Single-Variable-Calculus · Unit 16 · Video 1 · Interactive Practice

Average Rate Times Time: FTC Beats the Mean Value Theorem

IKey Formulas

Statement	Formula	What it says
Average value of $f$ on $[a, b]$	$\displaystyle \frac{1}{b-a}\int_a^b f(x)\,dx$	the height of the rectangle on $[a, b]$ with the same area as the region under $f$
FTC 1, read backwards	$\Delta F = (\text{average of } F')\,\Delta x$	the factor is named exactly
Mean value theorem	$\Delta F = F'(c)\,\Delta x$ , $\ a < c < b$	the factor is $F'$ at an unknown point $c$
Bounds from either theorem	$\big(\min_{[a,b]} F'\big)\Delta x \le \Delta F \le \big(\max_{[a,b]} F'\big)\Delta x$	one-rectangle lower and upper Riemann sums

Key Insight: An average is never below the minimum and never above the maximum, so every bound the mean value theorem gives on $\Delta F$ the fundamental theorem gives too. With more rectangles, the integral gives sharper bounds.

IIVisualization 1 — The Chord Slope Is the Average of $F'$

For $F'(x) = \frac{1}{1+x}$ , $F(0) = 1$ : the slope of the chord of $F$ over $[0, b]$ equals the average height of $F'$ .

Right endpoint b 4

IIIVisualization 2 — More Rectangles, Sharper Bounds on $F(4)$

One rectangle reproduces the mean value bounds $\frac{9}{5} < F(4) < 5$ ; refining the Riemann sums squeezes $F(4)$ much harder.

Rectangles n 1

IVQuiz Questions

Problem 1 · Average Value

Find the average value of $f(x) = x^2$ on the interval $[0, 3]$ .

✅ Correct!

\frac{1}{3}\int_0^3 x^2\,dx = \frac{1}{3}\cdot 9 = 3

: a rectangle of height

3

[0, 3]

has the same area,

9

, as the region under the parabola.

❌ That averages only the two endpoint values.

\frac{f(0) + f(3)}{2} = \frac{9}{2}

ignores every value in between; the average value integrates over the whole interval.

❌ That is the integral itself.

\int_0^3 x^2\,dx = 9

is the area; divide by the length of the interval,

3

❌ Not quite. Compute

\frac{1}{b-a}\int_a^b f(x)\,dx

with

a = 0

b = 3

Show solution

\frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot 9 = 3.

The average value is $3$ . It lies between the minimum $f(0) = 0$ and the maximum $f(3) = 9$ , as every average must.

Problem 2 · Change Equals Average Rate Times Time

Given: the average value of $F'$ on $[2, 6]$ is $3$ , and $F(2) = 5$ . Find $F(6)$ .

✅ Correct!

\Delta F = 3 \cdot 4 = 12

, so

F(6) = 5 + 12 = 17

❌ That is $\Delta F$ , not $F(6)$ .

F(6) - F(2) = 12

; add the starting value

F(2) = 5

❌ The average rate must be multiplied by the elapsed $\Delta x$ . The interval

[2, 6]

has length

4

, so

\Delta F = 3 \cdot 4

❌ Not quite. Use

\Delta F = (\text{average of } F')\,\Delta x

, then

F(6) = F(2) + \Delta F

Show solution

By FTC 1, $\Delta F = (\text{average of } F')\,\Delta x$ with $\Delta x = 6 - 2 = 4$ :

F(6) - F(2) = 3 \cdot 4 = 12 \quad\Longrightarrow\quad F(6) = 5 + 12 = 17.

The average is of the derivative $F'$ ; $F$ itself enters only through its change.

Problem 3 · Bounds from the Mean Value Theorem

Given: $F'(x) = \frac{1}{x}$ for $x > 0$ and $F(1) = 2$ . Find the two numbers the mean value theorem says $F(3)$ lies between.

✅ Correct!

F(3) - F(1) = \frac{2}{c}

with

1 < c < 3

, so

\frac{2}{3} < F(3) - 2 < 2

❌ Those are bounds on $F(3) - F(1)$ . Add

F(1) = 2

to both ends.

❌ The interval has length $\Delta x = 2$ , not $1$ .

F'(c)\,\Delta x = \frac{1}{c}\cdot 2

❌ Not quite. Write

F(3) - F(1) = F'(c)(3 - 1)

and bound

\frac{1}{c}

for

1 < c < 3

Show solution

By the mean value theorem, for some $c$ with $1 < c < 3$ ,

F(3) - F(1) = F'(c)\,(3 - 1) = \frac{2}{c}.

Since $\frac{1}{c}$ decreases, $\frac{2}{c}$ lies between $\frac{2}{3}$ (at $c = 3$ ) and $2$ (at $c = 1$ ). Adding $F(1) = 2$ :

\frac{8}{3} < F(3) < 4.

The same bounds come from $\int_1^3 \frac{dx}{x}$ estimated by one rectangle of height $\frac{1}{3}$ below and one of height $1$ above.

Problem 4 · Two Rectangles Beat One

Given: $F'(x) = \frac{1}{1+x}$ and $F(0) = 1$ . Using lower and upper Riemann sums for $\int_0^4 \frac{dx}{1+x}$ with two rectangles of width $2$ , find the bounds on $F(4)$ .

✅ Correct! The bounds are

1 + \frac{16}{15}

and

1 + \frac{8}{3}

, already tighter than the mean value range

\frac{9}{5}

5

❌ That is the one-rectangle (mean value) bound. Split

[0, 4]

into

[0, 2]

and

[2, 4]

and use the smallest and largest value of

\frac{1}{1+x}

on each half.

❌ Those bound the integral, $F(4) - F(0)$ . Add

F(0) = 1

❌ Each rectangle has width $2$ . Multiply each height by

2

, not

1

❌ Not quite. On each half the integrand decreases: its minimum is at the right end and its maximum at the left end.

Show solution

The integrand $\frac{1}{1+x}$ decreases, so on $[0, 2]$ it lies between $\frac{1}{3}$ and $1$ , and on $[2, 4]$ between $\frac{1}{5}$ and $\frac{1}{3}$ . With width $2$ :

\text{lower} = 2\left(\tfrac{1}{3} + \tfrac{1}{5}\right) = \tfrac{16}{15}, \qquad \text{upper} = 2\left(1 + \tfrac{1}{3}\right) = \tfrac{8}{3}.

So $\frac{16}{15} < F(4) - F(0) < \frac{8}{3}$ , and adding $F(0) = 1$ :

\frac{31}{15} < F(4) < \frac{11}{3} \qquad (\approx 2.07 \text{ to } 3.67).

The true value is $F(4) = 1 + \ln 5 \approx 2.61$ .

Solved: 0 / 4

Average Rate Times Time: FTC Beats the Mean Value Theorem

IKey Formulas

IIVisualization 1 — The Chord Slope Is the Average of F′F'F′

IIIVisualization 2 — More Rectangles, Sharper Bounds on F(4)F(4)F(4)

IVQuiz Questions

IIVisualization 1 — The Chord Slope Is the Average of $F'$

IIIVisualization 2 — More Rectangles, Sharper Bounds on $F(4)$