Unit 6 · Topic 6.9 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 6.9: Integrating Using Substitution

Q: Do I have to change the limits, or can I go back to x?

Either works. You can change the limits to values of u and finish in u, or find the antiderivative, put g(x) back for u, and use the original limits. The one thing you cannot do is mix the two.

Q: What if the new lower limit is bigger than the new upper limit?

Keep them in the order the substitution gives them. 41 f(u)\,du is a perfectly good integral, equal to -14 f(u)\,du, and the order is what keeps the sign right.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 6: 15 to 20% of the multiple-choice section

What you'll be able to do

Find indefinite integrals and evaluate definite integrals whose integrands need a substitution, or a rewrite into an equivalent form, before a basic rule applies.

Key vocabulary

Term	What it means	The exam trap
Substitution of variables	A technique for finding antiderivatives that replaces an inside expression $g (x)$ with a new variable $u$ and $g^{'} (x) d x$ with $d u$ .	Before you integrate, look at the rewritten integral: if any $x$ is still in it, the rewrite is not finished.
Indefinite integral, $\int f (x) d x$	The family of all antiderivatives of $f$ , written $F (x) + C$ where $F^{'} (x) = f (x)$ .	Write the $+ C$ on every indefinite integral, and write the answer in the original variable.
Limits of integration	The numbers $a$ and $b$ in $\int_{a}^{b} f (x) d x$ , the starting and ending values of the variable in the differential.	Ask which variable each limit belongs to every time you change variables.
Differential, $d u$	For $u = g (x)$ , the expression $d u = g^{'} (x) d x$ , which tells you what a substitution replaces.	Compare $d u$ with the factor that is actually in the integrand; they can differ by a constant.

Two graphs side by side. On the left, y = 2x/(x squared + 1) is shaded from x = 0 to x = 2, an area of 1.609. On the right, y = 1/u is shaded from u = 1 to u = 5, also an area of 1.609, because the substitution u = x squared + 1 carries the limits 0 and 2 to 1 and 5.

Part 1 Think first How much water in the first minute?

Make a guess before the lesson starts. Committing to an answer, even a wrong one, helps the right one stick.

Predict first

A pump starts filling an empty tank. After $t$ minutes, water flows in at $r (t) = 6 t (t^{2} + 1)^{2}$ liters per minute. The rate is 0 liters per minute when the pump starts and 24 liters per minute at $t = 1$ . How much water flows into the tank during the first minute?

In short. Integration by substitution renames an inside function $u = g (x)$ , so that $d u = g^{'} (x) d x$ and an integral of the form $\int f (g (x)) g^{'} (x) d x$ becomes $\int f (u) d u$ , which a basic rule can finish.

Substitution reverses the chain rule, so look for an inside function whose derivative, up to a constant factor, is also in the integrand.
Before integrating, every $x$ and $d x$ must be replaced, which may mean solving $u = g (x)$ for $x$ or moving a constant outside the integral.
In a definite integral the limits are values of the variable being integrated, so after substituting they become $g (a)$ and $g (b)$ . An indefinite integral ends in $+ C$ .

Part 2 Learn it Three short ideas, each checked as you go

6a. Substitution runs the chain rule backward

The chain rule from Unit 3 says $\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)$ . Read backward, it hands you an antiderivative:

\int f^{'} (g (x)) g^{'} (x) d x = f (g (x)) + C .

So when an integrand is some function of an inside expression, multiplied by the derivative of that inside expression, you can undo the chain rule.

Substitution of variables makes this routine. Call the inside expression $u = g (x)$ . Then $\frac{d u}{d x} = g^{'} (x)$ , which is written $d u = g^{'} (x) d x$ . Replace $g (x)$ by $u$ and $g^{'} (x) d x$ by $d u$ , integrate in $u$ , and then put $g (x)$ back in place of $u$ .

Take $\int 3 x^{2} \cos (x^{3}) d x$ . The inside expression is $x^{3}$ , and its derivative, $3 x^{2}$ , is sitting in front. Let $u = x^{3}$ , so $d u = 3 x^{2} d x$ , and

\int 3 x^{2} \cos (x^{3}) d x = \int \cos u d u = \sin u + C = \sin (x^{3}) + C .

Check it by differentiating: the chain rule turns $\sin (x^{3})$ back into $\cos (x^{3}) \cdot 3 x^{2}$ . An indefinite integral stands for every antiderivative at once, so it always ends in $+ C$ .

Quick check · Match

No calculator. Each integrand is a function of an inside expression times that expression's derivative. Tap an integral, then its antiderivative.

6b. Fixing a constant, and rewriting what is left

Often $d u$ is in the integral only up to a constant factor. In $\int x (x^{2} + 5)^{3} d x$ , let $u = x^{2} + 5$ . Then $d u = 2 x d x$ , but the integral has $x d x$ , not $2 x d x$ . Solve for what you have: $x d x = \frac{1}{2} d u$ . A constant can move outside an integral, so

\int x (x^{2} + 5)^{3} d x = \frac{1}{2} \int u^{3} d u = \frac{1}{8} u^{4} + C = \frac{1}{8} (x^{2} + 5)^{4} + C .

Only a constant can be fixed this way. If $d u$ calls for a factor of $x$ that the integrand does not have, as in $\int (x^{2} + 5)^{3} d x$ , this substitution does not work, and you cannot divide by $x$ to rescue it.

Sometimes the leftover $x$ can be rewritten in terms of $u$ instead. In $\int x \sqrt{x + 1} d x$ , let $u = x + 1$ , so $d u = d x$ and $x = u - 1$ :

\int x \sqrt{x + 1} d x = \int (u - 1) u^{1 / 2} d u = \int (u^{3 / 2} - u^{1 / 2}) d u,

which is $\frac{2}{5} (x + 1)^{5 / 2} - \frac{2}{3} (x + 1)^{3 / 2} + C$ .

Either way, the integral must hold only $u$ and $d u$ before you integrate. A stray $x$ means the rewrite is not finished.

Quick check · Calculate

No calculator. For some positive integer $n$ , $\int x^{3} (x^{4} + 1)^{4} d x = \frac{1}{n} (x^{4} + 1)^{5} + C$ . What is $n$ ?

Your answer

6c. Definite integrals: the limits change too

The limits of a definite integral are values of the variable in its differential: in $\int_{1}^{2} \dots d x$ , $x$ runs from 1 to 2. After a substitution the integral is in $u$ , so its limits must be values of $u$ . If $u = g (x)$ , then $x = a$ gives $u = g (a)$ and $x = b$ gives $u = g (b)$ :

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u .

Take $\int_{1}^{2} 2 x (x^{2} - 1)^{2} d x$ . Let $u = x^{2} - 1$ , so $d u = 2 x d x$ . At $x = 1$ , $u = 0$ , and at $x = 2$ , $u = 3$ :

\int_{1}^{2} 2 x (x^{2} - 1)^{2} d x = \int_{0}^{3} u^{2} d u = \frac{27}{3} - 0 = 9.

Keeping the old limits would give $\int_{1}^{2} u^{2} d u = \frac{7}{3}$ , the area of a different region.

You may instead go back to $x$ first, $\frac{1}{3} (x^{2} - 1)^{3}$ , and use the original limits: $\frac{1}{3} (27) - 0 = 9$ . Both routes work. What never works is a mix: a $u$ expression with $x$ limits, or an $x$ expression with $u$ limits.

Keep the new limits in the order the substitution gives them, even when $g (a)$ is larger than $g (b)$ . The graphs at the top show one substitution: two curves on two intervals, with one area.

Quick check · Multiple choice

No calculator. Using the substitution $u = x^{3} + 1$ , which integral is equal to $\int_{0}^{2} x^{2} \sqrt{x^{3} + 1} d x$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Keeping the $x$ limits after rewriting a definite integral in $u$ .	The limits are values of the variable being integrated. After the substitution that variable is $u$ , so $a$ and $b$ describe the wrong interval.	Work out $u = g (a)$ and $u = g (b)$ in the same step where you write $d u$ , and put them on the new integral at once.
Writing $d u$ in place of $x d x$ when $d u = 2 x d x$ .	The integral then holds twice what it should. Only the factor that is really in the integrand can be replaced.	Solve for what the integrand has, $x d x = \frac{1}{2} d u$ , and move the $\frac{1}{2}$ outside the integral.
Integrating while some $x$ is still left in a $u$ integral.	$x$ is not a constant with respect to $u$ , so it cannot be pulled out or ignored.	Solve $u = g (x)$ for $x$ and replace it, as $x = u - 1$ replaces $x$ when $u = x + 1$ . If that is not possible, choose a different $u$ .

Part 3 Watch it done A definite integral by substitution

Watch each step, then finish the last one yourself.

Evaluate $\int_{0}^{π / 2} 4 \cos x (1 + \sin x)^{3} d x$ .

Step 1 · Choose u
The inside expression is $1 + \sin x$ , and its derivative, $\cos x$ , is a factor of the integrand. Let $u = 1 + \sin x$ , so $d u = \cos x d x$ .
Step 2 · Change the limits
When $x = 0$ , $u = 1 + \sin 0 = 1$ . When $x = \frac{π}{2}$ , $u = 1 + \sin \frac{π}{2} = 2$ .
Why this step?
Why work out new limits instead of keeping 0 and $\frac{π}{2}$ ?
After the substitution the integral is in $u$ , and its limits have to be values of $u$ . The numbers 0 and $\frac{π}{2}$ are values of $x$ ; putting them on a $u$ integral would add up a different region.
Step 3 · Rewrite and find an antiderivative
$\int_{0}^{π / 2} 4 \cos x (1 + \sin x)^{3} d x = \int_{1}^{2} 4 u^{3} d u$ , and an antiderivative of $4 u^{3}$ is $u^{4}$ .
Your step · Calculate

No calculator. Finish it. The integral is now $\int_{1}^{2} 4 u^{3} d u$ , and an antiderivative of $4 u^{3}$ is $u^{4}$ . What is the value of the integral?

Your answer

Part 4 Try it A logarithm and a leftover x

Try it · Calculate

No calculator. Evaluate $\int_{1}^{e^{2}} \frac{\ln x}{x} d x$ .

Your answer

Try it · Multiple choice

No calculator. Which integral is equal to $\int x (x - 4)^{7} d x$ after the substitution $u = x - 4$ ?

Part 5 Explain it Write it as a free-response answer

Free response on the exam is written by hand, so do this one on paper. Nothing you type here leaves the page. When you are done, open the model answer and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 5, from free-response Part B (calculator not permitted).

Free-response practice · Question 5 style

No calculator. Water flows into a tank at a rate of $W (t) = 3 t \sqrt{t^{2} + 9}$ liters per minute, where $t$ is measured in minutes, for $0 \leq t \leq 4$ .

(a) Find $\int 3 t \sqrt{t^{2} + 9} d t$ . Show the work that leads to your answer.

(b) Using the substitution $u = t^{2} + 9$ , write $\int_{0}^{4} W (t) d t$ as an integral in terms of $u$ , with limits of integration in terms of $u$ . Do not evaluate it.

(c) Evaluate $\int_{0}^{4} W (t) d t$ . Using correct units, explain the meaning of this value in the context of the problem.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Let $u = t^{2} + 9$ , so $d u = 2 t d t$ and $3 t d t = \frac{3}{2} d u$ . Then $\int 3 t \sqrt{t^{2} + 9} d t = \frac{3}{2} \int u^{1 / 2} d u = \frac{3}{2} \cdot \frac{2}{3} u^{3 / 2} + C = (t^{2} + 9)^{3 / 2} + C$ .

(b) When $t = 0$ , $u = 9$ , and when $t = 4$ , $u = 25$ . So $\int_{0}^{4} W (t) d t = \frac{3}{2} \int_{9}^{25} u^{1 / 2} d u$ .

(c) $\frac{3}{2} \int_{9}^{25} u^{1 / 2} d u = {[u^{3 / 2}]}_{9}^{25} = 125 - 27 = 98$ . From $t = 0$ to $t = 4$ minutes, 98 liters of water flowed into the tank.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Names $u = t^{2} + 9$ with $d u = 2 t d t$ , and rewrites $3 t d t$ as $\frac{3}{2} d u$ .
In (a), gets $(t^{2} + 9)^{3 / 2} + C$ , with the $+ C$ .
In (b), writes $\frac{3}{2} \int_{9}^{25} u^{1 / 2} d u$ , with the limits changed to 9 and 25.
In (c), gets 98 and says 98 liters of water flowed into the tank from $t = 0$ to $t = 4$ minutes.

Part 6 Check yourself Two functions in a table

Both case questions use this table. Read every value you need from it.

The functions $f$ and $g$ are differentiable for all real numbers, and $f^{'}$ and $g^{'}$ are continuous. The table gives some of their values.

Values of f, its derivative, g and its derivative
$x$	$f (x)$	$f^{'} (x)$	$g (x)$	$g^{'} (x)$
1	5	$- 4$	2	3
2	2	2	4	1
3	8	1	3	$- 2$
4	3	$- 5$	1	$- 1$

Case question 1 · Calculate

No calculator. What is the value of $\int_{1}^{3} f^{'} (g (x)) g^{'} (x) d x$ ?

Your answer

Case question 2 · Decide and justify

No calculator. A classmate writes $\int_{2}^{4} f^{'} (g (x)) g^{'} (x) d x = f (4) - f (2) = 1$ . Is the classmate right?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student evaluates $\int_{0}^{2} x e^{x^{2}} d x$ by substitution. Tap the first line with a mistake.

Check yourself · Fill the statement

No calculator. Tap a blank, then tap an answer.

To find $\int e^{5 x} d x$ , let $u =$ . Then $d u = 5 d x$ , so $d x =$ , and $\int e^{5 x} d x =$ .

Part 7 Keep it Bring back earlier ideas

Substitution leans on the chain rule and on what an indefinite integral means. Pulling both back up is what keeps them.

Bring it back · Multiple choice

No calculator. If $h (x) = e^{\cos x}$ , what is $h^{'} (x)$ ?

Bring it back · Calculate

No calculator. If $\int f (x) d x = x^{3} - 5 x + C$ , what is $f (2)$ ?

Your answer

In three lines

Substitution reverses the chain rule: name the inside function u and replace its derivative and dx by du.
Fix a constant factor by moving it outside, and rewrite any leftover x in terms of u before integrating.
In a definite integral, change the limits to values of u, or go back to x and use the original limits; an indefinite integral ends in + C.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a free-response question, write $u$ , $d u$ and, for a definite integral, the new limits on their own line before you integrate, and give the antiderivative before any arithmetic. Put $+ C$ on every indefinite integral, and never join a $u$ expression and an $x$ expression with an equals sign unless they really are equal.

Questions students ask

How do I know what to choose for u?

Look for an inside expression whose derivative, or a constant multiple of it, is also a factor of the integrand: the expression under a root, inside a power, in an exponent, or inside a trig function. If your first choice leaves an $x$ you cannot rewrite, try another.

Do I have to change the limits, or can I go back to x?

Either works. You can change the limits to values of $u$ and finish in $u$ , or find the antiderivative, put $g (x)$ back for $u$ , and use the original limits. The one thing you cannot do is mix the two.

What if the new lower limit is bigger than the new upper limit?

Keep them in the order the substitution gives them. $\int_{4}^{1} f (u) d u$ is a perfectly good integral, equal to $- \int_{1}^{4} f (u) d u$ , and the order is what keeps the sign right.

How can I check an answer to an indefinite integral?

Differentiate it. By the chain rule you should get the original integrand back exactly; a missing constant factor shows up at once.

Why can I move a constant outside the integral but not an x?

A constant multiple of a function has the same constant multiple of its antiderivative. An $x$ changes as you integrate, so it is not a constant multiple; it has to be written in terms of $u$ first.

Course alignment, for teachers

AP Calculus AB topic 6.9, Unit 6: Integration and Accumulation of Change.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.