Unit 6 · Topic 6.4 Beta

The Fundamental Theorem and Accumulation Functions

An accumulation function g(x) = ∫ from a to x of f(t) dt starts at 0 and grows at the rate f(x) : for continuous f , g′(x) = f(x) , so g is an antiderivative of f .

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

Picture a paint roller moving along a wall, painting everything below a curve. Stop it at any position

x

and you can measure the painted area so far. Now ask how fast that area grows as the roller moves. The answer is startlingly simple, and it ties the two halves of calculus together: the area grows at a rate equal to the height of the curve.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\int_0^3 2\,dx$ equals

Show the answer

The region is a rectangle of width 3 and height 2.

Correct: 6:
2:
3:
0:

2. If $y = \sin(x^2)$ , then $\frac{dy}{dx} =$

$2x\cos(x^2)$
$\cos(x^2)$
$\cos(2x)$
$2x\sin(x^2)$

Show the answer

Chain rule: the derivative of the outside at the inside, times the derivative of the inside.

Correct: $2x\cos(x^2)$ :
$\cos(x^2)$ :
$\cos(2x)$ :
$2x\sin(x^2)$ :

Part 4 · See it

See it first

Top: the graph of f(t) = 0.5t + 1 on [0, 6], with the area under it from 0 to x = 3 shaded; that area is g(3) = 5.25 and the height f(3) is 2.5. Bottom: the accumulation function g(x) = x squared over 4 + x, with the point (3, 5.25) and a dashed tangent line there of slope g prime of 3 = f(3) = 2.5. — Top: the area under f from 0 to x is g(x). Bottom: the slope of g at x is the height of f at x. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Define $g(x)$ $= \int_a^x f(t)\,dt$ $g(x)$ is the area accumulated under $f$ from $a$ up to $x$ , and $g(a) = 0$
Move $x$ a tiny step $h$ to the rightthe area grows by a thin strip about $f(x)$ tall and $h$ wide, so $\frac{g(x + h) - g(x)}{h}$ $\approx f(x)$
Let $h \to 0$ $g'(x) = f(x)$ : the Fundamental Theorem of Calculus, so $g$ is an antiderivative of $f$
The upper limit is a function $u(x)$ the chain rule adds a factor: $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt$ $= f(u(x))\,u'(x)$

Part 6 · Key ideas

Key ideas

An accumulation function $g(x)$ $= \int_a^x f(t)\,dt$ is the area so far; $g(a) = 0$ .
Fundamental Theorem of Calculus: if $f$ is continuous, $g'(x) = f(x)$ . The constant lower limit drops out.
An antiderivative of $f$ is any $F$ with $F' = f$ . Every accumulation function of $f$ is one.
Upper limit $u(x)$ : $\frac{d}{dx}\int_a^{u(x)} f(t)\,dt$ $= f(u(x)) \cdot u'(x)$ . Forgetting $u'(x)$ is the classic slip.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{d}{dx}\int_2^x f(t)\,dt$ $= f(x) - f(2)$ , because the lower limit must be subtracted.

What actually happens: The lower limit is a constant: it decides where the accumulation starts, not how fast it grows. The derivative is just $f(x)$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f$ and $g(x)$ $= \int_0^x f(t)\,dt$

The graph of the continuous function

f

[0, 6]

is made of three line segments. Let

g(x)

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

for

0 \le x \le 6

1. Find $g(4)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$g(4)$ $= \int_0^4 f(t)\,dt$ : trapezoid from 0 to 2, $\frac{1 + 3}{2} \cdot 2 = 4$ , plus rectangle from 2 to 4, $3 \cdot 2 = 6$ . So $g(4) = 10$ .

Answer: 10

2. Find $g'(1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

By the Fundamental Theorem of Calculus, $g'(x) = f(x)$ because $f$ is continuous. So $g'(1) = f(1) = 2$ .

Answer: 2

3. Find $g''(3)$ .

7
$-1.5$
0
3

Show the answer

$g'(x) = f(x)$ , so $g''(x) = f'(x)$ . On $(2, 4)$ the graph of $f$ is horizontal, so $g''(3) = f'(3) = 0$ .

7: That is $g(3)$ , the area from 0 to 3. Two derivatives of $g$ give the slope of $f$ .
$-1.5$ : That is the slope of $f$ on $(4, 6)$ . At $x = 3$ the graph of $f$ is flat.
Correct: 0: Correct: $g'' = f'$ , and $f$ is constant on $(2, 4)$ , so $f'(3) = 0$ .
3: That is $f(3) = g'(3)$ . One more derivative gives $f'(3)$ , the slope of $f$ at 3.

4. Find $\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\,dt$ .

Type an expression for $\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\,dt =$ .

Show the answer

By the Fundamental Theorem of Calculus, with a continuous integrand and upper limit $x$ : $\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\,dt$ $= \sqrt{1 + x^3}$ .

Answer: $\frac{d}{dx}\int_1^x \sqrt{1 + t^3}\,dt =$ $\sqrt{1 + x^{3}}$

5. Find $\frac{d}{dx}\int_2^{5x} \cos(t^2)\,dt$ .

Type an expression for $\frac{d}{dx}\int_2^{5x} \cos(t^2)\,dt =$ .

Show the answer

With $u = 5x$ : $\frac{d}{dx}\int_2^{u} \cos(t^2)\,dt$ $= \cos(u^2)\cdot u'$ $= 5\cos(25x^2)$ .

Answer: $\frac{d}{dx}\int_2^{5x} \cos(t^2)\,dt =$ $5\,\cos(25x^{2})$

6. $f$ is continuous and $g(x)$ $= \int_2^x f(t)\,dt$ . Which is $g'(x)$ ?

$f(x) - 2$
$f(x)$
$f(x) - f(2)$
$f'(x)$

Show the answer

For continuous $f$ and constant $a$ , $\frac{d}{dx}\int_a^x f(t)\,dt$ $= f(x)$ . The lower limit only fixes where $g = 0$ .

$f(x) - 2$ : The 2 is a limit of integration, not something subtracted from $f$ .
Correct: $f(x)$ : Correct: the Fundamental Theorem gives the integrand at the upper limit; the constant lower limit drops out.
$f(x) - f(2)$ : The lower limit is a constant. It shifts the value of $g$ but not its derivative: no $f(2)$ term.
$f'(x)$ : That is $g''(x)$ . One derivative of $g$ gives $f$ itself.

Part 9 · Summary

Summary

An accumulation function

g(x)

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

starts at 0 and grows at the rate

f(x)

: for continuous

f

g'(x) = f(x)

, so

g

is an antiderivative of

f

. With an upper limit

u(x)

, the derivative is

f(u(x))\,u'(x)

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of ff and g(x)g(x)=∫0xf(t)dt= \int_0^x f(t)\,dt

Summary

What comes next

Connections

The graph of $f$ and $g(x)$ $= \int_0^x f(t)\,dt$