Calculus Third Edition

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4.2.4How can I prove and apply the FTC?

The Fundamental Theorem of Calculus

4-78.

In Lesson 4.2.3, you discovered the Fundamental Theorem of Calculus, which describes the connection between the two major branches of calculus: derivatives and integrals.

Carefully read the Math Note box below. Then, in your own words, explain how each part of the Fundamental Theorem of Calculus illustrates the connection between derivatives and integrals.

4-79.

When applying the Fundamental Theorem of Calculus to evaluate a definite integral like $\int _ { a } ^ { b } f ( x ) d x$ , a common way to show your steps is:

$\left. \begin{array}{l}{ \int _ { a } ^ { b } f ( x ) d x }\\{ = F ( x ) | _ { a } ^ { b } }\\{ = F ( b ) - F ( a ) }\end{array} \right.$

Evaluate $\int _ { 4 } ^ { 9 } ( 2 x + 5 ) d x$ , showing your steps as outlined above.

4-80.

Evaluate the following definite integrals by applying the Fundamental Theorem of Calculus. Check your answers by using the integration function on your calculator.

$\int _ { 1 } ^ { 8 } 2 x ^ { - 3 } d x$

$\int _ { \pi / 4 } ^ { \pi / 2 } \operatorname { sin } ( x ) d x$

$\int _ { 4 } ^ { 9 } 3 \sqrt { x } d x$

$\int _ { 0 } ^ { 2 } ( 3 x ^ { 2 } - 6 x + 2 ) d x$

$2 \int _ { 1 } ^ { 3 } \frac { 3 } { x ^ { 2 } } d x$

$\int _ { - 1 } ^ { 8 } x ^ { 1 / 3 } d x$

$\int _ { 3 } ^ { 9 } ( 2 x - 20 ) d x + 2 \int _ { - 2 } ^ { 3 } ( x - 10 ) d x$

$\int _ { 1 } ^ { 27 } ( 9 m ^ { - 2 / 3 } - 2 m ^ { - 3 } ) d m$

4-81.

Use the Fundamental Theorem of Calculus to evaluate each integral expression.

$\frac { d } { d x } \int _ { 3 } ^ { x } ( 3 t - 5 ) d t$

$\int _ { 3 } ^ { x } ( \frac { d } { d t } ( 3 t - 5 ) ) d t$

$\frac { d } { d x } \int _ { 3 } ^ { 5 } ( 3 x - 5 ) d x$

$\int ( \frac { d } { d x } ( 3 x - 5 ) ) d x$

$\frac { d } { d x } \int \operatorname { cos } ( x ^ { 2 } ) d x$

$\int _ { 1 } ^ { 4 } ( \frac { d } { d x } \operatorname { cos } ( x ^ { 2 } ) ) d x$

Compare your methods and results for the various parts.

4-82.

Katherine is babysitting her calculus teacher’s daughters Natalie, Morgan, and Lydia. Katherine’s pay depends on the rate function $g\left(x\right) = 4x + 3$ where $x$ is the number of hours spent babysitting and $g\left(x\right)$ is the rate of pay in dollars per hour. Her calculus teacher is strict about sending text messages while babysitting and adds a hole to the pay rate function for every instance a message is sent. For example, if Katherine sends a text message two hours into babysitting her new rate function will be $h ( x ) = \frac { ( 4 x + 3 ) ( x - 2 ) } { ( x - 2 ) }$ .

Assume Katherine worked for $0 ≤ x ≤ 3$ hours. Use your calculator’s graphical integration feature to determine the total amount of money Katherine earns for $g\left(x\right) = 4x + 3$ and $h ( x ) = \frac { ( 4 x + 3 ) ( x - 2 ) } { ( x - 2 ) }$ .
Should Katherine worry about losing pay when she sends a text message? Explain your reasoning.
How can this discontinuous situation be correctly represented using integrals?
What would happen if Katherine sent text messages the entire time she was babysitting?

4-83.

Chang Young is attempting to evaluate the following integral: $\int _ { 1 } ^ { 5 } ( 5 x + 2 ) d x$ . Homework Help ✎
He writes the following steps:

$\left. \begin{array} { l } { \frac { 5 } { 2 } 5 ^ { 2 } + 2 ( 5 ) + C - \frac { 5 } { 2 } 1 ^ { 2 } + 2 ( 1 ) + C } \\ { \frac { 5 } { 2 } \cdot 25 + 10 + C - \frac { 5 } { 2 } + 2 + C } \\ { \frac { 125 } { 2 } - \frac { 5 } { 2 } + 12 + 2 C } \\ { \frac { 120 } { 2 } + 12 + 2 C } \\ { 60 + 12 + 2 C } \\ { 72 + 2 C } \end{array} \right.$

He knows that this is a definite integral and there should not be any $C$ s in his answer. Also, the answer key says the answer is $68$ . He needs your help to find his error and figure out how to eliminate his $+ 2C$ .

4-84.

Evaluate each of the following integrals. Homework Help ✎

$\int ( 6 x ^ { 3 } - 2 x + 5 ) d x$
$\int _ { 2 } ^ { 4 } ( 6 x ^ { 3 } - 2 x + 5 ) d x$
$\int ( 9 t ^ { 2 } - 1 ) d t$
$\int _ { - 2 } ^ { 2 } ( 9 t ^ { 2 } - 1 ) d t$
$\int ( \operatorname { sin } ( m ) + \frac { 1 } { 3 } m ^ { 2 } ) d m$
$\int _ { - \pi } ^ { \pi } ( \operatorname { sin } ( m ) + \frac { 1 } { 3 } m ^ { 2 } ) d m$

4-85.

Rewrite each of the following integral expressions as single integrals. Homework Help ✎

$\int _ { - 3 } ^ { - 5 } f ( x ) d x + \int _ { - 5 } ^ { - 3 } g ( x ) d x$
$3 \int _ { 1 } ^ { 6 } f ( x ) d x + 5 \int _ { 1 } ^ { 6 } g ( x ) d x$
$\int _ { 6 } ^ { 11 } f ( x ) d x + \int _ { 11 } ^ { 6 } f ( x ) d x$
$\int _ { 7 } ^ { 10 } f ( t ) d t - \int _ { 7 } ^ { 9 } f ( t ) d t$

4-86.

Review Hanah’s method for setting up a derivative. Use Hanah’s definition of derivative from Lesson 3.2.1 to differentiate $f(x)=4x^2-9x-1.$ Homework Help ✎

4-87.

Differentiate. Homework Help ✎

$\frac{d}{dx}(\sin(x−5))$
$\frac{d}{dx}(x^{82}−6)$
$\frac { d } { d m } ( - 2 m ^ { 3 / 2 } )$
$\frac{d}{dx}(x+6)^3$

4-88.

Without a calculator, describe the graph of $f(x) = x^3 +12x^2 + 36x - 6$ . A complete answer states where $f$ is increasing, decreasing, concave up, concave down, and any points of inflection. Homework Help ✎

Compute without a calculator

4-89.

Write the equation of the line tangent to $f(x)=x^3+12x^2+36x-6$ at its point of inflection. Homework Help ✎

4-90.

Earlier in this chapter, it was discovered that $f ( x ) = \sqrt [ 3 ] { x }$ was not differentiable at $x = 0$ . Homework Help ✎

Why does the derivative of $f ( x ) = \sqrt [ 3 ] { x }$ not exist at $x = 0$ ?
Is $g ( x ) = \sqrt [ 3 ] { x ^ { 2 } }$ differentiable at $x = 0$ ? Why or why not?
Is $h ( x ) = \sqrt [ 3 ] { x ^ { 3 } }$ differentiable at $x = 0$ ? Why or why not?
Explain why there is a point of inflection at $x = 0$ for $f ( x ) = \sqrt [ 3 ] { x }$ .

4-91.

If $\int _ { 2 } ^ { 4 } g ( x ) d x = 6$ , evaluate: Homework Help ✎

$\int _ { 2 } ^ { 2 } ( g ( x ) + 3 ) d x$
$\int _ { 4 } ^ { 2 } ( 3 - 5 g ( x ) ) d x$