Day 19: U-Substitution & Accumulation Functions

U-Substitution: Reversing the Chain Rule

The integration counterpart of the chain rule. When the integrand contains a function and (a multiple of) its derivative, substitute $u$ for the inner function to collapse the integral into a basic form.

Core Theorem

u = g(x)

so that

du = g'(x)\,dx

, then

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

For a definite integral, either convert the limits to

u

-values or convert back to

x

before evaluating.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Evaluate

\int 2x\,(x^2 + 1)^5 \, dx

How to think about it

u-substitution works when the integrand contains an inner function *and* (a multiple of) its derivative. Set

u

to the inner function; if its

du

is sitting there, the integral collapses to a basic one.

Key idea: Let $u = x^2 + 1$ . Then $du = 2x\,dx$ , which is already present.

Full solution

1. Choose u

u = x^2 + 1 \implies du = 2x\,dx

2. Rewrite in u

\int u^5 \, du

3. Integrate

\frac{u^6}{6} + C

4. Substitute back

\frac{(x^2 + 1)^6}{6} + C

Worked Example 02Medium

Evaluate

\int x\sqrt{x^2 + 4} \, dx

How to think about it

Key idea: Let $u = x^2 + 4$ , so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}\,du$ .

Full solution

1. Choose u and solve for the piece you need

u = x^2 + 4 \implies du = 2x\,dx \implies x\,dx = \tfrac{1}{2}\,du

2. Rewrite in u

\int \sqrt{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}\int u^{1/2}\,du

3. Integrate

\tfrac{1}{2}\cdot\frac{u^{3/2}}{3/2} = \frac{u^{3/2}}{3}

4. Substitute back

\frac{(x^2 + 4)^{3/2}}{3} + C

Step-by-Step SOP

1
Pick u
Choose $u$ as the inner function — often what sits inside a power, root, exponential, or denominator.
2
Compute du
Differentiate: $du = g'(x)\,dx$ . Solve for whatever piece the integrand needs.
3
Substitute
Rewrite the whole integral in $u$ and $du$ — no $x$ should remain.
4
Integrate
Evaluate the simpler integral in $u$ .
5
Convert back
Replace $u$ with $g(x)$ for an indefinite integral, or change the limits for a definite one.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Evaluate

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

Need a hint?

Let

u = x^2 + 1

and change the limits:

x=0 \Rightarrow u=1

x=1 \Rightarrow u=2

Show solution

1. Substitute and change the limits

u = x^2 + 1

du = 2x\,dx

; the limits become

u: 1 \to 2

2. Rewrite in u

\int_1^2 \frac{1}{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}\int_1^2 \frac{1}{u}\,du

3. Integrate and evaluate

\tfrac{1}{2}\Big[\ln|u|\Big]_1^2 = \tfrac{1}{2}(\ln 2 - \ln 1) = \tfrac{1}{2}\ln 2

Practice 02Hard

Evaluate

\int_0^{\pi/2} \cos(x)\,e^{\sin(x)} \, dx

Need a hint?

Let

u = \sin(x)

, so

du = \cos(x)\,dx

. Change the limits.

Show solution

1. Substitute

u = \sin(x)

; the limits

x: 0 \to \pi/2

become

u: 0 \to 1

2. Rewrite and integrate

\int_0^1 e^u \, du = \Big[e^u\Big]_0^1 = e - 1

Common Pitfalls

⚠
Not changing the limits on a definite integralIf you switch to $u$ but keep the original $x$ -limits, the answer is wrong. Either convert the limits to $u$ -values, or convert back to $x$ before evaluating.
⚠
Leftover x termsAfter substituting, every $x$ must be gone. If an $x$ remains, the substitution is wrong or you also need to solve $u = g(x)$ for $x$ and substitute that.

Accumulation Functions: Analyzing g(x) = ∫ f(t) dt

When a function is defined as $g(x) = \int_a^x f(t)\,dt$ , the FTC turns every question about $g$ into a question about $f$ : $g' = f$ and $g'' = f'$ . Given the graph of $f$ , you can find $g$ 's values, extrema, concavity, and inflection points — a staple AP free-response question.

Core Theorem

For

g(x) = \displaystyle\int_a^x f(t)\,dt

with

f

continuous:
•

g(a) = 0

;

g(x)

is the signed area under

f

from

a

x

.
•

g'(x) = f(x)

g

increases where

f > 0

, decreases where

f < 0

.
•

g

has a local max where

f

changes

+\to-

, a local min where

f

changes

-\to+

.
•

g''(x) = f'(x)

g

is concave up where

f

is increasing; inflection points of

g

occur where

f

has a local extremum.
• Absolute extrema of

g

[a, b]

: compare

g

at the critical numbers (where

f = 0

with a sign change) and at the endpoints.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Let

g(x) = \int_0^x f(t)\,dt

, where the graph of

f

[0, 6]

is the line segments through

(0, 3)

(2, 3)

(4, -1)

(6, -1)

. Find

g(2)

and

g(4)

How to think about it

By the FTC,

g' = f

and

g'' = f'

, so every question about

g

becomes a question about

f

. Read

g

's increase and extrema off the *sign* of

f

; read

g

's concavity off whether

f

is *increasing*.

Key idea: $g$ is signed area under $f$ . On $[0, 2]$ it is a rectangle; on $[2, 4]$ it is a trapezoid that crosses the axis.

Full solution

1. g(2): area on [0, 2]

f = 3

(constant) on

[0, 2]

g(2) = 3 \times 2 = 6

2. g(4): add the area on [2, 4]

[2, 4]

f

drops linearly from

3

-1

, crossing zero at

t = 3.5

. Positive part (a triangle, base

1.5

, height

3

+2.25

. Negative part (base

0.5

, height

1

-0.25

. So

g(4) = 6 + 2.25 - 0.25 = 8

Worked Example 02Medium

For the same

g

and

f

as above, on what interval is

g

increasing, and where does

g

have its absolute maximum on

[0, 6]

? Justify.

How to think about it

Key idea: $g' = f$ . $g$ increases where $f > 0$ ; the maximum is where $f$ changes from $+$ to $-$ (compare with endpoints).

Full solution

1. Where g increases

f > 0

(0, 3.5)

and

f < 0

(3.5, 6)

, so

g

is increasing on

(0, 3.5)

and decreasing on

(3.5, 6)

(because

g' = f

2. Absolute maximum

g' = f

changes from positive to negative at

x = 3.5

, and

g

increases before it and decreases after, so the absolute maximum of

g

[0, 6]

is at

x = 3.5

Step-by-Step SOP

1
Write down g' and g''
$g'(x) = f(x)$ and $g''(x) = f'(x)$ by the FTC — now every question is about $f$ .
2
Values of g: accumulate signed area
$g(x) = \int_a^x f$ — add up areas of triangles/rectangles under the graph of $f$ , subtracting where $f < 0$ .
3
Increasing / extrema of g: sign of f
Sign chart for $f$ : $+$ ⇒ $g$ up, $-$ ⇒ $g$ down; sign change ⇒ local extremum of $g$ .
4
Concavity / inflection of g: behavior of f
$f$ increasing ⇒ $g$ concave up; $f$ 's local extrema ⇒ $g$ 's inflection points.
5
Absolute extrema of g
Compare $g$ at the interior critical numbers and at both endpoints.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Let

h(x) = \int_1^x f(t)\,dt

, where

f

is differentiable and its graph on

[1, 5]

is concave down with a single maximum at

t = 3

. On what interval is the graph of

h

concave up, and where is its inflection point? Justify.

Need a hint?

h'' = f'

. Concave up means

h'' > 0

, i.e.

f

is increasing.

Show solution

1. Translate concavity of h to behavior of f

h''(x) = f'(x)

h

is concave up where

f' > 0

, i.e. where

f

is increasing.

2. Read off f's behavior

f

increases on

(1, 3)

and decreases on

(3, 5)

(it peaks at

t = 3

3. Conclude

The graph of

h

is concave up on

(1, 3)

and concave down on

(3, 5)

, with an inflection point at

x = 3

because

f'

(which is

h''

) changes sign there.

Practice 02Medium

g(x) = \int_0^x f(t)\,dt

and

\int_0^3 f(t)\,dt = -4

\int_3^7 f(t)\,dt = 10

. Find

g(7)

and

\int_7^3 f(t)\,dt

Need a hint?

Use additivity for

g(7)

, and the reversal property for the last integral.

Show solution

1. g(7) by additivity

g(7) = \int_0^7 f = \int_0^3 f + \int_3^7 f = -4 + 10 = 6

2. Reverse the limits

\int_7^3 f(t)\,dt = -\int_3^7 f(t)\,dt = -10

Common Pitfalls

⚠
Confusing the graph of f with the graph of gA peak on the graph of $f$ is an inflection point of $g$ (where $g'' = f'$ changes sign), not a maximum of $g$ .
⚠
Forgetting g(a) = 0 $g(x) = \int_a^x f$ always starts at $0$ when $x = a$ . Use that as the anchor when accumulating.

U Substitution And Accumulation Functions

U-Substitution: Reversing the Chain Rule

Worked Examples

1. Choose u

2. Rewrite in u

3. Integrate

4. Substitute back

1. Choose u and solve for the piece you need

2. Rewrite in u

3. Integrate

4. Substitute back

Pick u

Compute du

Substitute

Integrate

Convert back

Practice — Your Turn

1. Substitute and change the limits

2. Rewrite in u

3. Integrate and evaluate

1. Substitute

2. Rewrite and integrate

Accumulation Functions: Analyzing g(x) = ∫ f(t) dt

Worked Examples

1. g(2): area on [0, 2]

2. g(4): add the area on [2, 4]

1. Where g increases

2. Absolute maximum

Write down g' and g''

Values of g: accumulate signed area

Increasing / extrema of g: sign of f

Concavity / inflection of g: behavior of f

Absolute extrema of g

Practice — Your Turn

1. Translate concavity of h to behavior of f

2. Read off f's behavior

3. Conclude

1. g(7) by additivity

2. Reverse the limits

Gary Chang