CC The Second Fundamental Theorem of Calculus

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Section 5.4 The Second Fundamental Theorem of Calculus

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Motivating Questions

How does the integral function $A (x) = \int_{1}^{x} f (t) d t$ define an antiderivative of $f ?$
What is the statement of the Second Fundamental Theorem of Calculus?
How do the First and Second Fundamental Theorems of Calculus enable us to formally see how differentiation and integration are almost inverse processes?

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Supplemental Videos.

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The main topics of this section are also presented in the following videos:

Introduction
¹⁸
unl.yuja.com/V/Video?v=7114356&node=34303251&a=165316552&autoplay=1
Examples Part 1
¹⁹
unl.yuja.com/V/Video?v=7114355&node=34303321&a=140598179&autoplay=1
Examples Part 2
²⁰
unl.yuja.com/V/Video?v=7114354&node=34303210&a=55461230&autoplay=1

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In Section 4.4, we learned the Fundamental Theorem of Calculus (FTC), which from here forward will be referred to as the First Fundamental Theorem of Calculus, as in this section we develop a corresponding result that follows it. Recall that the First FTC tells us that if

f

is a continuous function on

[a, b]

and

F

is any antiderivative of

f

(that is,

F^{'} = f

), then

\int_{a}^{b} f (x) d x = F (b) - F (a) .

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We have used this result in two settings:

If we have a graph of $f$ and we can compute the exact area bounded by $f$ on an interval $[a, b],$ we can compute the change in an antiderivative $F$ over the interval.
If we can find an algebraic formula for an antiderivative of $f,$ we can evaluate the integral to find the net signed area bounded by the function on the interval.

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For the former, see Example 5.1 or Example 5.3. For the latter, we can easily evaluate definite integrals such as

\int_{1}^{4} x^{2} d x,

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since we know that the function

F (x) = \frac{1}{3} x^{3}

is an antiderivative of

f (x) = x^{2} .

Thus,

\begin{aligned} \int_{1}^{4} x^{2} d x & = {\frac{1}{3} x^{3} |}_{1}^{4} \\ = \frac{1}{3} (4)^{3} - \frac{1}{3} (1)^{3} \\ = 21 . \end{aligned}

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Thus, the First FTC can used in two ways. First, to find the difference

F (b) - F (a)

for an antiderivative

F

of the integrand

f,

even if we may not have a formula for

F

itself. To do this, we must know the value of the integral

\int_{a}^{b} f (x) d x

exactly, perhaps through known geometric formulas for area. In addition, the First FTC provides a way to find the exact value of a definite integral, and hence a certain net signed area exactly, by finding an antiderivative of the integrand and evaluating its total change over the interval. In this case, we need to know a formula for the antiderivative

F .

Both of these perspectives are reflected in Figure 5.27.

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Figure 5.27. At left, the graph of $f (x) = x^{2}$ on the interval $[1, 4]$ and the area it bounds. At right, the antiderivative function $F (x) = \frac{1}{3} x^{3},$ whose total change on $[1, 4]$ is the value of the definite integral at left.

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The value of a definite integral may have additional meaning depending on context: as the change in position when the integrand is a velocity function, the total amount of pollutant leaked from a tank when the integrand is the rate at which pollution is leaking, or other total changes if the integrand is a rate function. Also, the value of the definite integral is connected to the average value of a continuous function on a given interval:

f_{AVG [a, b]} = \frac{1}{b - a} \int_{a}^{b} f (x) d x .

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In the last part of Section 5.1, we studied integral functions of the form

A (x) = \int_{c}^{x} f (t) d t .

Figure 5.8 is a particularly important image to keep in mind as we work with integral functions, and the corresponding java applet at gvsu.edu/s/cz

²¹

gvsu.edu/s/cz

can help us understand the function

A .

In what follows, we use the First FTC to gain additional understanding of the function

A (x) = \int_{c}^{x} f (t) d t,

where the integrand

f

is given (either through a graph or a formula), and

c

is a constant.

Example 5.28.

Consider the function

A

defined by the rule

A (x) = \int_{1}^{x} f (t) d t,

where

f (t) = 4 - 2 t .

Compute $A (1)$ and $A (2)$ exactly.
Use the First Fundamental Theorem of Calculus to find a formula for $A (x)$ that does not involve integrals. That is, use the first FTC to evaluate $\int_{1}^{x} (4 - 2 t) d t .$
Observe that $f$ is a linear function; what kind of function is $A ?$
Using the formula you found in (b) that does not involve integrals, compute $A^{'} (x) .$
While we have defined $f$ by the rule $f (t) = 4 - 2 t,$ it is equivalent to say that $f$ is given by the rule $f (x) = 4 - 2 x .$ What do you observe about the relationship between $A$ and $f ?$

Hint.

Plug in 1 and 2 for $x$ in the integral, then use the First Fundamental Theorem of Calculus to solve.
Find an antiderivative for $f (t)$ and then plug in $x$ and 1 using the First FTC.
What is the highest power of $A (x) ?$
Take the derivative of the function you found in part (b).
What does part (d) tell you?

Answer.

$A (1) = 0$ and $A (2) = 1 .$
$A (x) = 4 x - x^{2} - 3$
Quadratic.
$A^{'} (x) = 4 - 2 x$
$f = A^{'}$

Solution.

We want to plug in 1 and 2 for $x$ in the formula $A (x) = \int_{1}^{x} (4 - 2 t) d t :$

$\begin{aligned} A (1) & = \int_{1}^{1} (4 - 2 t) d t \\ = 0, \end{aligned}$

$\begin{aligned} A (2) & = \int_{1}^{2} (4 - 2 t) d t \\ = {(4 t - t^{2}) |}_{1}^{2} \\ = 4 (2) - (2)^{2} - (4 (1) - (1)^{2}) \\ = 1 \end{aligned}$
We will do the same as we did in part (a),

$\begin{aligned} A (x) & = \int_{1}^{x} (4 - 2 t) d t \\ = {(4 t - t^{2}) |}_{1}^{x} \\ = 4 x - x^{2} - (4 (1) - (1)^{2}) \\ = 4 x - x^{2} - 3 \end{aligned}$
$A (x)$ is a polynomial of degree 2, so $A (x)$ is a quadratic function.
Starting with $A (x) = 4 x - x^{2} - 3,$ we take the derivative of each term to find $A^{'} (x) = 4 - 2 x .$
From part (d), we have that $A^{'} (x) = f (x) .$

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Subsection 5.4.1 The Second Fundamental Theorem of Calculus

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The result of Example 5.28 is not particular to the function

f (t) = 4 - 2 t,

nor to the choice of “

1

” as the lower bound in the integral that defines the function

A .

For instance, if we let

f (t) = \cos (t) - t

and set

A (x) = \int_{2}^{x} f (t) d t,

we can determine a formula for

A

by the First FTC. Specifically,

\begin{aligned} A (x) & = \int_{2}^{x} (\cos (t) - t) d t \\ = {\sin (t) - \frac{1}{2} t^{2} |}_{2}^{x} \\ = \sin (x) - \frac{1}{2} x^{2} - (\sin (2) - 2) . \end{aligned}

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Differentiating

A (x),

since

\sin (2) - 2

is a constant, it follows that

A^{'} (x) = \cos (x) - x,

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and thus we see that

A^{'} (x) = f (x);

A

is an antiderivative of

f .

Also, since

A (2) = \int_{2}^{2} f (t) d t = 0,

A

is the only antiderivative of

f

for which

A (2) = 0 .

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In general, if

f

is any continuous function, and we define the function

A

by the rule

A (x) = \int_{c}^{x} f (t) d t,

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where

c

is an arbitrary constant, then we can show that

A

is an antiderivative of

f .

To see why, let’s demonstrate that

A^{'} (x) = f (x)

by using the limit definition of the derivative. Doing so, we observe that

\begin{aligned} A^{'} (x) & = lim_{h \to 0} \frac{A (x + h) - A (x)}{h} \\ = lim_{h \to 0} \frac{\int_{c}^{x + h} f (t) d t - \int_{c}^{x} f (t) d t}{h} \\ (5.3) & = lim_{h \to 0} \frac{\int_{x}^{x + h} f (t) d t}{h}, \end{aligned}

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where Equation (5.3) follows from the fact that

\int_{c}^{x} f (t) d t + \int_{x}^{x + h} f (t) d t = \int_{c}^{x + h} f (t) d t .

Now, observe that for small values of

h,

\int_{x}^{x + h} f (t) d t \approx f (x) \cdot h,

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by a simple left-hand approximation of the integral. Thus, as we take the limit in Equation (5.3), it follows that

A^{'} (x) = lim_{h \to 0} \frac{\int_{x}^{x + h} f (t) d t}{h} = lim_{h \to 0} \frac{f (x) \cdot h}{h} = f (x) .

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Hence,

A

is indeed an antiderivative of

f .

In addition,

A (c) = \int_{c}^{c} f (t) d t = 0 .

The preceding argument demonstrates the truth of the Second Fundamental Theorem of Calculus, which we state as follows.

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The Second Fundamental Theorem of Calculus.

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f

is a continuous function and

c

is any constant, then

f

has a unique antiderivative

A

that satisfies

A (c) = 0,

and that antiderivative is given by the rule

A (x) = \int_{c}^{x} f (t) d t .

Example 5.29.

Suppose that

f

is the function given in Figure 5.30 and that

f

is a piecewise function whose parts are either portions of lines or portions of circles, as pictured.

Figure 5.30. At left, the graph of $y = f (x) .$ At right, axes for sketching $y = A (x) .$

In addition, let

A

be the function defined by the rule

A (x) = \int_{2}^{x} f (t) d t .

What does the Second FTC tell us about the relationship between $A$ and $f ?$
Compute $A (1)$ and $A (3)$ exactly.
Sketch a precise graph of $y = A (x)$ on the axes at right that accurately reflects where $A$ is increasing and decreasing, where $A$ is concave up and concave down, and the exact values of $A$ at $x = 0, 1, \dots, 7 .$
How is $A$ similar to, but different from the function $F$ that you found in Example 5.3?
With as little additional work possible, sketch precise graphs of the functions $B (x) = \int_{3}^{x} f (t) d t$ and $C (x) = \int_{1}^{x} f (t) d t .$ Justify your results with at least one sentence of explanation.

Hint.

If you don’t recall it, review the statement of the Second FTC above.
Note that $A (1) = \int_{2}^{1} f (t) d t .$
Don’t miss our key conclusion from (a).
Compare the values of $A (1)$ and $F (1) .$
What does the Second FTC tell us about the relationship between $B$ and $f ?$

Answer.

$A^{'} (x) = f (x) .$
$A (1) = - \frac{π}{4} .$
$A$ is increasing wherever $f$ is positive; $A$ is CCU wherever $f$ is increasing. $A (2) = 0,$ $A (3) = - 0.5,$ $A (4) = - 1.5,$ $A (5) = - 2,$ $A (6) = - 2 + \frac{π}{4},$ and $A (7) = - 2 + \frac{π}{2} .$
$F$ and $A$ differ by the constant $\frac{π}{4} - \frac{1}{2} .$
$B$ and $C$ have the same shape as $A$ and $F,$ and differ from $A$ by a constant. Observe that $B (3) = 0$ and $C (1) = 0 .$

Solution.

By the Second FTC, $A^{'} (x) = f (x) .$
Since $A (1) = \int_{2}^{1} f (t) d t = - \int_{1}^{2} f (t) d t,$ it follows $A (1) = - \frac{π}{4} .$
Note that $A$ is increasing wherever $f$ is positive, and $A$ is CCU wherever $f$ is increasing. Similar conclusions follow for $A$ being decreasing and/or concave down. Moreover, $A (2) = 0,$ $A (3) = - 0.5,$ $A (4) = - 1.5,$ $A (5) = - 2,$ $A (6) = - 2 + \frac{π}{4},$ and $A (7) = - 2 + \frac{π}{2} .$
In our current example, $A$ is an antiderivative of $f$ that satisfies $A (0) = - \frac{1}{2} - \frac{π}{4} .$ Our earlier work with $F$ showed that $F$ is an antiderivative of $F$ that satisfied $F (0) = - 1 .$ Since $F$ and $A$ are both antiderivatives of $f,$ they differ by a constant, and that constant is $- 1 - (- \frac{1}{2} - \frac{π}{4}) = \frac{π}{4} - \frac{1}{2} .$
The Second FTC tells us that $B^{'} = f$ and $C^{'} = f .$ Thus, $B$ and $C$ are each antiderivatives of $f,$ have the same shape as $A$ and $F,$ and each differ from $A$ by just a constant. Observing that $B (3) = 0$ and $C (1) = 0$ enables us to easily sketch these shifted versions of $A .$

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Subsection 5.4.2 Understanding Integral Functions

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The Second FTC provides us with a way to construct an antiderivative of any continuous function. In particular, if we are given a continuous function

g

and wish to find an antiderivative of

G,

we can now say that

G (x) = \int_{c}^{x} g (t) d t

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provides the rule for such an antiderivative, and moreover that

G (c) = 0 .

Note especially that we know that

G^{'} (x) = g (x),

\begin{matrix} (5.4) & \frac{d}{d x} [\int_{c}^{x} g (t) d t] = g (x) . \end{matrix}

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This result is useful for understanding the graph of

G .

Example 5.31.

Investigate the behavior of the integral function

E (x) = \int_{0}^{x} e^{- t^{2}} d t .

Solution.

E

is closely related to the well known error function

²²

The error function is defined by the rule

\erf (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t

and has the key property that

0 \leq \erf (x) < 1

for all

x \geq 0

and moreover that

lim_{x \to \infty} \erf (x) = 1 .

in probability and statistics. It turns out that the function

e^{- t^{2}}

does not have an elementary antiderivative.

While we cannot evaluate

E

exactly for any value other than

x = 0,

we still can gain a tremendous amount of information about the function

E .

By applying the rule in Equation (5.4) to

E,

it follows that

E^{'} (x) = \frac{d}{d x} [\int_{0}^{x} e^{- t^{2}} d t] = e^{- x^{2}},

so we know a formula for the derivative of

E,

and we know that

E (0) = 0 .

This information is precisely the type we were given in Example 3.7, where we were given information about the derivative of a function, but lacked a formula for the function itself.

Using the first and second derivatives of

E,

along with the fact that

E (0) = 0,

we can determine more information about the behavior of

E .

First, we note that for all real numbers

x,

e^{- x^{2}} > 0,

and thus

E^{'} (x) > 0

for all

x .

Thus

E

is an always increasing function. Further, as

x \to \infty,

E^{'} (x) = e^{- x^{2}} \to 0,

so the slope of the function

E

tends to zero as

x \to \infty

(and similarly as

x \to - \infty

). Indeed, it turns out that

E

has horizontal asymptotes as

x

increases or decreases without bound.

In addition, we can observe that

E^{″} (x) = - 2 x e^{- x^{2}},

and that

E^{″} (0) = 0,

while

E^{″} (x) < 0

for

x > 0

and

E^{″} (x) > 0

for

x < 0 .

This information tells us that

E

is concave up for

x < 0

and concave down for

x > 0

with a point of inflection at

x = 0 .

The only thing we lack at this point is a sense of how big

E

can get as

x

increases. If we use a midpoint Riemann sum with 10 subintervals to estimate

E (2),

we see that

E (2) \approx 0.8822;

a similar calculation to estimate

E (3)

shows little change (

E (3) \approx 0.8862

), so it appears that as

x

increases without bound,

E

approaches a value just larger than

0.886,

which aligns with the fact that

E

has horizontal asymptotes. Putting all of this information together (and using the symmetry of

f (t) = e^{- t^{2}}

), we see the results shown in Figure 5.32.

Figure 5.32. At left, the graph of $f (t) = e^{- t^{2}} .$ At right, the integral function $E (x) = \int_{0}^{x} e^{- t^{2}} d t,$ which is the unique antiderivative of $f$ that satisfies $E (0) = 0 .$

Because

E

is the antiderivative of

f (t) = e^{- t^{2}}

that satisfies

E (0) = 0,

values on the graph of

y = E (x)

represent the net signed area of the region bounded by

f (t) = e^{- t^{2}}

from 0 up to

x .

We see that the value of

E

increases rapidly near zero but then levels off as

x

increases, since there is less and less additional accumulated area bounded by

f (t) = e^{- t^{2}}

x

increases.

Example 5.33.

Suppose that

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

and

F (x) = \int_{0}^{x} f (t) d t .

On the axes at left in Figure 5.34, plot a graph of $f (t) = \frac{t}{1 + t^{2}}$ on the interval $- 10 \leq t \leq 10 .$ Clearly label the vertical axes with appropriate scale.
What is the key relationship between $F$ and $f,$ according to the Second FTC?
Use the first derivative test to determine the intervals on which $F$ is increasing and decreasing.
Use the second derivative test to determine the intervals on which $F$ is concave up and concave down. Note that $f^{'} (t)$ can be simplified to be written in the form $f^{'} (t) = \frac{1 - t^{2}}{(1 + t^{2})^{2}} .$
Using technology appropriately, estimate the values of $F (5)$ and $F (10)$ through appropriate Riemann sums.
Sketch an accurate graph of $y = F (x)$ on the righthand axes provided, and clearly label the vertical axes with appropriate scale.

Figure 5.34. Axes for plotting $f$ and $F .$

Hint.

Use computing technology appropriately to generate the desired graph.
Again, recall the statement of the Second FTC.
Where is $F^{'}$ positive? $F^{'}$ negative?
Note that $F^{″} = f^{'} .$
Remember that $F (5) = \int_{0}^{5} \frac{t}{1 + t^{2}} d t .$
Don’t forget that $F (0) = 0 .$

Answer.

See the plot at below left.
$F^{'} = f .$
$F$ is increasing for all $x > 0;$ $F$ is decreasing for $x < 0$
$F$ is CCU on $- 1 < x < 1$ and CCD for $x < - 1$ and $x > 1 .$
$F (5) \approx 2.35973;$ $F (10) \approx 2.35973 .$
See the graph shown below on the right.

Solution.

See the plot at below left.
$F^{'} = f,$ by the Second FTC.
$F$ is increasing wherever $F^{'} = f$ is positive, so for all $x > 0 .$ Similarly, $F$ is decreasing for $x < 0$
$F$ is concave up wherever $F^{'} = f$ is increasing or wherever $F^{″} = f^{'}$ is positive. It is straightforward to show that $f^{″}$ is positive for $- 1 < x < 1$ and negative otherwise, thus $F$ is concave up on $- 1 < x < 1$ and concave down for $x < - 1$ and $x > 1 .$
$F (5) = \int_{0}^{5} \frac{t}{1 + t^{2}} d t \approx 2.35973,$ using a midpoint Riemann sum with 10 subintervals. Similarly, $F (10) = \int_{0}^{10} \frac{t}{1 + t^{2}} d t \approx 2.35973 .$
Recalling that $F (0) = 0$ and using the values and information we’ve found in (b)-(e), we arrive at the graph at below right.

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Subsection 5.4.3 Differentiating an Integral Function

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We have seen that the Second FTC enables us to construct an antiderivative

F

for any continuous function

f

as the integral function

F (x) = \int_{c}^{x} f (t) d t .

If we have a function of the form

F (x) = \int_{c}^{x} f (t) d t,

then we know that

F^{'} (x) = \frac{d}{d x} [\int_{c}^{x} f (t) d t] = f (x) .

This shows that integral functions, while perhaps having the most complicated formulas of any functions we have encountered, are nonetheless particularly simple to differentiate. For instance, if

F (x) = \int_{π}^{x} \sin (t^{2}) d t,

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then by the Second FTC, we know immediately that

F^{'} (x) = \sin (x^{2}) .

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In general, we know by the Second FTC that

\frac{d}{d x} [\int_{a}^{x} f (t) d t] = f (x) .

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This equation says that “the derivative of the integral function whose integrand is

f,

f .

” We see that if we first integrate the function

f

from

t = a

t = x,

and then differentiate with respect to

x,

these two processes “undo” each other.

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What happens if we differentiate a function

f (t)

and then integrate the result from

t = a

t = x ?

That is, what can we say about the quantity

\int_{a}^{x} \frac{d}{d t} [f (t)] d t ?

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We note that

f (t)

is an antiderivative of

\frac{d}{d t} [f (t)]

and apply the First FTC. We see that

\begin{aligned} \int_{a}^{x} \frac{d}{d t} [f (t)] d t & = {f (t) |}_{a}^{x} \\ = f (x) - f (a) . \end{aligned}

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Thus, we see that if we first differentiate

f

and then integrate the result from

a

x,

we return to the function

f,

minus the constant value

f (a) .

So the two processes almost undo each other, up to the constant

f (a) .

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The observations made in the preceding two paragraphs demonstrate that differentiating and integrating (where we integrate from a constant up to a variable) are almost inverse processes. This should not be surprising; integrating involves antidifferentiating, which reverses the process of differentiating. On the other hand, we see that there is some subtlety involved, because integrating the derivative of a function does not quite produce the function itself. This is because every function has an entire family of antiderivatives, and any two of those antiderivatives differ only by a constant.

Example 5.35.

Evaluate each of the following derivatives and definite integrals. Clearly cite whether you use the First or Second FTC in so doing.

$\frac{d}{d x} [\int_{4}^{x} e^{t^{2}} d t]$
$\int_{- 2}^{x} \frac{d}{d t} [\frac{t^{4}}{1 + t^{4}}] d t$
$\frac{d}{d x} [\int_{x}^{1} \cos (t^{3}) d t]$
$\int_{3}^{x} \frac{d}{d t} [\ln (1 + t^{2})] d t$
$\frac{d}{d x} [\int_{4}^{x^{3}} \sin (t^{2}) d t]$ .

Hint.

Second FTC.
First FTC.
$\int_{x}^{1} g (t) d t = - \int_{1}^{x} g (t) d t .$
First FTC.
Let $F (x) = \int_{4}^{x} \sin (t^{2}) d t$ and observe that this problem is asking you to evaluate $\frac{d}{d x} [F (x^{3})]] .$

Answer.

$\frac{d}{d x} [\int_{4}^{x} e^{t^{2}} d t] = e^{x^{2}} .$
$\int_{- 2}^{x} \frac{d}{d t} [\frac{t^{4}}{1 + t^{4}}] d t = \frac{x^{4}}{1 + x^{4}} - \frac{8}{9} .$
$\frac{d}{d x} [\int_{x}^{1} \cos (t^{3}) d t] = - \cos (x^{3}) .$
$\int_{3}^{x} \frac{d}{d t} [\ln (1 + t^{2})] d t = \ln (1 + x^{2}) - \ln (4) .$
$\frac{d}{d x} [\int_{4}^{x^{3}} \sin (t^{2}) d t] = \sin (x^{6}) \cdot 3 x^{2} .$

Solution.

By the Second FTC, $\frac{d}{d x} [\int_{4}^{x} e^{t^{2}} d t] = e^{x^{2}} .$
By the First FTC, $\int_{- 2}^{x} \frac{d}{d t} [\frac{t^{4}}{1 + t^{4}}] d t = {[\frac{t^{4}}{1 + t^{4}}] |}_{- 2}^{x} = \frac{x^{4}}{1 + x^{4}} - \frac{16}{17} .$
Since $\int_{x}^{1} g (t) d t = - \int_{1}^{x} g (t) d t,$ it follows by this fact and the Second FTC that $\frac{d}{d x} [\int_{x}^{1} \cos (t^{3}) d t] = - \frac{d}{d x} [\int_{1}^{x} \cos (t^{3}) d t] = - \cos (x^{3}) .$
By the First FTC, $\int_{3}^{x} \frac{d}{d t} [\ln (1 + t^{2})] d t = {\ln (1 + t^{2}) |}_{3}^{x} = \ln (1 + x^{2}) - \ln (10) .$
Letting $F (x) = \int_{4}^{x} \sin (t^{2}) d t$ it follows that we need to compute $\frac{d}{d x} [F (x^{3})]] .$ By the Chain Rule, $\frac{d}{d x} [F (x^{3})]] = F^{'} (x^{3}) \cdot 3 x^{2} .$ By the Second FTC, we know that $F^{'} (x) = \sin (x^{2}),$ and thus $\frac{d}{d x} [\int_{4}^{x^{3}} \sin (t^{2}) d t] = \frac{d}{d x} [F (x^{3})]] = \sin (x^{6}) \cdot 3 x^{2} .$

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Subsection 5.4.4 Summary

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For a continuous function $f,$ the integral function $A (x) = \int_{1}^{x} f (t) d t$ defines an antiderivative of $f .$
The Second Fundamental Theorem of Calculus is the formal, more general statement of the preceding fact: if $f$ is a continuous function and $c$ is any constant, then $A (x) = \int_{c}^{x} f (t) d t$ is the unique antiderivative of $f$ that satisfies $A (c) = 0 .$
Together, the First and Second FTC enable us to formally see how differentiation and integration are almost inverse processes through the observations that

$\int_{c}^{x} \frac{d}{d t} [f (t)] d t = f (x) - f (c)$

and

$\frac{d}{d x} [\int_{c}^{x} f (t) d t] = f (x) .$

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Exercises 5.4.5 Exercises

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1. A definite integral starting at 3.

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Let

g (x) = \int_{3}^{x} f (t) d t,

where

f (t)

is given in the figure below.

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(Click on the graph for a larger version.)

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Find each of the following:

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g (3) =

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g^{'} (4) =

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C. The interval (with endpoints given to the nearest 0.25) where

g

is concave down:

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interval =

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(Give your answer as an interval or a list of intervals, e.g., (-infinity,8] or (1,5),(7,10), or enter nonefor no intervals.)

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D. The value of

x

where

g

takes its maximum on the interval

0 \leq x \leq 8 .

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x =

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2. Variable in the lower limit.

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Find the derivative:

\frac{d}{d x} \int_{x}^{a} \tan (\ln (t)) d t =

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3. Approximating a function with derivative $e^{- \frac{x^{2}}{5}}$ .

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Find a good numerical approximation to

F (3)

for the function with the properties that

F^{'} (x) = e^{- x^{2} / 5}

and

F (0) = 4 .

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F (3) \approx

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4. Sketching an antiderivative function based on definite integral values.

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Let

g

be the function pictured at left in Figure 5.36, and let

F

be defined by

F (x) = \int_{2}^{x} g (t) d t .

Assume that the shaded areas have magnitudes

A_{1} = 4.29,

A_{2} = 12.75,

A_{3} = 0.36,

and

A_{4} = 1.79 .

Assume further that the portion of

A_{2}

that lies between

x = 0.5

and

x = 2

6.06 .

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Sketch a carefully labeled graph of

F

on the axes provided, and include a written analysis of how you know where

F

is zero, increasing, decreasing, CCU, and CCD.

Figure 5.36. At left, the graph of $g .$ At right, axes for plotting $F .$

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5. Sand on the Beach.

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The tide removes sand from the beach at a small ocean park at a rate modeled by the function

R (t) = 2 + 5 \sin (\frac{4 π t}{25})

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A pumping station adds sand to the beach at rate modeled by the function

S (t) = \frac{15 t}{1 + 3 t}

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Both

R (t)

and

S (t)

are measured in cubic yards of sand per hour,

t

is measured in hours, and the valid times are

0 \leq t \leq 6 .

At time

t = 0,

the beach holds 2500 cubic yards of sand.

What definite integral measures how much sand the tide will remove during the time period $0 \leq t \leq 6 ?$ Why?
Write an expression for $Y (x),$ the total number of cubic yards of sand on the beach at time $x .$ Carefully explain your thinking and reasoning.
At what instantaneous rate is the total number of cubic yards of sand on the beach at time $t = 4$ changing?
Over the time interval $0 \leq t \leq 6,$ at what time $t$ is the amount of sand on the beach least? What is this minimum value? Explain and justify your answers fully.

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6. Altitude Changes.

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When an aircraft attempts to climb as rapidly as possible, its climb rate (in feet per minute) decreases as altitude increases, because the air is less dense at higher altitudes. Given below is a table showing performance data for a certain single engine aircraft, giving its climb rate at various altitudes, where

c (h)

denotes the climb rate of the airplane at an altitude

h .

Table 5.37. Data for the climbing aircraft.

$h$ (feet)	$0$	$1000$	$2000$	$3000$	$4000$	$5000$	$6000$	$7000$	$8000$	$9000$	$10,000$
$c$ (ft/min)	$925$	$875$	$830$	$780$	$730$	$685$	$635$	$585$	$535$	$490$	$440$

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Let a new function

m,

that also depends on

h,

(say

y = m (h)

) measure the number of minutes required for a plane at altitude

h

to climb the next foot of altitude.

Determine a similar table of values for $m (h)$ and explain how it is related to the table above. Be sure to discuss the units on $m .$
Give a careful interpretation of a function whose derivative is $m (h) .$ Describe what the input is and what the output is. Also, explain in plain English what the function tells us.
Determine a definite integral whose value tells us exactly the number of minutes required for the airplane to ascend to 10,000 feet of altitude. Clearly explain why the value of this integral has the required meaning.
Determine a formula for a function $M (h)$ whose value tells us the exact number of minutes required for the airplane to ascend to $h$ feet of altitude.
Estimate the values of $M (6000)$ and $M (10000)$ as accurately as you can. Include units on your results.

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Section 5.4 The Second Fundamental Theorem of Calculus

Motivating Questions

Supplemental Videos.

Example 5.28.

Subsection 5.4.1 The Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus.

Example 5.29.

Subsection 5.4.2 Understanding Integral Functions

Example 5.31.

Example 5.33.

Subsection 5.4.3 Differentiating an Integral Function

Example 5.35.

Subsection 5.4.4 Summary

Exercises 5.4.5 Exercises

1. A definite integral starting at 3.

2. Variable in the lower limit.

3. Approximating a function with derivative e−x25.

4. Sketching an antiderivative function based on definite integral values.

5. Sand on the Beach.

6. Altitude Changes.

3. Approximating a function with derivative $e^{- \frac{x^{2}}{5}}$ .