end to return to the original variable $x$. For definite integrals, there is an alternative: convert the limits of integration from $x$-values to $u$-values at the same time you substitute, and then evaluate without ever returning to $x$. Both methods give the same answer, but the “change the bounds” approach is usually cleaner.
Quick Reference
Substitution Rule for definite integrals:
If $g'$ is continuous on $[a, b]$ and $f$ is continuous on the range of $g$, then $$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.$$
Method 1 -- Change the bounds (preferred):
- Set $u = g(x)$, compute $du = g'(x)\,dx$.
- Convert the limits: when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$.
- Evaluate $\displaystyle\int_{g(a)}^{g(b)} f(u)\,du$.
Method 2 -- Back-substitute (also correct):
- Find the indefinite integral $F(u) + C = \int f(u)\,du$, back-substitute to get $F(g(x)) + C$.
- Evaluate $F(g(x))\Big|_a^b = F(g(b)) - F(g(a))$.
Key Concepts
1. Changing the Limits
When $u = g(x)$:
- The lower limit $x = a$ becomes $u = g(a)$.
- The upper limit $x = b$ becomes $u = g(b)$.
The new limits are the $u$-values corresponding to the original $x$-values. After changing the limits, the integral is entirely in $u$ -- no back-substitution is needed.
Important detail: The new limits can be “reversed” (i.e., $g(a) > g(b)$ even if $a < b$). This is fine; the sign is absorbed into the evaluation of $F(g(b)) - F(g(a))$.
2. When the New Bounds Are Reversed
If $g(a) > g(b)$ after substitution (which happens when $g$ is decreasing on $[a, b]$), the new integral has a larger lower limit and a smaller upper limit. Do not swap them; simply evaluate as written. The negative sign from $g(b) < g(a)$ accounts for the orientation.
Example. $\displaystyle\int_2^0 f(u)\,du = -\int_0^2 f(u)\,du$ by the property of reversed limits. This is a consequence of the substitution, not an error.
3. Equivalence of Both Methods
Both methods produce $F(g(b)) - F(g(a))$:
- Changing bounds: $\int_{g(a)}^{g(b)} f(u)\,du = F(g(b)) - F(g(a))$.
- Back-substitute then evaluate: $F(g(x))\Big|_a^b = F(g(b)) - F(g(a))$.
“Keep the original bounds when substituting.” If you substitute $u = g(x)$ but leave the limits as $x = a$ and $x = b$, the integral $\int_a^b f(u)\,du$ is incorrect because $a$ and $b$ are $x$-values, not $u$-values. You must either change the limits or back-substitute before evaluating.
Worked Example
Method 1 (change bounds): Evaluate $\displaystyle\int_0^2 x(x^2+1)^3\,dx$.
Step 1 -- Substitute. $u = x^2 + 1$, $du = 2x\,dx$, so $x\,dx = \dfrac{du}{2}$.
Step 2 -- Change limits. $x = 0 \Rightarrow u = 1$; $x = 2 \Rightarrow u = 5$.
Step 3 -- Rewrite. $$\int_0^2 x(x^2+1)^3\,dx = \int_1^5 u^3\,\frac{du}{2} = \frac{1}{2}\cdot\frac{u^4}{4}\Big|_1^5 = \frac{1}{8}(625 - 1) = \frac{624}{8} = 78.$$
Boxed answer: $78$.
Method 2 (back-substitute): Evaluate $\displaystyle\int_0^{\pi/4} \sec^2(2x)\,dx$.
Indefinite integral first. $u = 2x$, $du = 2\,dx$:
$\displaystyle\int \sec^2(2x)\,dx = \frac{1}{2}\int\sec^2 u\,du = \frac{1}{2}\tan u + C = \frac{1}{2}\tan(2x) + C$.
Evaluate at original bounds.
$\dfrac{1}{2}\tan(2x)\Big|_0^{\pi/4} = \dfrac{1}{2}\tan\!\left(\dfrac{\pi}{2}\right) - \dfrac{1}{2}\tan(0)$.
Wait -- $\tan(\pi/2)$ is undefined. Check the domain: $2x = \pi/2$ when $x = \pi/4$, i.e., at the upper bound. The integrand $\sec^2(2x)$ has a vertical asymptote at $x = \pi/4$, so this integral diverges. The calculation has found an improper integral (covered in a later section).
Lesson: Always check whether the integrand has singularities on the integration interval before evaluating.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Keeping $x$-bounds with a $u$-integral | $\int_0^1 f(u)\,du$ when bounds were for $x \in [0,1]$ | Convert bounds: $u = g(0)$ and $u = g(1)$ |
| Swapping reversed bounds | Changing $\int_5^1$ to $\int_1^5$ and forgetting the sign | Reversed bounds contribute a sign; do not swap without negating the integral |
| Evaluating a divergent integral without checking | Blindly evaluating at an asymptote | Check that the integrand is continuous on the closed interval $[a,b]$ before applying FTC |
Leveled Practice
Level 1 -- Change the Bounds
Problem 1. Evaluate $\displaystyle\int_0^1 2x e^{x^2}\,dx$.
Show answer
$u = x^2$, $du = 2x\,dx$. Bounds: $x=0 \Rightarrow u=0$; $x=1 \Rightarrow u=1$.
$\displaystyle\int_0^1 e^u\,du = e^u\Big|_0^1 = e - 1$.
Level 2 -- Reversed Bounds
Problem 2. Evaluate $\displaystyle\int_1^0 \frac{1}{(2x-1)^2}\,dx$.
Show answer
Note the reversed limits: lower bound $x = 1$, upper bound $x = 0$. The integrand has a singularity at $x = 1/2$ which is inside $[0, 1]$, so first check: this is an improper integral. Here we assume for practice purposes that $1/2 \notin [1, 0]$ -- but actually the integration interval is from 1 to 0 with $[0,1]$ reversed.
Since $1/2$ lies in $(0,1)$, this integral diverges. (The student should recognize the singularity at $x = 1/2$ and not proceed to evaluate.)
Level 2 -- Standard Problem
Problem 3. Evaluate $\displaystyle\int_0^{\sqrt{\pi}} x\cos(x^2)\,dx$.
Show answer
$u = x^2$, $du = 2x\,dx$. Bounds: $x=0 \Rightarrow u=0$; $x=\sqrt{\pi} \Rightarrow u=\pi$.
$\displaystyle\int_0^{\pi}\cos u\,\frac{du}{2} = \frac{1}{2}\sin u\Big|_0^{\pi} = \frac{1}{2}(0 - 0) = 0$.
Mastery Checklist
Mental Model
Changing the bounds is like converting a recipe from metric to imperial: once you decide to work in grams instead of ounces, you also need to convert the ingredient quantities. If the recipe says “150 g,” you convert to ounces at the start and never switch back. Similarly, once you substitute $u = g(x)$, you convert everything -- including the bounds -- to $u$-units, and you never need to return to $x$.
Connections
Looking back
- Substitution for indefinite integrals (Section 4.5): The indefinite version; Method 2 uses it as a sub-step.
- Properties of integrals (Section 4.2): The reversed-limits convention $\int_b^a = -\int_a^b$ appears when $g(a) > g(b)$.
Looking ahead
- Symmetry (Section 4.5): Substitution is the tool used to prove the symmetry formulas for even and odd functions.
- Improper integrals (Chapter 7): Integrands with singularities on the interval require a separate treatment.
Back to Integration Foundations | Next: Symmetry and Integration