TBIL-Cal Average Value (AI1)

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Section 6.1 Average Value (AI1)

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Learning Outcomes

Compute the average value of a function on an interval.

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Subsection 6.1.1 Activities

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Activity 6.1.1.

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Suppose a car drives due east at 70 miles per hour for 2 hours, and then slows down to 40 miles per hour for an additional hour.

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(a)

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How far did the car travel in these 3 hours?

$110$ miles
$150$ miles
$180$ miles
$220$ miles

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(b)

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What was its average velocity over these 3 hours?

$55$ miles per hour
$60$ miles per hour
$70$ miles per hour
$75$ miles per hour

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Activity 6.1.2.

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Suppose instead the car starts with a velocity of

30

miles per hour, and increases velocity linearly according to the function

v (t) = 30 + 20 t

so its velocity after three hours is

90

miles per hour.

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(a)

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How can we model the car’s distance traveled using calculus?

Integrate velocity, because position is the rate of change of velocity.
Integrate velocity, because velocity is the rate of change of position.
Differentiate velocity, because position is the rate of change of velocity.
Differentiate velocity, because velocity is the rate of change of position.

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(b)

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Then, which of these expressions is a mathematical model for the car’s distance traveled after 3 hours?

$\int (30 + 20 t) d t$
$\int (30 t + 10 t^{2}) d t$
$\int_{0}^{3} (30 + 20 t) d t$
$\int_{0}^{3} (30 t + 10 t^{2}) d t$

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(c)

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How far did the car travel in these 3 hours?

$110$ miles
$150$ miles
$180$ miles
$220$ miles

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(d)

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Thus, what was its average velocity over three hours?

$55$ miles per hour
$60$ miles per hour
$70$ miles per hour
$75$ miles per hour

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Observation 6.1.3.

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To obtain the average velocity of an object traveling with velocity

v (t)

for

a \leq t \leq b,

we may find its distance traveled by calculating

\int_{a}^{b} v (t) .

Thus, the average velocity is obtained by dividing by the time

b - a

elapsed:

\frac{1}{b - a} \int_{a}^{b} v (t) d t .

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For example, the following calculuation confirms the previous activity:

\frac{1}{3 - 0} \int_{0}^{3} (30 + 20 t) d t .

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Definition 6.1.4.

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Given a function

f (x)

defined on

[a, b],

it’s average value is defined to be

\frac{1}{b - a} \int_{a}^{b} f (x) d x .

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Activity 6.1.5.

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(a)

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Which of the following expressions represent the average value of

f (x) = - 12 x^{2} + 8 x + 4

over the interval

[- 1, 2] ?

$\frac{1}{3} \int_{- 1}^{2} (- 12 x^{2} + 8 x + 4) d x$
$\frac{- 1}{1} \int_{1}^{2} (- 12 x^{2} + 8 x + 4) d x$
$\frac{1}{2} \int_{1}^{2} (- 12 x^{2} + 8 x + 4) d x$
$\frac{- 1}{4} \int_{- 1}^{2} (- 12 x^{2} + 8 x + 4) d x$

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(b)

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Show that the average value of

f (x) = - 12 x^{2} + 8 x + 4

over the interval

[- 1, 2]

- 4 .

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Activity 6.1.6.

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(a)

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Which of the following expressions represent the average value of

f (x) = x \cos (x^{2}) + x

on the interval

[π, 4 π] ?

$\frac{1}{3 π} \int_{0}^{4 π} (x \cos (x^{2}) + x) d x$
$\frac{1}{4 π} \int_{0}^{4 π} (x \cos (x^{2}) + x) d x$
$\frac{1}{3 π} \int_{π}^{4 π} (x \cos (x^{2}) + x) d x$
$\frac{1}{4 π} \int_{π}^{4 π} (x \cos (x^{2}) + x) d x$

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(b)

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Find the average value of

f (x) = x \cos (x^{2}) + x

on the interval

[π, 4 π]

using the chosen expression.

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Activity 6.1.7.

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Find the average value of

g (t) = \frac{t}{t^{2} + 1}

on the interval

[0, 4] .

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Activity 6.1.8.

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A shot of a drug is administered to a patient and the quantity of the drug in the bloodstream over time is

q (t) = 3 t e^{- 0.25 t},

where

t

is measured in hours and

q

is measured in milligrams. What is the average quantity of this drug in the patient’s bloodstream over the first 6 hours after injection?

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Activity 6.1.9.

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Which of the following is the average value of

f (x)

over the interval

[0, 8] ?

A plot of f(x). — Figure 124. Plot of $f (x) .$

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Note

f (x) = {\begin{cases} 1, & 0 \leq x \leq 3 \\ 4, & 3 < x \leq 6 \\ 2, & 6 < x \leq 8 \end{cases} .

$4$
$2$
$\frac{7}{3}$
$19$
$2.375$

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Subsection 6.1.2 Videos

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Figure 125. Video: Compute the average value of a function on an interval

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Subsection 6.1.3 Exercises

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Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/AI1/

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