THE INDEfinite INTEGRAL AN INTRODUCTION TO THE INDEFINITE INTEGRAL (PART I) Before you begin this section, I will advise that you review the topic: .ANTIDERIVATIVES So far, we have dealt with definite integrals and we've seen how the Fundamental Theorem relates definite integrals with antiderivatives. Provided that we can find an antiderivative of a function, the definite integral of that function can be easily evaluated. To that effect, we will require a more convenient notation for antiderivatives. This is where the Indefinite Integral comes in. The conventional notation for an indefinite integral has been derived from the fundamental theorem. Let's take a quick flashback:

∫

b a

f(x) dx

=

1

F(b – a)

(Where F is an antiderivative of f, that is, F' = f) If y = f(x), then is

∫

x a

2

f(t) dt

is an antiderivative of F. Based on (1) and (2), we express the indefinite integral as

∫

f(t) dt

Essentially, we are implying that

∫ f(t) dt

=

F(x)

because F'(x) = f(x). Note that there is no interval in the notation for an indefinite integral, which is probably one reason why it's called “indefinite”. That's a personal observation anyway. When we talk about an indefinite integral, what we are really referring to is a GENERAL indefinite integral, which is why we introduce the arbitrary constant C in the evaluation of an indefinite integral. Because the constant C can hold just about any value, we end up with a family of functions; one antiderivative for each constant. Thus, an indefinite integral can also be regarded as a family of functions. When we evaluate a definite integral, we always end up with a NUMBER. My point here is that a definite integral is simply a number, while an indefinite integral is a function/a family of functions. THIS DISTINCTION

IS

CRITICALLY IMPORTANT. There is, however, a connection between the two, which is given by part 2 of the Fundamental Theorem. In the section titled: The Fundamental Theorem – An Introduction, recall that

∫

b

f( x) dx

=

∫ f(t) dt

=

a

F(x)] a

b

3

and now, we are saying that

F(x)

4

Thus, the relationship between the definite integral and indefinite integral can be expressed as

∫

b a

f( x) dx

=

∫

f(x)] a

5

b

To this end, we have come up with a table of indefinite integrals which is basically a modified version of the table of antiderivatives given in page 1 of the section titled: The Fundamental Theorem – Illustrations and Examples. (Additional indefinite integral functions are given in the table on the following page). There's one more concept about antiderivatives we must understand. We generally define the most general antiderivative F of a continuous function f on an interval I as F(x) + C. In other words, the general antiderivative of f on a given interval is F WHICH IS VALID ONLY ON AN INTERVAL. The same is true for an indefinite integral. When we define an indefinite integral, we do so with the understanding that such integral is defined on a specific interval.

Integrand y = f(x)

Indefinite Integral

c ∫ f(x) dx

y = c f(x) y = k

kx + c

y = sin x

– cos x + C

y = sec2 x

tan x + C

y = sec x tan x

sec x + C

y = cosec x cot x

– cosec x + C

y = cosec2 x

– cot x + C

y = cos x

sin x + C

f(x) + g(x)

∫ f(x) xn+1 n+1

y = xn

dx + + C

∫ g(x)

dx

(n ≠ -1)

Table 1: TABLE OF INDEFINITE INTEGRALS

APPLICATIONS OF THE INTEGRAL A vital application of the integral is seen in a theorem known as the Total Change Theorem, which has been derived from FTC 2, the fact that the function F' is the rate of change of y = F(x) with respect to x, and that F(b) – F(a) is the change in y when x changes from a to b. Mathematically, this means

∫

b a

f(x) dx

=

F(b) – F(a)

6

F '(x) dx

=

F(b) – F(a)

7

and since F' = f, we have

∫

b a

This is the idea of the Total Change Theorem , which simply states that the integral of a rate of change is the total change. This definition is illustrated in equation (7). This theorem will also be illustrated in the examples that follow. Please study each example one after the other.

EXAMPLE 1 Verify by differentiation that the formula below is correct.

x

∫

√x2 + 1

dx

=

√x2 + 1

+

C

Solution To verify the equation above, we differentiate the RHS of the equation. If the equation is correct, we should get back the integrand on the LHS. To differentiate

y

=

√x2 + 1

+

C

we use the chain rule. Let u = x2 + 1 , so that

y

√u + C

=

This gives

du/dx = 2x

and so

dy/du = 1/2√u

and results in

dy/dx = du/dx x dy/dx = 2x

x

dy/du 1/2√u

dy/dx = x/√u which equals

dy/dx

=

x √x2 + 1

THIS PROVES THE FORMULA IS CORRECT

EXAMPLE 2 Verify by differentiation that the formula below is correct.

∫

1 x2 √x2 + 1

dx

=

– √x2 + a2 a2 x

+ C

Solution Notice that the RHS of the equation is a rational function, which means we will differentiate using the quotient rule. So, if

f(x)

– √x2 + a2

=

+ C

a2 x

Then, let u(x) = – √x2 + a2 , and v(x) = a2 x . Recall the Quotient Formula:

f ' (x)

v(x) u'(x)

=

–

u(x) v'(x)

[v(x)]

2

Using the chain rule to differentiate u(x), we have

u'(x)

–x √x + a2

=

v'(x)

and

2

=

a2

Therefore,

a 2x f'(x)

–x

–

√x + a2 2

a2

– √x2 + a2

= (a2x) 2

We simplify the numerator to get

f'(x)

– a2x2

=

+ a2(x2 + a2) √x2 + a2

f'(x)

– a2x2

=

f'(x)

f'(x)

x

+ a2x2 +

x

2

a4

=

a4 x2 √x2 + a2

EXAMPLE 3 Verify by differentiation:

x cos x dx

=

x sin x

+

cos x + C

Solution Let f(x)

=

x sin x + cos x + C. Therefore, f'(x)

=

d/dx(x sin x)

+

1 a4 x2

a4 x2

THE FORMULA IS CORRECT

∫

x

1

a4

=

(a2x )2

√x2 + a2

a4 √x + a2

=

1

d/dx(cos x) + d/dx(C)

1 x2 √x2 + 1

f'(x)

(sin x + x cos x)

=

+

(– sin x) + 0

f'(x)

=

sin x + x cos x – sin x

f'(x)

=

sin x – sin x

f'(x)

=

+ x cos x

x cos x

THE FORMULA IS CORRECT.

EXAMPLE 4 Find the general indefinite integral:

∫

x

-¾

dx

Solution Using table 1, we have

∫ ∫ ∫

x

-¾

dx

=

x

-¾

dx

=

x

-¾

dx

=

x¼

x-¾ + 1 -¾ + 1 4 x

=

¼

x¼

4x ¼ + C

EXAMPLE 5 Find the general indefinite integral:

∫ (x

3

+ 6x + 1) dx

Solution We apply the general indefinite integral formula, i.e:

xn+1 n+1

+

C

to each element of the integrand f(x) = x3 + 6x + 1. This gives

∫ (x

3

+ 6x + 1) dx

=

x3+1 3+1

=

¼ x4

=

x4 4

6x1+1 1+1

+ +

3x2

+

+

x

+

6 x2 2

1x0+1 0+1

+

C

+

x 1

Therefore,

∫ (x

3

+ 6x + 1) dx

=

+

¼ x4

3x2

+

x

+

C

EXAMPLE 6 Find the general indefinite integral:

∫

sin x

dx

1 – sin 2x

Solution We need to simplify the integrand as far as possible before its indefinite integral can be calculated. Recall that

sin 2 x + cos 2 x

=

i

1

We plug (i) into the denominator of the integrand to get

∫

sin x sin x + cos 2 x 2

– sin 2 x

dx

Canceling out the like terms in the denominator results in

∫

sin x cos 2 x

dx

We then break the integrand apart to give

∫

1 cos x

sin x cos x

ii

dx

Recall that

1 cos x

=

sec x

and

sin x cos x

=

tan x

Thus, (ii) becomes

∫

(sec x tan x) dx

From the table of indefinite integrals, we have

∫

=

(sec x tan x) dx

sec x

+ C

In other words,

∫

sin x 1 – sin 2x

dx

=

∫

(sec x tan x) dx

=

sec x

+ C

EXAMPLE 7 Find the general indefinite integral:

∫

(sin θ + 3cos θ) dθ

Solution Using the addition rule of integrals, we have

∫ sin θ

+

dθ

∫ cos θ

3

dθ

From the table of indefinite integrals,

∫ sin θ

dθ

+

3

∫ cos θ

dθ

=

– cos θ

=

3sin θ

+ 3sin θ + –

C

cos θ + C

EXAMPLE 8 Find the general indefinite integral. Illustrate by graphing several members of the family on the same screen.

∫

x√ x dx

Solution The integrand can also be expressed as x3/2.So the integral becomes

∫

x 3/2 dx

Using the power rule, we have,

∫

=

x 3/2 dx

=

x(3/2) +1 (3/2) + 1 x5/2

x5/2

=

x

5/2 2/5

So,

∫

x 3/2 dx

=

0.4 x5/2 + C

By assigning specific values to the constant C, what we have is a family of functions whose curves are basically translates of one another. The graphs are shown below. In this case, the assigned values of C are 3, 5, 7, 10, 13, 15, 17, 20, 23, 25, 27, and 30. Observe how each curve translates vertically as the value of C increases.

EXAMPLE 9 Find the general indefinite integral. Illustrate by graphing several members of the family on the same screen.

∫ (cos x

– 2sin x) dx

Solution Using table 1 and the subtraction rule for integrals, we have

∫ (cos x

– 2sin x) dx

=

∫

cos x dx

–

2

=

sin x – 2(– cos x) + C

=

sin x + 2cos x + C

∫

sin x dx

By assigning various values to C, we obtain a family of functions in the graph below. In this case, the assigned values of C are -3, -2, -1, 1, 2, 3, 4, 5, 6, and 7. The thickest curve is that of f(x) = sin x + 2cos x (see the graph above). The others are simply vertical translates.

EXAMPLE 10 Evaluate the integral

∫

1 0

(1 – 2x – 3x2 )

dx

Solution Observe how the integral will be evaluated; the procedure is slightly different from the usual method: The integrand here is f (x) = 1 – 2x – 3x2 . Using the general antidifferentiation formula, we have

F (x)

=

x – x2 – x3

This means

∫

1 0

(1 – 2x – 3x2 )

dx

x – x2 – x3

=

]

1 0

From here we evaluate F at the endpoints 0 and 1:

[(1) – (1)2 – (1)3] – [(0) – (0)2 – (0)3]

=

=

-1

As you can see, we have slightly deviated from the usual method, which, normally should be

∫ =

∫

1 0

1 0

(1)

(1)

dx

dx

–

∫

–

2

1 0

(2x)

∫

1 0

dx

(x) dx

– –

∫

1 0

(3x2)

3

∫

1 0

dx

(x2)

dx

Using the properties of integrals, the answer becomes

1 – 2(1/2) – 3(1/3)

-1

=

See the difference!!! Usually, the integrand is first split into three integrals, which are then evaluated individually to produce an answer. On the other hand, the first step was quite different; we obtain an antiderivative of the integrand, and then evaluate the antiderivative at the endpoints a and b. Clearly, both processes illustrate the relationship between definite and indefinite integrals.

EXAMPLE 11 Evaluate the integral

∫

3 1

1 t2

–

1 t4

dt

Solution For easy evaluation, we express the integral as

∫

3

(t – 2 – t– 4)

1

dt

The antiderivative of the integrand f(t)

=

t–

2

–

∫

(–t – 1) – ( t

F(t)

=

(–1/t ) – ( –1/3 × 1/t3)

–3

dt

=

–1/t + 1/3t3

/-3)

]

Evaluating F(t) at endpoints a = 1 and b = 3 equals

=

[ –1/3

+ 1/3(3)3

=

[ –1/3

+

=

28/81

] – [ –1/1 + 1/3(1) ] 1/81] – [ –1/1 + 1/3] 3

Therefore,

∫

3 1

1 t2

, equals

=

(t – 2 – t– 4)

1

4

F(t)

So, 3

t–

–

1 t4

dt

=

28/81

1 0

EXAMPLE 12 Evaluate the integral

∫1

2

2

t6 – t t4

dt

Solution First, we simplify the integrand to get

2

∫1

t6 t4

t2 t4

–

2

∫1

=

dt

t2

–

Which equals

∫

2

(t 2 – t–2)

1

dt

So, the integrand is f(t)

=

t2 – t –

F(t)

=

F(t)

2

, whose antiderivative is

(t 3/3) – (–1/t)

=

(t 3/3) + (1/t)

This means

∫

2

(t 2 – t–2)

1

dt

t 3/3 + 1/t

=

]

2 1

Which equals

=

[

]

[ 8/3

=

–

(2) 3/3 + 1/(2)

] –

[

[ 1/3

+ 1/2

[ 19/6] – [4/3]

=

]

(1) 3/3 + 1/(1)

]

+ 1

11/6

=

That is

∫

2 1

(t 2 – t–2)

dt

=

11/6

EXAMPLE 13 Evaluate the integral

∫1

2

x2 + 1 √x

dx

1 t2

dt

Solution Again, we have to simplify the integrand before we can proceed any further.

∫

2 x2 + 1 √x 1

dx

∫ (x 2

=

1

(2 - 0.5)

=

∫

+

x (0 - ½)

x2

2 1

)

x0

+

x1/2

∫

=

dx

dx

x1/2

2 1

(x

3/2

+

)

x -0.5

dx

So, we'll be dealing with

∫ (x 2

1

3/2

)

+

x -0.5

∫ (√x 2

OR

dx

1

+

3

1/√x

)

dx

The antiderivative of the integrand is

F(x)

Which means that

∫ (√x 2

1

3

+

1/√x

=

[0.4√(2)

=

[0.4×4√2 +

=

[1.6√2 +

5

=

)

=

0.4√x5 +

=

dx

]

2√2

2√2

]

]

0.4√x5 + 2√x

[ 0.4√(1)

–

+ 2√(2)

5

]

2 1

]

+ 2√(1)

[0.4×1 + 2]

– –

2√x

[2.4]

3.6√2 – 2.4

So,

∫ (√x 2

1

3

+

1/√x

)

dx

=

3.6√2 – 2.4

The value 3.6√2 – 2.4 is the EXACT value of the integral. However, if we want a decimal approximation, we could write 2.691169.

EXAMPLE 14 Evaluate the integral

2

∫1

x

+

1 x

2

dx

Solution First, we expand the integrand

f(x)

=

+

x

2

1 x

to get f(x) = x2 + 2 + x–2 . Hence, the integral becomes

∫ (x

)

2

1

+ 2 + x–2

2

dx

The antiderivative of the integrand is

F(x)

=

(x 3/3)

+

(2x ) + ( –1/x)

F(x)

=

(x 3/3)

+

(2x ) – (1/x)

This means

∫ (x

)

2

1

2

+ 2 + x–2

=

dx

(x 3/3) + (2x ) – (1/x)

]

2 1

Which equals

[([2] /3) + (2[2] ) – (1/[2])] – [([1] /3) + (2[1] ) – (1/[1])] [(8/3) + (4 ) – (/[2)] – [(1/3) + (2) – (1)]

=

3

= =

37/6

3

=

– 4/3

29/6

Therefore,

∫

2 1

+

x

1 x

2

dx

=

29/6

EXAMPLE 15 Evaluate the integral

∫

4 1

√

5 x

dx

Solution We simplify the integrand

f(x)

=

√

5 x

into

f(x)

= √5 × 1/√x

So that the integral becomes

∫

4

(√5

1

× 1/√x) dx

Using the properties of definite integrals, we re-write the integral as

∫ (1/√x) 4

√5

dx

1

So, the integrand is f(x) = 1/√x , whose antiderivative is F(x) = 2√x . Therefore,

√5

∫ (1/√x) 4

1

= =

=

dx

√5

√5(2√(4) –

2√(1))

=

√5((2×2) –

(2×1))

=

2√x

] 14

√5((2×2) –

(2×1))

2√5

In other words,

∫

4 1

√

5 x

dx

=

2√5

EXAMPLE 16 Evaluate the integral

∫

4

√t

1

–

2 √t

dt

Solution The integral can be re-written as

∫ (t 4

1

½

– 2t - ½

The integrand is f(t)

∫ (t 4

1

½

)

dx

= t ½ – 2t - ½ , and its antiderivative is F(t) = (2/3)√t3 – 4√t . This means that

– 2t - ½

)

dx

=

(2/3)√t

3

– 4√t

] 14

This equals

[(2/3)√(4) – 4√(4)] – [(2/3)√(1) – 4√(1)] = [(2/3)(2) – 4(2)] – [(2/3)(1) – 4(1)] = [(8/3) – (8)] – [(2/3) – 4] = 2/3 =

3

3

So,

∫

4

√t

1

2

–

=

dt

√t

2/3

EXAMPLE 17 Evaluate the integral

-2

∫ -5

x4 – 1 x2 + 1

dx

Solution We simplify the integral:

-2

∫ -5

x4 – 1 x2 + 1

dx

-2 (x2 – 1)(x2 + 1)

∫ -5

=

x2 + 1

dx

Which becomes

∫

-2

(x

−5

2

– 1) dx

The integrand is f(x) = x2 – 1, and its antiderivative is F(x) = (x3/3) – x.. This gives

∫

-2 −5

=

(x

– 1) dx

2

[((-2) /3) 3

=

[(–8/3)

=

–2/3

]

–

+ 2

=

(-2)

]

–

–

]

[((-5) /3)

–

3

[(–125/3) =

– (–110/3)

x

(x3/3) –

+ 5

-2 -5

]

(-5)

]

36

Thus,

-2

∫ -5

x4 – 1 x2 + 1

dx

=

36

EXAMPLE 18 Evaluate the integral

∫

π/3 π/6

cosec 2 x

dx

Solution We find that the integrand is f(x) = cosec

2

x, and from the table of antiderivatives, we see that F(x)= – cot x.

This means

∫

π/3

cosec 2 x

π/6

dx

=

– cot x

]π/6π/3

Since

cot x

cos x sin x

=

Then, we have

∫

π/3 π/6

cosec 2 x

=

– –

=

–

dx

½

cos (π/3) = sin (π/6)

½

+

√3/2

–

cos (π/3) sin (π/6)

– √3/2

–

√3/2

–

½

√3/2 ½

=

2√3

So,

∫

π/3 π/6

cosec 2 x

dx

=

2√3

If we choose to rationalize the result, we have

∫

π/3 π/6

cosec 2 x

dx

=

(2√3)/3

Note some textbooks use csc to represent the cosecant identity. In these examples, I use cosec.

EXAMPLE 19 Evaluate the integral

∫

π/2 0

(cos θ + 2sin θ)

dθ

Solution The integrand here is f(θ) = cos θ + 2sin θ, and its antiderivative is

This means

F(θ)

=

sin θ + 2(– cos θ)

F(θ)

=

sin θ – 2cos θ

∫

π/2

(cos θ + 2sin θ)

0

sin θ – 2cos θ

=

dθ

] π/2 0

Which equals

=

[sin (π/2)

– 2cos (π/2)

=

[2

]

–

– 2(0)

[sin (0)

]

–

[0

– 2(1)

]

=

]

– 2cos (0)

3

Therefore,

∫

π/2 0

(cos θ + 2sin θ)

dθ

=

3

EXAMPLE 20 Evaluate the integral

π/4

∫0

1 + cos2θ cos2θ

dθ

Solution We need to simplify the integrand first. Therefore, we have

π/4

∫0

1 cos2θ

cos2θ cos2θ

+

dθ

π/4

∫0

=

1 + cos2θ

1

dθ

and from the basic trigonometric identities, the integral becomes

∫

π/4 0

(sec θ 2

+ 1)

dθ

Thus,the integrand we'll be dealing with is f(θ) = sec2θ + 1 , and its antiderivative is F(θ) = tanθ + θ. This means

=

[tan(π/4)

=

[1

]

+

(π/4)

]

–

+ (π/4)

–

[tan(0)

[0

+ 0

=

∫

]

+

=

]

(0)

1 + (π/4)

So,

π/4

∫0

1 + cos2θ cos2θ

dθ

π/4 0

(sec θ 2

+ 1)

dθ

=

1 + (π/4)

EXAMPLE 21 Evaluate the integral

∫

8 1

x – 1 3 √x2

dx

Solution We simplify the integral to give

∫ ∫

=

8 1

8 1

(x

x – 1 3 √x2

dx

)

–

(1 - 2/3)

x(0 - 2/3

1/3

So, the integrand is f(x) = x

F(x)

)

=

∫

dx

=

-2/3

– x

=

8 1

x x2/3

∫

8 1

x0

– (x

x2/3

–

1/3

dx

x -2/3) dx

, and its antiderivative is

¾ [3√x]4

–

3 [3√x]

This gives

∫

8 1

(x

=

[ [

=

[12

=

–

1/3

x -2/3) dx

¾ [3√(8)]4 ¾ [2]4

– 6

–

=

¾ [3√x]4

]

–

3 [2]

]

–

–

–

[

[¾

]

=

–

]

3 [1] 8.25

Thus,

∫

8 1

x – 1 3 √x2

=

dx

8.25

EXAMPLE 22 Evaluate the integral

∫

3 −2

| x2

– 1 | dx

8 1

–

¾ [3√(1)]4

¾ [1]4

– 3

]

3 [3√x]

[

]

3 [3√(8)]

–

]

3 [3√(1)]

Solution The integrand is involves absolute value. So, based on the definition of absolute value,

f(x)

=

| x – 1| 2

=

x2 – 1

if x ≥ 0

– ( x2 – 1)

if x < 0

Before we proceed, carefully compare the graphs of these two functions. More importantly, take a close look at that part of the curve that lies in the region [-1, 1] (i.e., the region labeled B above.

We can gather from both graphs, that Area under y = x2 – 1

on [-2, 3]

Area under y = |x – 1| 2

on [-2, 3]

=

A–B+C

=

A+B+C

So, based on the definition of the absolute function,

∫

3

| x2

−2

– 1 | dx

=

A + B + C

Where

A

=

∫

B

=

∫

C

∫

=

-1 −2 1 −1

(x

– 1) dx

2

– (x2 – 1) dx

3

(x

−1

– 1) dx

2

After evaluating each integral, we find that

A

=

4/3, B

=

4/3, C

| x2

– 1 | dx

=

20/3

Summing up the three areas gives

∫

3 −2

=

28/3

EXAMPLE 23 Evaluate the integral

∫ | x – x | dx 2

2

−1

Solution The integrand is f(x) = | x – x2 | The graph of f is shown below. From the definition of the absolute function,

f(x)

=

|x –x| 2

=

2

x – x2

if x ≥ 0

– ( x – x2)

if x < 0

Also, from the graph, we see that f(x) ≥ 0 on[0,1] and f(x) ≤ 0 on [-1,0] and [1,2]. This means

∫ | x – x | dx 2

=

2

−1

∫

0 −1

–(x – x2) dx

+

∫

1 0

(x

– x2) dx

When we evaluate each integral on the RHS individually using FTC 2, we have

∫ | x – x | dx 2

−1

2

= =

5/6 + 1/6 + 5/6 11/6

+

∫

2 1

–(x – x2) dx

EXAMPLE 24 Evaluate the integral

∫

2 −1

(x

– 2|x|) dx

Solution From the definition of the absolute function,

f(x)

=

x – 2|x|

=

x – 2|x|

=

x – 2x

if x ≥ 0

x – 2(–x)

if x < 0

x – 2x

if x ≥ 0

x + 2x

if x < 0

Which equals

f(x)

=

The function f(x) = x – 2|x|, is graphed below. From the graph, we see that

f(x)

=

x – 2|x|

=

x – 2x

on [0, 2]

x + 2x

on [-1, 0]

Therefore,

∫

2 −1

(x

– 2|x|) dx

=

∫

2 0

(x

– 2x) dx

+

∫

0 −1

(x

+ 2x) dx

Using FTC 2,

∫

2 −1

(x

– 2|x|) dx

= =

– 2 + (– 1.5) – 3.5

This tutorial is concluded in PART II, where we look at more examples and practical applications of the Indefinite Integral.

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