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Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch7/

http://matrix.skku.ac.kr/Cal-Book1/Ch8/

http://matrix.skku.ac.kr/Cal-Book1/Ch9/

http://matrix.skku.ac.kr/Cal-Book1/Ch10/

http://matrix.skku.ac.kr/Cal-Book1/Ch11/

http://matrix.skku.ac.kr/Cal-Book1/Ch12/

Chapter 5.

5.1 Areas and Distances   http://youtu.be/mT_oxlD6RSA

문제풀이 by 남택현  http://youtu.be/Y_nCn76RPmY

5.2 The Definite Integral     http://youtu.be/GIm3Oz58Ti8

문제풀이 by 남택현  http://youtu.be/iUsf1h_hTAE

5.3 The Fundamental Theorem of Calculus     http://youtu.be/Zf1HT2H2fbA

문제풀이 by 정승찬 & Kim  http://youtu.be/Pa4Z38KkDVY

5.4 Indefinite Integrals and the Net Change Theorem

5.5 The Substitution Rule     http://youtu.be/h7tmvmNOliU

문제풀이 by 이한울 http://youtu.be/0TMbpCPO4Uc

5.6 The Logarithm Defined as an Integral     http://youtu.be/kD0Z9PqetsA

문제풀이 by 이한울 http://youtu.be/ymDImdIQ90c

미적분학 Midterm Exam    http://youtu.be/QAEI7A2DMMM

5.1 Areas and Distances

Area Problem: The area of a region is computed as the limit of the sum of areas of rectangles, as the number of rectangles become large.

Let be the area of the region that lies under the curve of a continuous function [where ] and bounded by the vertical lines and , and the -axis as in Figure 1.

Figure 1

For regions with straight sides such as a rectangle, the area is defined as the product of the length and the width. The area of a triangle is half of the base times the height. The area of a polygon is found by dividing it into triangles (as in Figure 2) and summing the areas of the triangles.

Figure 2

In the case of area of a region with curved sides, we first approximate the region by rectangles and then take the limit of the sum of areas of these rectangles as the number of rectangles increase.

Consider a general region . Subdivide into strips , , ..., of equal width as in Figure 3 with subintervals

, , , ..., ,

where and , and with the right endpoints of the subintervals given by

, , , ...

Figure 3                        Figure 4

The th strip is approximated by , which is the area of the th rectangle with width and height . (See Figure 4.)

We approximate the area of by the sum of the areas of these rectangles, which is

.

As , this approximation appears to become better and better. Therefore, the area of the region is defined as follows: The area of the region that lies under the graph of the continuous function is the limit of the sum of the areas of approximating rectangles:

1.         .

It can also be shown that we get the same value if we use left endpoints provided the right hand limit exists:

2.         .

Instead of using left endpoints or right endpoints, the height of the th rectangle can be taken as the value of at any sample point in the th subinterval . Figure 5 shows approximating rectangles when the sample points are not chosen to be endpoints.

A more general expression for the area of is

3.

Sums with many terms are written more compactly using sigma notation. For example,

Then, 1., 2. and 3. can be rewritten as

, (on right endpoints)

, (on left endpoints)

. (some specific point in each subinteravls)

Later, these areas, represent the Riemann sum in section 5.2 of right endpoints, left endpoints and midpoints in each subinterval, respectively.

5.1 EXERCISES (Areas and Distances)

1. Find the area under the curve from 0 to 2.

Solution.  We divide the interval into n equal parts. Thus the length of each sub-interval is and the th sub-interval is given by Now we apply the right end formula to find required area.

.

plot(x^5, 0, 2, ymax=33, fill="axis")+plot(x^5, 0, 2.2, ymax=33, color='red')

CAS 2. Find the area of the region under the graph of from 0 to 2.

http://matrix.skku.ac.kr/cal-lab/cal-5-1-exs-2.html

Solution.

var('i,n')

A=sum(2/n*exp(-2*i/n), i, 1, n)

A.simplify()

l=limit(A.simplify(), n=oo)

l.expand()

3. Find the area under the curve from to , where .

Solution.  (Since , the value what we evaluate is equal to the area.)

4. (a) Let be the area of a polygon with equal sides inscribed in a circle with radius .

By dividing the polygon into congruent triangles with central angle , show that

.

(b)Show that . [Hint: Use Equation 3.4.2]

Solution.

(a) The area of one piece of

consists of such pieces so the area of

To estimate area under the graph of , one can take the sample point as the

point such that ,. Similarly we can choose the sample point as the point such that .

(b) Using ,.

(Let . Then implies .)

Think of the lower and upper sums of the graph.

5.2 The Definite Integral  http://youtu.be/GIm3Oz58Ti8

Certain summation formulas will be helpful when using Riemann sums to evaluate definite integrals:

1.

2.

3.

4.

5.      , where is a constant.

6.

7.

DEFINITION1

If is a function defined for , divide the interval into subintervals of equal width . Let , be the endpoints of these sub-intervals and we let be any sample points in these subintervals, so lies in the th sub-interval, . Then the definite integral of from to is

provided the right-hand limit exists. The function is called integrable on if exists.

Note 1  Definition 1 is valid even if has a finite number of discontinuities the limit still exists.

Note 2  The symbol is called an integral sign and was introduced by Leibniz. Here is called the integrand and and are called the limits of integration; is the lower limit and is the upper limit. The in is the variable of integration; is all one symbol. Integration is the process of evaluating an integral.

Note 3  The definite integral is a number; it is independent of the dummy variable . In fact, any letter can be used in place of without changing the value of the integral:

Note 4

B. Riemann (1826~1866)

The sum is called a Riemann sum (after the German mathematician Bernhard Riemann (1826–1866)). If is positive, then the Riemann sum can be interpreted as a sum of areas of appro-ximating rectangles. (See Figure 1.) The definite integral can be interpreted as the area under the curve from to . (See Figure 2.)

If , the Riemann sum  is                                      If , the integral

the sum of areas of rectangles                                  is the area under the curve from to

is an approximation to the net area.                                   is the net area.

EXAMPLE 1

(a) Evaluate the Riemann sum for taking the sample points to be right endpoints and , , and .

(b) Evaluate .

Solution. (a) With the interval width is  and the right endpoints are , , , , , and . So the Riemann sum is

.

.

If takes on both positive and negative values, as in Figure 3, then the Riemann sum is the sum of the areas of the rectangles that lie above the -axis and the negatives of the areas of the rectangles that lie below the -axis (the areas of the gold rectangles minus the areas of the blue rectangles). The limit of such Riemann sums is shown in Figure 4. A definite integral can be interpreted as a net area, that is, a difference of areas:

where is the area of the region above the -axis and below the graph of , and is the area of the region below the -axis and above the graph of .

Note 5  A definite integral can be approximated to within any desired degree of accuracy by a Riemann sum as follows: For every number there is an integer such that

for every integer and for every choice of in .

Note 6  We can use unequal subintervals , , ..., of . We construct the limit so the width of the widest rectangle, , approaches zero. So in this case,

8.                              .

The expression on the left side of this equation is called the definite integral of from to .

In general, we write    .

EXAMPLE 2

(a) Evaluate the Riemann sum for taking the sample points to be right endpoints and , , and .

(b) Evaluate .

http://matrix.skku.ac.kr/cal-lab/cal-5-2-3.html

Solution.

(a) With the interval width is     and the right endpoints are , , , , , and . So the Riemann sum is

.

Notice that is not a positive function and so the Riemann sum does not represent a sum of areas of rectangles. But it does represent the sum of the areas of the gold rectangles (above the -axis) minus the sum of the areas of the blue rectangles (below the -axis) in Figure 5.

(b) With sub-intervals we have

Thus , , , , and, in general,

. Since we are using right endpoints, we can use

2. :

(by Equation 5. with

(by Equations 5 and 7)

.

Since takes on both positive and negative values this integral cannot be interpreted as an area. But it can be interpreted as the difference in areas , where and are shown  in Figure 6.

p1=plot(4-x^2,  x,  0,  2, fill="axis", fillcolor='goldenrod')

p2=plot(4-x^2,  x,  2,  3, fill="axis", fillcolor='blue')

p1+p2

EXAMPLE 3

Evaluate

Solution.

Figure 7

Since is continuous on , we know from Definition 1 that the definite integral exists. Choosing    and

we have

.

EXAMPLE 4

Evaluate the following integrals by interpreting each in terms of areas.

(a)                        (b)

Solution. (a) Since , this integral can be  interpreted  as the area under the curve from to . But, since , we get , which shows that the graph of is the quarter-circle with radius in Figure 8. Therefore        .

(b) The graph of is the line with slope 1 shown in Figure 9. The integral is computed as the area of trapezoid.  .

The Midpoint Rule

In approximating a definite integral by Riemann sums, if we denote the sample points to be the middle points of the subinterval , then the corresponding method of approximation is called the Midpoint Rule.

Midpoint Rule

where and = midpoint of

EXAMPLE 5

Use the Midpoint Rule with to approximate .

Solution.  The endpoints of the five subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0, so the midpoints are 1.1, 1.3, 1.5, 1.7, and 1.9. The width of the subintervals is , so the  Midpoint Rule gives

.

Since for , the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles.     ■

Properties of the Definite Integral

DEFINITION2

If is integrable on , then

.

If is in the domain of , then

.

Properties of the Definite Integral

If and are integrable on and if is a constant, then , , and are integrable on and

1.

2.

3.

4.

Figure 10

From Property 1 we have that the integral of a constant function is the constant times the length of the interval. If and , then is the area of the shaded rectangle in Figure 10.

From Property 3 we have that the integral of a sum is the sum of the integrals. Since the limit of a sum is the sum of the limits, the area under is the area under plus the area under , for positive functions.

.

EXAMPLE 6

Evaluate .

Solution.  Using Properties 2 and 3 of integrals, we have

.

We know from Property 1 that   .

Since , .

The next property tells us how to combine integrals of the same function over adjacent intervals:

Figure 11

If is an integrable function on a closed interval containing the numbers , and , then

5. . (See Figure 11.)

EXAMPLE 7

If and , find .

Solution.  By Property 5, we have  .

So,   .

Comparison Properties of the Integral

Figure 12

6. If for , then .

7. If for , then .

8. If for , then .

(See Figure 12.)

Proof of Property 8. Since , Property 7 gives

.

Using Property 1 to evaluate the integrals on the left and right sides, we obtain

.

EXAMPLE 8

Find a lower bound and an upper bound for the integral   .

Solution.

on

Since is an increasing function on , its absolute maximum value is and its absolute minimum value is . Thus, by Property 8,

.

The integral is greater than the area of the lower small rectangle (square) and less than the area of the large rectangle. (See Figure 13.)

CAS EXAMPLE 9

Evaluate .

http://matrix.skku.ac.kr/cal-lab/cal-5-2-Exm9.html

Solution.

Figure 14

var('x')

f=1/(sin(x)^2)

show(plot(f,  (x, pi/4, pi/3)))

print integral(f, (x, pi/4, pi/3))

print integral(f, (x, pi/4, pi/3)).n()

Answer : -1/3*sqrt(3) + 1  0.422649730810374

5.2 EXERCISES (The Definite Integral)

1-4. Find the Riemann sum by using the Midpoint Rule with the given value of to approximate the integral.

1. ,

Solution.

Let . With the interval width is and midpoints are for . So the Riemann sum is

a=9*sqrt(2)+3*sqrt(6)+14

a.n()

2. ,

Solution. Let . With the interval width is and midpoints are for . So the Riemann sum is

3. ,

Solution.

CAS 4. ,

Solution.  var('i');

f(x)=1/sqrt(1+4*x^2);

sum(f(0.2*i-0.1)*0.2, i, 1, 10)

5-8. Express the limit as a definite integral on the given interval.

5. ,

Solution.  .

6. ,

Solution.  .

7. ,

Solution. .

8.,

Solution.  .

9-18. Determine whether the statement is true or false. If it is true, explain why.

If it is false, give a counterexample.

9. If and are continuous on , then  .

Solution. True by Definition

10. If and are continuous on , then  .

Solution. False: A counterexample is  and .

11. If and are continuous on and for all , then

.

Solution.   False: counterexample. ,

and .

12. If is continuous on , then  .

Solution.   True: Property 2

13. If is continuous on , then  .

Solution. False: Counterexample

14. If is continuous on and , then .

Solution.  False

Let .  Then and .

Hence .

15. If then for all .

Solution.  False.

16. If and are continuous and and then .

Solution.  True: Let , , and    ().

Then.

Because , , ..., .

So .   Therefore .

17. If and are differentiable and for , then for   .

Solution.  False:

18. All continuous functions are integrable.

Solution.  True

19-22. Evaluate the integral. (You should mention which method you are using.)

19.

Solution.  .

var('x');

f(x)=2+5*x

p1=plot(f(x),  (x,  -3,  5))

p2=parametric_plot((-2,  x),  (x,  -14,  30),  linestyle='--')

p3=parametric_plot((4,  x),  (x,  -14,  30),  linestyle='--')

show(p1+p2+p3)

20.

Solution.  .

var('x,  t');

f(x)=3*x^2-2*x+3

p1=plot(f(x),  (x,  -1,  1))+plot(f(x),  (x,  1,  3),  fill="axis")+plot(f(x),(x,  3,  4))

p2=parametric_plot((1,  t),  (t,  0,  45),  linestyle='--')

p3=parametric_plot((3,  t),  (t,  0,  45),  linestyle='--')

show(p1+p2+p3)

print integral(f(x),x,1,3)

21.

Solution.   .

var('x');

f(x)=4-x^2

print integral(f(x),  x,  0,  3)

plot(f(x),  (x,  -1,  0))+plot(f(x),  (x,  0,  3), fill=true)+plot(f(x),(x,3,  5))

22.

Solution.

f(x)=x^(3)-1

print integral(f(x),  x,  1,  2)

plot(f(x),  0,  1,  ymax=2)+plot(f(x),  1,  2,  ymax=10, fill="axis")+plot(f(x),  2,  5/2, ymax=20)

23-26. Evaluate the integral by interpreting it as a sum of the areas.

23.

Solution.

24.

Solution.

25.

Solution.   Let . Then .

Since and , we have .

26.

Solution.  Let . Then .

Since and , we have .

CAS 27. Prove that .

Solution.  By using the endpoint rule,

.

Hence, .

28. Prove that .

Solution.  By using the endpoint rule,

Hence, .

29.If and , find .

Solution.

30. If and , find .

Solution.

31. Find if

Solution.  Since , is continuous.

32-35. Verify the inequality without evaluating the integrals.

32.

Solution.  Since for , we have .

Hence, .

var('x, y')

p1=plot(sin(x)^3, (x, 0, 2), rgbcolor=(1, 0, 0));

p2=plot(sin(x)^2, (x, 0, 2), rgbcolor=(0, 0, 1));

show(p1+p2)

33.

Solution.  Since for ,we obtain .

Hence .

34.

Solution.  Since for , we obtain

.

Hence, .

35.

Solution.  Since and for , we obtain

.

5.3 The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is the most important theorem in calculus and, is ranked as one of the greatest accomplishments of the human mind. The FTC gives the precise inverse relationship between the derivative and the integral, establishing the connection between differential calculus (which arose from the tangent problem) and integral calculus (which arose from the area problem). The two parts of the theorem say that differentiation and integration are inverse processes. Each undoes what the other does. This fact enables us to compute areas and integrals very easily without having to compute them as limits of sums. The second part of the FTC provides a very powerful and much simpler method for evaluating the definite integral of a function.

Antiderivatives

Let be a function defined on an interval . Suppose we can find a function such that . If such a function exists, it is called an antiderivative of . In each case, we are required to find a function, given knowledge about its derivatives.

DEFINITION1

A function is called an antiderivative of on an interval if for all in .

For example, an antiderivative of is , since .

Recall that if two functions have identical derivatives on an interval, then they must differ by a constant. To see this, suppose and are any two antiderivatives of , then . This implies on the interval . Hence is a constant function. Thus .

THEOREM 1  Antiderivaitves Differ by a Constant

If is an antiderivative of on an interval , then the most general antiderivative of on is

,

where is an arbitrary constant. By assigning specific values to the constant , we obtain a family of functions whose graphs are vertical translates of one another.

The Fundamental Theorem of Calculus

We now recall the following facts:

1.If is continuous on , then exists.

2. If is continuous on and , then

.

3. If on , then

1. .

THEOREM 2  Mean Value Theorem for Integration

If is continuous on , then there exists a such that

.

Proof  Since  is continuous on , there exist a maximum and minimum (Extreme Value Theorem)1). So implies

and MVT .

.

By the Mean Value Theorem, there exist in such that

.

Let be a continuous function on . For define a new function by

2. .

If is a fixed number, then the integral is a definite number. If happens to be

a positive function, then can be interpreted as the area under the graph of from to , where can vary from to . Thus is the “area so far” function; (See Figure 2.)

Figure 2

The Fundamental Theorem of Calculus has two parts:

THEOREM 3  The First Part of the Fundamental Theorem of Calculus

If is a continuous function on , then

is continuous on , differentiable on , and satisfies

.

Proof  Let be a point in .

By Mean Value Theorem for integration, there exist in such that

.

Since as and is continuous

■

An alternate notation for the derivative portion is,

Figure 3

3. .

At the end points, has a one-sided derivative, and the same formula holds. That is, the right-handed derivative of at is , and the left-handed derivative of at is .

That is if we first integrate and then differentiate the result, we get back to the original function . This first part of the Fundamental Theorem of Calculus says that our earlier observation is correct. Namely if is a continuous function and , then .

The first part of the Fundamental Theorem of Calculus establishes a remarkable relationship between the operation of differentiation and integration, the two processes initially appear to be unrelated. The second part of the Fundamental Theorem of Calculus is given below.

THEOREM 4  The Second Part of the Fundamental Theorem of Calculus

Let be a continuous function on . Suppose that is any antiderivative of that satisfy Then.

Proof

Figure 4

Consider the function . Since is differentiable on , it is continuous

there (including at the end points, where the one-sided derivatives exist). We also know

that and are differentiable on , and by the first Fundamental Theorem of Calculus their derivatives are equal. Recall that we had (as a consequence of the mean value theorem for derivatives) that and differ by a constant. That is, there is a number such that for all . Then

Visual Proof : http://archives.math.utk.edu/visual.calculus/4/ftc.9/

Because , Part 2 can be rewritten as

.

Recall that is a number, not a function.

The derivative of the function is by part 1 of the FTC

with the non-negative continuous function . So can be

interpreted as the area under the graph of from to , as in Figure 4.

EXAMPLE 1

Find the derivative of the function

Solution.  Since is continuous, the First Part of the Fundamental Theorem of  Calculus gives .

EXAMPLE 2

Find .

Solution.  Let . Then

(by the Chain Rule)

.

EXAMPLE 3

Compute .

Solution.  The function is continuous everywhere. An antiderivative is . So from part 2 of the FTC we have

.

EXAMPLE 4

Determine the area under the parabola from 0 to 2.

Solution.  An antiderivative of is . The required area is found using Part 2 of the FTC.

.

EXAMPLE 5

Find .

Solution. An antiderivative of is since , we can write

. So .

EXAMPLE 6

Compute the area under the sine curve from 0 to , where .

Solution.  Since is an antiderivative of , so

.

With , the area under the sine curve from 0 to is . (See Figure 5.)

Figure 5

CAS EXAMPLE 7

Estimate the value of the integral.

Solution.  var('x')

f=ln(x)^2

show(plot(f,  x, 1, 2))

print integral(ln(x)^2,  x,  1,  2)

print numerical_integral(f,  1,  2)

The integrand : [ ln(x)^2 - 2ln(x) + 2 ] x

(2.0907420860874118e-15, error bound)

Figure 6

5.3 EXERCISES (The Fundamental Theorem of Calculus)

1. Let , where is the function whose graph is shown below.

(a)Evaluate and .

(b)Estimate , and .

(c) On what interval is increasing?

(d) Where does have a maximum value?

(e) Sketch a rough graph of .

(f) Use the given graph to sketch the graph of . Compare with the graph of .

Solution.  (a)

(b)

(c)

(d)

(e) ( is blue, is red)

(f)

2-3. Draw the area represented by . Then find in two ways:

(a)by using Part 1 of the FTC and (b)by evaluating the integral using Part 2 and then differentiating.

2.

Solution.   (a)  (b)

3.

Solution.  (a) (b)

4-7. Find the derivative of the function using part 1 of the FTC.

4.

Solution.

5.    [Hint: ]

Solution.

CAS 6.

Solution.

7.

Solution.

8-10. Evaluate the integral using Part 2 of the FTC.

8.

Solution.

CAS 9.

Solution.

.

10.

Solution.

11. Let . Use Part 1 of the FTC to find .

Solution.  so .

var('t,x ');

g(x)=integral(t*e^(2*t), t, 1, x)

f(x)=diff(g(x), x)

print g(x)

print f(x)

print f(1)

12. Give a non-polynomial function () such that and .

Solution.  For any function set .

Then clearly and so

For example .

13. Let and . Find .

Solution.  and

Hence , so .

14. Let defined on . Find .

Solution.  .  So .

Note  . and for .

Hence should be . Therefore .

15. Let .  Find .

Solution.  Differentiate both sides to get  .

.

Hence,.

16-17. Evaluate the integral and interpret it as a difference of areas.

16.

Solution. = since .

CAS 17.

Solution.

18.  If , where , find .

Solution.    Therefore, .

CAS 19. Find the value of if , is continuous, and .

Solution.

20. If is continuous and and are differentiable functions, find a formula for .

Solution.   .

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5.4 Indefinite Integrals and the Net Change Theorem

Indefinite Integrals

The FTC connects antiderivatives and definite integrals. Part 1 says that if is continuous, then is an antiderivative of . Part 2 says that can be found by evaluating , where is an antiderivative of .

If is an anti-derivative of , then the indefinite integral is

where is an arbitrary constant.

means

Thus an indefinite integral represents an entire family of functions . They are related by the FTC

.

The “richness” of FTC depends on the “availability” of antiderivatives of function.

Table of Indefinite Integrals

, where is a constant.

for nonzero .

, .

Remember that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. A formula for a general indefinite integral is assumed to be valid only on an interval (finite or infinite).

The following example shows that the general antiderivative of the function , , is     , and are constants.

EXAMPLE 1

Compute .

Solution.  We have

.

EXAMPLE 2

Obtain the general indefinite integral .

Solution.  We have

.

EXAMPLE 3

Evaluate .

Solution.

EXAMPLE 4

Find .

Solution.  Rewrite the function and integrate:

.

Applications of The Net Change Theorem

Here represents the rate of change of with respect to and is the change in when changes from to . Although might change in both directions, increase or decrease, represents the net change in . So the FTC 2 can restated in words as follows, the integral of a rate of change is the net change.

1. If is the rate of change of quantity , then

= Net Change over .

2. If gives the velocity of a moving object whose position function is , then  = Displacement over  is the net change of between time and .

3.(a) If an object moves along a straight line with position function , then its velocity is , so

1. is the net change of position, or displacement, of the particle during the time period from to .

(b) To calculate the distance traveled during the time interval,  consider the intervals when (the particle moves to the right) and also the intervals when (the particle moves to the left). In both cases the distance is computed by integrating the speed. Therefore

2.  total distance traveled

Both displacement and distance traveled can be interpreted in terms of areas under a velocity curve. (See Figure 1.)

Figure 1

EXAMPLE 5

A particle moves along a line so that its velocity at time is (measured in meters per second).

(a) Find the displacement of the particle during the time period .

(b) Find the distance traveled during this time period.

Solution.  (a) The displacement is

.

This means that the particle moved 4.5 m toward the left.

(b) Note that and so on the interval and on . Thus, the distance traveled is

.

Note  The unit of measurement for is the product of the units for and the units for .

CAS EXAMPLE 6

Find the general indefinite integral     .

http://matrix.skku.ac.kr/cal-lab/cal-5-4-exm-7.html

Solution.  x=var('x')

f=ln(tan(x))/sin(x)^2

integral(f,x)

Mobile mathematics

5.4 EXERCISES (Indefinite Integrals and the Net Change Theorem)

1-4. Verify by differentiation that the formula is correct.

1.

Solution.  We take derivative of the right hand side and show that it is integrand.

2.

Solution.  We take derivative of the right hand side and show that it is integrand.

CAS 3.

Solution. Let . Then and   (because and )

But

Therefore.

4.

Solution.  Let . Then and      .

Substitute , then .

Therefore, .

5-13. Find the general indefinite integral.

5.

Solution.

6.

Solution.

CAS 7.

Solution.

var('x');

f(x)=x^3+6*x+1

F(x)=integral(f(x),  x)

print F(x)

8.

Solution.

CAS 9.

Solution.

10.

Solution.

11.

Solution.

12.

Solution.

CAS 13.

Solution.

14-22. Evaluate the integral.

14.

Solution.

CAS 15.

Solution.

16.

Solution.

17.

Solution.

CAS 18.

Solution.

19.

Solution.

var('x');

f(x)=(4*x^(2)+2*x+1)/x

print integral(f(x),  x)

print integral(f(x),  x,  1,  e^2)

plot(f(x), (x,  0,  1),

ymax=50)+plot(f(x),  (x,  1,  e^2),

fill=true)+plot(f(x),  (x,  e^2,  10))

CAS 20.

Solution.  integral(sqrt(2/x+2),x,0,2)

21.

Solution.

22.

Solution.

23. Estimate the area of the region that lies under the curve and above the -axis.

Solution.  It is easy to see that the graph has a real root 1.1914.

f(x)=x^3+2*x^7 –2*x^9

c=find_root(f(x),  1,  2);

print integral(f(x),  x,  0,  c)

plot(f(x),  (x,  -0.5,  0),

ymax=2,  ymin=-2)+plot(f(x),  (x,  0,  c),

fill="axis")+plot(f(x),  (x,  c,  2))

24. Water is being added to a tank at a rate of per minute. How much water is added to the tank from to ?

http://matrix.skku.ac.kr/cal-lab/cal-5-4-24.html

Solution.  The amount of water added for is

where  and .

x=var('x')

f(x)=100-4*x-x^2

g= integral(f(x), x)

R=integral(f(x), (x, 1, 6))

show (g)

show (R)

p1=plot(f(x),  0,  7)

p2=parametric_plot((1,  x),  (x,  -14,  105),  linestyle='--')

p3=parametric_plot((6,  x),  (x,  -14,  100),  linestyle='--')

p4=plot(f(x),  1,  6,  ymax=100,  fill="axis")

show(p1+p2+p3+p4)

plot(f(x),1,6,ymax=110,fill="axis")+plot(f(x),1,6,ymax=110,color='blue')

25. Find the area of the region that lies to the right of the -axis and to the left of the parabola .

Solution.  Since the parabola meets the -axis at and , the area is given by .

26. A particle is moving along a line with the velocity function . Find the dis-placement and the distance traveled by the particle during the time .

Solution.  The displacement,

The distance.

27. A honeybee population starts with 30 bees and increases at a rate of bees per week. How many honeybees are there after 10 weeks?

Solution.  Since the net change in population during 10 weeks is , the total number of honeybees after 10 weeks is .

28. The acceleration function (in ) of a particle is given by and the initial velocity is  . Find the velocity of the particle at time and determine the total distance traveled for .

Solution.  Since , the velocity at time is .

And since , .

Therefore the distance traveled for is .

5.5 The Substitution Rule    http://youtu.be/h7tmvmNOliU

The Substitution Rule replaces a relatively complicated integral by a simpler integral, by changing from the original variable to a new variable that is a function of . The main challenge lies in finding an appropriate substitution. At first you can try choosing to be some function in the integrand or some complicated part of the integrand.

For example, to evaluate integral

1. ,  introduce a new variable with . So . Now we can rewrite the given integral as a new integral in the new variable as:

2.  .

A simple check reveals that the above method of “substitution” works well:

.

In general, this method works whenever the given integral can be written in the form . Since , and since by the Chain Rule,

then

3.  .

With the “change of variable” or “substitution” , we have . Then using the Chain Rule for differentiation 3. reduces to

or, alternatively writing

, we get .

The Substitution Rule

If is a differentiable function whose range is an interval and is continuous on , then .

Note  At the final step, make sure to return to the original variable .

EXAMPLE 1

Find .

Solution.  The integrand suggests the substitution . Then . Now using and the Substitution Rule, we have

.

EXAMPLE 2

Find .

Solution.  1.Put . Then , so . Thus, the Substitution Rule gives

.

2.Alternatively take . Then  so .

Therefore,

.

EXAMPLE 3

Find .

Solution.  Let , then .

Therefore, .

EXAMPLE 4

Evaluate .

Solution.  If we multiply the numerator and denominator by then we obtain

.  Put , then . Therefore,

.      ■.

EXAMPLE 5

Find .

Solution. Since , this suggests the substitution . Then .

Since , so this result can also be written as

4.

EXAMPLE 6

Find .

Solution. Put , then .

Therefore,

.

EXAMPLE 7

Evaluate .

Solution. Since , we have

.

The substitution in the substitution rule is not unique. For example refer to example 2.

Using Substitution Rule for evaluating Definite Integrals

A definite integral can be evaluated by substitution, in two ways. One way is to evaluate the indefinite integral first and then use the FTC. Another way is to put everything in terms of the new variable . This will change the integrand and the limits of integration. The new limits of integration are the values of obtained from and .

The Substitution Rule for Definite Integrals

If is continuous on and is continuous on the range of , then

5. .

Proof  Let be an antiderivative of . Then, by   , is an antiderivative of , so by Part 2 of the Fundamental Theorem, we have

But, applying the FTC 2 a second time, we also have

.

EXAMPLE 8

Find .

Solution. Put and . Now when , and when , so the new limits of integration are to . Therefore,

.

EXAMPLE 9

Calculate .

Solution. Put . Then , so . When , and when , . Thus

.

EXAMPLE 10

Find .

Solution.

Figure 1

The integrand suggests using , so then . Now when , ; when , . Thus .

Symmetry

When the graph of a function is symmetric with respect to either the -axis (even function) or the origin (odd function), then the definite integral of on a symmetric interval , that is , can be evaluated by means of a shortcut.

Integrals of Symmetric Functions

(a) Even Function Rule

If is an even integrable function on , then .

(b) Odd Function Rule

If is an odd integrable function on , then.

Proof  Split the given integral into two:

5.

In the first integral on the far right side put . Then and when , . Therefore .

So 5. becomes

6.  .

(a) If is even, then so Equation 6. gives

(b) If is odd, then so Equation 6. gives

.

For the case where is positive and even, part (a) says that the area under from to is twice the area from 0 to because of symmetry.

(a)                             (b)

Recall that an integral can be expressed as the area above the - axis and below , minus the area below the -axis and above . Thus, part (b) says the integral is 0 because the areas cancel. (See Figure 2.)

EXAMPLE 11

Evaluate .

Solution.  In this case is an even function on the symmetric interval . (See Figure 2 (a).).

EXAMPLE 12

Evaluate .

Solution.  In this case is an odd function on the symmetric interval . (See Figure 2 (b).)

var('x')

f=x^3 + x

plot(f)

integral(f, x, -1, 1)

5.5 EXERCISES (The Substitution Rule)

1-3. Evaluate the given indefinite integral using an appropriate substitution.

1.

Solution.  ,

2.

Solution.  ,

3.

Solution. ,

4-10. Find the indefinite integral.

CAS 4.

http://matrix.skku.ac.kr/cal-lab/cal-5-5-4.html

Solution. Let then .

@interact

def _(c=((1,  10,  0.1)))

f(x)=x/(2*x^2+1)

F=diff(f(x),  x)+c

P=plot(F)

show(P)

5.

Solution.

6.

Solution.

CAS 7.

Solution.

8.

Solution.  ,

9.

Solution.

CAS 10.

Solution.

11-23. Evaluate the definite integral, if it exists.

11.

Solution.  (Because is an odd function)

12.

Solution.

CAS 13.

Solution.

CAS 14.

Solution.

CAS 15.

Solution.  4

CAS 16.

Solution.  .  (First of all we substitute .)

17.

Solution.  ,

18.

Solution.  Since , first we have

Then,

(Because is an even function.) Hence

CAS 19.

Solution.  because is odd function.

CAS 20.

Solution.

CAS 21.

http://matrix.skku.ac.kr/cal-lab/cal-5-5-21.html

Solution.

var('x, a')

f(x)=x*sqrt(a^2 – x^2)

integral(f, x)

22.

Solution.

23.

Solution.  because is odd function.

24-25. Find the exact area.

24. ,

Solution.

CAS 25. ,

Solution.  4

26. Let be a function symmetrical with respect to . Show , when .

Solution.  Since .

Hence   because is a function symmetrical with respect to .

5.6 The Logarithm Defined as an Integral

The Natural Logarithm

In this section, first of all we will use the FTC to define as an integral and then define the exponential function as its inverse.

DEFINITION1

The natural logarithmic function is the function defined by . Note that it

exists since is a continuous function. If , then can be interpreted geometrically as the area under the hyperbola from to . (See Figure 1.) For , we have

For , and so is the negative

of the area shown in Figure 2.

Figure 1                   Figure 2

EXAMPLE 1

(a) By comparing areas, show that .

(b) Use the Midpoint Rule with to estimate the value of .

Solution.  (a) We can interpret as the area under the curve from 1 to 2. From Figure 3 we see that this area is larger than the area of rectangle BCDE and smaller than the area of trapezoid ABCD. Thus, we have

(b) If we use the Midpoint Rule with , , and , we get

.

Using part 1 of the FTC we have and so

1.

2. Laws of Logarithms

If and are positive numbers and is a rational number.

1.     2.      3.

Proof  1. Let , where is a positive constant. Then, using 2. and the Chain Rule, we have   .

Therefore, and have the same derivative and so they must differ by a constant, that is, .

Putting in this equation, we get . Thus  .

If we now replace the constant by any number , then we have .

2. Using Law 1 with , we have and so .

Using Law 1 again, we have .

The proof of Law 3 is left as an exercise.

In order to graph , we first determine its limits:

3.  (a)          (b)

Proof  (a) Using Law 3 with and (where is any positive integer), we have . Now , so this shows that as . is an increasing function since its derivative is . Therefore, as .

(b) If we let , then as . Thus, using (a), we have

.

If , then and which shows that is increasing and concave downward on . With this information, we draw the graph of . (See Figure 4.)

Since and is an increasing continuous function that takes on arbitrarily large values, the Intermediate Value Theorem shows that there is a number where ln takes on the value 1. (See Figure 5.) This important number is denoted by .

Figure 4                       Figure 5

DEFINITION2

is the number such that .

The Natural Exponential Function

The natural exponential function is one of the most frequently occurring functions in calculus and its applications. We defined before and can extend continuously to all real value .

Since the logarithm is an increasing function, it is one-to-one and therefore has an inverse function, which we denote by exp. Thus, according to the definition of an inverse function,

4.

and the cancellation equations are

5.  and

In particular, we have

since

since

⇔

We obtain the graph of by reflecting the graph of about the line . (See Figure 6.) The domain of exp is the range of the logarithm. . The range of exp is the domain of log, , which is the domain of the logarithm is the range of exp.

If is any rational number, then the third law of logarithms gives i.e.

.

Therefore, by 4, .

Thus, whenever is a rational number. This leads us to write , even for irrational values of , by the equation

In other words, for the reasons given, we define to be the inverse of the function . In this notation 4 becomes

6.

and the cancellation equations 5 become

7.

8.    for all

Figure 7 The exponential function

Properties of the Exponential Function

The exponential function is an increasing continuous function with domain and range . Thus, for all . Also

So the -axis is a horizontal asymptote of .

We now verify where has the other properties expected of an exponential function.

9     Laws of Exponents

If and are real numbers and is any real number, then

1.             2.              3.

Proof  1. Using the first law of logarithms and Equation 8, we have

Since ln is a one-to-one function, it follows that .

THEOREM 3  Prove that

Proof  The function is differentiable because it is the inverse function of . To find its derivative, we use the inverse function method. Let . Then and differentiating this latter equation implicitly with respect to , we get

.

General Exponential Functions

If and is any rational number, then by 7 and 9, .

Therefore, even for irrational numbers , we define

10

Thus, for instance, .

The function is called the exponential function with base . Notice that is positive for all because is positive for all .

Definition 10 allows us to extend one of the laws of logarithms. We already know that where is rational. But if we now let be any real number we have, from Definition 10,

Thus

11              for any real number

The general laws of exponents follow from Definition 10 together with the laws of exponents for .

12 Properties of Exponential Functions

If and are real numbers and , then

1.                2.

3.                 4.

Proof 1. Using Definition 10 and the properties of exponential functions  , we have

3. Using Equation 11 we obtain

.

The remaining proofs are left as exercises.

The differentiation formula for exponential functions is also a consequence of Definition 10 :

13.

Proof

If , then , so , which shows that is increasing (see Figure 8). If , then and so is decreasing (see Figure 9).

Figure 8                            Figure 9

http://matrix.skku.ac.kr/2011-BK-Coll/coll_20110512.htm

General Logarithmic Functions

If and , then is a one-to-one function. Its inverse function is called the logarithmic function with base and is denoted by .

Thus

14    ⇔

In particular, we see that .

The properties of logarithms are similar to those for the natural logarithm and can be deduced from the corresponding properties of exponential functions.

To differentiate , we write the equation as . From Equation 14 we have , so

Since is a constant, we can differentiate as follows:

15

The Number Expressed as a Limit

In this section is defined as the number such that . It can also be defined as

16

Proof  Let . Then , so . But, by the definition of derivative,

.

Since , we have .

Hence .

5.6 EXERCISES (The Logarithm Defined as an Integral)

CAS 1. (a) By comparing areas, show that .

(b) Use the Midpoint Rule with to estimate .

http://matrix.skku.ac.kr/cal-lab/cal-RiemannSum.html

Solution.  (a)

From above figure, we have .

Since , and ,

we have .(b) Let . Then we get

2.  By comparing areas, show that

Solution.   Note that . Let . If we use the left endpoint rule with subintervals to estimate , then the th height is . If we use the right endpoint rule with subinterval to estimate , the th height is.   Since for both rules is 1 and , we obtain   the desired expressions.

3. (a) By comparing areas, show that .

(b) Deduce that .

Solution.  (a) By similar approach as Example 1,

and .

Hence .

(b) Since is an increasing function and , we have .

Therefore  .

4. Deduce the following laws of logarithms.

(a)

(b)

(c)

Solution. Let and .

Then and .

(a)

(b)

(c)

5. Show that by using an integral.

Solution.  (See exercise 2.)

6. Evaluate

Solution.  .

CAS 7. Evaluate .

Solution.  Let . Then .

Hence we have .

8. Evaluate .

Solution.  The above limit can be written as a definite integral, namely . To evaluate the integral, we first use the substitution . Then

The last integral above is computed as follows:

Hence

9. Find .

Solution. We know that as and as while the second factor approaches , which is 0.

The integral has as its upper limit of integration(Part 1 the FTC) suggesting that differentiation might be involved. Then the denominator reminds us the definition of the derivative with . So this becomes . Thus define then . Now

(By the FTC3)

Note  Alternatively you may use L’Hospital’s Rule.

Calculus