Chapter 5 Integrationmathstat.sci.tu.ac.th/~archara/MA111/MA111-118/... · Deﬁnition 5.1. For...

Chapter 5

Integration

5.1 Antiderivatives; The Indefinite Integral

Antiderivatives

A function F is called an antiderivative of f on an interval I if F ′(x) = f(x)for all x in I.

Definition 5.1.

For instance, let f(x) = x2. It isn’t difficult to discover an antiderivative F (x) = 13x3

because F ′(x) = x2 = f(x). But the function G(x) = 13x3+100 also satisfies G′(x) = x2.

Therefore, both F and G are antiderivatives of f . Indeed, any function of the formH(x) = 1

3x3 + C, where C is a constant, is an antiderivative of f .

Question! Are there any others?

Answer. No.

Thus, if F and G are any two antiderivatives of f , then

F ′(x) = f(x) = G′(x)

so G(x)− F (x) = C, where C is a constant. We can write this as G(x) = F (x) + C, sowe have the following result.

If F is an antiderivative of f on an interval I, then the most general antideriva-

tive of f on I isF (x) + C

where C is an arbitrary constant.

Theorem 5.1.

112

MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 113

The Indefinite Integral

The process of finding antiderivatives is called antidifferentiation or integration.Thus, if

d

dx[F (x)] = f(x) (5.1)

then integrating (or antidifferentiating) the function f(x) produces an antiderivative ofthe form F (x) + C. To emphasize this process, Equation (5.1) is recast using integral

notation.∫

f(x) dx = F (x) + C (5.2)

where C is an arbitrary constant. For example,

∫

x2 dx = 13x3 + C is equivalent to

d

dx[13x3] = x2

Note that if we differentiate an antiderivative of f(x), we obtain f(x) back again. Thus,

d

dx

[∫

f(x) dx

]

= f(x) (5.3)

The expression∫

f(x) dx is called an indefinite integral. The “elongated s” thatappears on the left side of (5.2) is called an integral sign, the function f(x) is calledthe integrand, and the constant C is called the constant of integration.

The differential symbol, dx, in the differentiation and antidifferentiation operations

d

dx[ ] and

∫

[ ] dx

serves to identify the independent variable. If an independent variable other than x isused, say t, then the notation must be adjusted appropriately. Thus,

d

dx[F (t)] = f(t) and

∫

[f(t)] dx = F (t) + C

are equivalent statements. Here are some examples of derivative formulas and theirequivalent integration formulas:

Derivative EquivalentFormula Integration Formula

d

dx[x3] = 3x2

∫

3x2 dx = x3 + C

d

dx[√x] =

1

2√x

∫

1

2√xdx =

√x+ C

d

dt[tan t] = sec2 t

∫

sec2 t dt = tan t+ C


Integration Formulas

Some of the most important integration formulas are given in the following Table.

∫

xn dx =xn+1

n+ 1+ C (n 6= −1)

∫

1

xdx = ln |x|+ C

∫

ex dx = ex + C

∫

ax dx =ax

ln a+ C, a > 0, a 6= 1

∫

sin x dx = − cos x+ C

∫

cosx dx = sin x+ C∫

tanx dx = ln | sec x| + C

∫

cotx dx = ln | sinx| + C∫

sec x dx = ln | sec x+ tan x|+ C

∫

csc x dx = − ln | csc x+ cotx|+ C∫

sec2 x dx = tan x+ C

∫

csc2 x dx = − cot x+ C∫

sec x tan x dx = sec x+ C

∫

csc x cot x dx = − csc x+ C∫

1√1− x2

dx = sin−1 x+ C

∫

1

1 + x2dx = tan−1 x+ C

= − cos−1 x+ C = − cot−1 x+ C∫

1

|x|√x2 − 1

dx = sec−1 x+ C

∫

sinh x dx = cosh x+ C

= − csc−1 x+ C∫

cosh x dx = sinh x+ C

∫

sech 2x dx = tanhx+ C∫

csch 2x dx = − coth x+ C

∫

sech x tanh x dx = −sech x+ C∫

csch x coth x dx = −csch x+ C

∫

1√1 + x2

dx = sinh−1 x+ C∫

1√x2 − 1

dx = cosh−1 x+ C

∫

1

1− x2dx = tanh−1 x+ C, |x| < 1

= coth−1 x+ C, |x| > 1∫

1

x√1− x2

dx = −sech −1x+ C

∫

1

|x|√x2 + 1

dx = −csch −1x+ C

Properties of the Indefinite Integral

Our first properties of antiderivatives follow directly from the simple constant factor,sum, and difference rules for derivative.


Let f and g have antiderivatives (indefinite integrals) and let c be a constant.Then

(i)

∫

cf(x) dx = c

∫

f(x) dx

(ii)

∫

[

f(x) + g(x)]

dx =

∫

f(x) dx+

∫

g(x) dx

(iii)

∫

[

f(x)− g(x)]

dx =

∫

f(x) dx−∫

g(x) dx

Theorem 5.2.

Evaluate

∫

(3ex + 5x2/3) dx.

Example 5.1.

Solution

Evaluate

∫

(3 cosx+ 2 sec2 x) dx.

Example 5.2.

Solution


Evaluate

(a)

∫

sin 2x

cosxdx (b)

∫

2t4 − t2√t− 1

t4dt

Example 5.3.

Solution

Exercise 5.1

1. Find the most general antiderivative of the function.

(a) f(x) = 5 (b) f(x) = x2 + π

(c) f(x) = x5/4 (d) f(x) = 1/3√x2

(e) f(x) = x2 − x (f) f(x) = 4x5 − x3

(g) f(x) = 27x7 + 3x5 − 45x3 +√2x

(h) f(x) =3

x2− 2

x3(i) f(x) =

4x6 + 3x4

x3


2. Evaluate the integral and check your answer by differentiating.

(a)

∫

3x4 dx (b)

∫

(x2 + x) dx

(c)

∫

(3x4 − 3x) dx (d)

∫

(x+ 1)2 dx

(e)

∫

3√x dx (f)

∫(

3− 1

x4

)

dx

(g)

∫

x1/3 − 3

x2/3dx (h)

∫

(x2 + 1)2√x

dx

(i)

∫

(sin x− cosx) dx (j)

∫

2 sec x tan x dx

(k)

∫

5 sec2 x dx (l)

∫

(3ex − 2) dx

(m)

∫

(3 cosx− 1/x) dx (n)

∫(

5x− 3

ex

)

dx

(o)

∫

ex + 3

exdx (p)

∫

x1/4(x5/4 − 4) dx

Answer to Exercise 5.1

1. (a) 5x+ C (b) 13x3 + πx+ C (c) 4

9x9/4 + C (d) 3 3

√x+ C (e) 1

3x3 − 1

2x2 + C

(f) 23x6 − 1

4x4 + C (g) 27

8x8 + 1

2x6 − 45

4x4 +

√22x2 + C (h) − 3

x+ 1

x2 + C

(i) x4 + 32x2 + C

2. (a) 35x5 + C (b) 1

3x3 + x2

2+ C (c) 3

5x5 − 3

2x2 + C (d) 1

3(x+ 1)3 + C

(e) 2x3/2+C (f) 3x+ 13x−3+C (g) 3

2x2/3−9x1/3+C (h) 2

9x9/2+ 4

5x5/2+2x1/2+C

(i) − cos x− sin x+ C (j) 2 sec x+ C (k) 5 tanx+ C (l) 3ex − 2x+ C

(m) 3 sinx− ln |x|+C (n) 52x2 + 3e−x +C (o) x− 3e−x +C (p) 2

5x5/2 − 16

5x5/4

5.2 Integration by Substitution

In this section we will study a technique, called substitution, that can often be used totransform complicated integration problems into simpler ones.

u-substitution

The method of substitution can be motivated by the examining the chain rule from theviewpoint of antidifferentiation. For this purpose, suppose that F is an antiderivativeof f and that g is differentiable function. The chain rule implies that the derivative ofF (g(x)) can be expressed as

d

dx[F (g(x))] = F ′(g(x))g′(x)


which we can write in integral form as∫

F ′(g(x))g′(x) dx = F (g(x)) + C (5.4)

or since F is an antiderivative of f ,∫

f(g(x))g′(x) dx = F (g(x)) + C (5.5)

For our purpose it will be useful to let u = g(x) and to write du/dx = g′(x) in thedifferential form du = g′(x) dx. With this notation (5.5) can be express as

∫

f(u) du = F (u) + C (5.6)

The process of evaluating an integral of form (5.5) by converting it into form (5.6) withthe substitution

u− g(x) and du = g′(x) dx

is called the method of u-substitution. The following example illustrates how themethod works.

Evaluate

∫

3x2(x3 + 1)5 dx.

Example 5.4.

Solution

In general, there are no hard and fast rules for choosing u, and in some problems nochoice of u will work. In such cases other methods need to be used.

Guideline for u-Substitution

Step 1. Make a choice for u, say u = g(x).

Step 2. Compute du/dx.


Step 3. Make the substitution u = g(x) and du = g′(x) dx.

At this stage, the entire integral must be in terms of u; no x’s should remain. If this isnot the case, try a different choice of u.

Step 4. Evaluate the resulting integral.

Step 5. Replace u by g(x), so the final answer is in terms of x.

Evaluate

∫

x4 sin(x5 − 3) dx.

Example 5.5.

Solution

Evaluate

∫

cos 4x√2− sin 4x dx.

Example 5.6.

Solution


Evaluate

∫

x√4− 5x2

dx.

Example 5.7.

Solution

Evaluate

∫

sec2√x√

xdx.

Example 5.8.

Solution

Evaluate

∫

5√x2 + 1 · x5 dx.

Example 5.9.

Solution


Exercise 5.2

1. Evaluate the integral by making the given substitution.

(a)

∫

cos 3x dx, u = 3x

(b)

∫

x2(x3 + 2) dx, u = x3 + 2

(c)

∫

4

(1 + 2x)3dx, u = 1 + 2x

(d)

∫

(√x+ 2)3√

xdx, u =

√x+ 2

2. Evaluate the indefinite integral

(a)

∫

2x(x2 + 3)4 dx (b)

∫

(2x+ 1)(x2 + x)3 dx

(c)

∫ √x− 1 dx (d)

∫

x2

√x3 − 2

dx

(e)

∫

dx

5− 3x(f)

∫

2x+ 1

x2 + x− 1dx

(g)

∫

1 + 4x√1 + x+ 2x2

dx (h)

∫

1√x(√x+ 1)2

dx

(i)

∫

cos 2x dx (j)

∫

cosx√sin x+ 1 dx

(k)

∫

x sin(x2) dx (l)

∫

sin x√cosx

dx

(m)

∫

cos4 x sin x dx (n)

∫

sin x(cos x+ 3)3/4 dx

(o)

∫

sec x tanx√1 + sec x dx (p)

∫

cosxesinx dx

(q)

∫

ex√1 + ex dx (r)

∫

xex2+1 dx

(s)

∫

dx

x ln x(t)

∫

4

x(ln x+ 1)2dx

(u)

∫ √cot x csc2 x dx (v)

∫

sin x(cosx− 1)3 dx

(w)

∫

sec3 x tanx dx (x)

∫

ex − e−x

ex + e−xdx

(y)

∫

1 + x

1 + x2dx (z)

∫

2x+ 3

x+ 7dx



1. (a) 13sin 3x+C (b) 2

3(x3 + 2)3/2 +C (c) −1/(1 + 2x)2 +C (d) 1

2(√x+ 2)4 +C

2. (a) 15(x2 + 3)5 + C (b) 1

4(x2 + x)4 + C (c) 2

3(x− 1)3/2 + C (d) 2

3

√x3 − 2 + C

(e) −13ln |5− 3x|+ C (f) ln |x2 + x− 1|+ C (g) 2

√1 + x+ 2x2 + C

(h) −2(√x+1)−1+C (i) 1

2sin 2x+C (j) 2

3(sin x+1)3/2+C (k) −1

2cos(x2)+C

(l) −2√cosx+ C (m) −1

5cos5 x+ C (n) −4

7(cosx+ 3)7/4 + C

(o) 23(1 + sec x)3/2 + C (p) esinx + C (q) 2

3(1 + ex)3/2 + C (r) 1

2ex

2+1 + C

(s) ln | lnx|+C (t) −4(ln x+1)−1+C (u) −23(cotx)3/2+C (v) −1

4(cosx−1)4+C

(w) 13sec3 x+ C (x) ln(ex + e−x) + C (y) tan−1 x+ 1

2ln(1 + x2) + C

(z) 2(x+ 7)− 11 ln |x+ 7|+ C

5.3 The Definite Integral

Definition of Area

The first goal in this section is to give a mathematical definition of area. We begin byattempting to find the area of the region S that lies under the curve y = f(x) from a tob. In general, we start by subdividing S into n strips S1, S2, . . . , Sn of equal width as inFigure.

S1

S2

S3 Si Sn

a x1 x2 x3 . . . xi−1 xi . . . xn−1 b x

y

y = f(x)

The width of the interval [a, b] is b− a, so the width of each of the n strips is

∆x =b− a

n

These strips divide the interval [a, b] into n subintervals

[x0, x1], [x1, x2], [x2, x3], . . . , [xn−1, xn]

where x0 = a and xn = b. The right-hand endpoints of the subintervals are

x1 = a+∆x, x2 = a + 2∆x, x3 = a+ 3∆x, . . .


Let’s approximate the i th strip Si by a rectangle with width ∆x and height f(xi),which is the value of f at the right-hand endpoint. Then the area of the i th rectangleis f(xi)∆x. What we thing of intuitively as the area of S is approximated by the sumof the areas of these rectangles, which is

Rn = f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x =

n∑

i=1

f(xi)∆x

Notice that this approximation appears to become better and better as the number ofstrips increases, that is, as n → ∞. Therefore, we define the area A of the region S inthe following way.

The area A of the region S that lies under the graph of the continuous functionf is the limit of the sum of the areas of approximating rectangles:

A = limn→∞

Rn = limn→∞

n∑

i=1

f(xi)∆x

Definition 5.2.

Find the area under the curve y = 9− x2 over the interval [0, 3].

Example 5.10.

Solution


We now have that a limit of the form

limn→∞

n∑

i=1

f(xi)∆x = limn→∞

[

f(x1)∆x+ f(x2)∆x+ · · ·+ f(xn)∆x]

arises when we compute an area. However this type of limit occurs in a wide variety ofsituations even when f is not necessarily a positive function. In the next Chapter wewill see that limit of this form also arise in finding volumes of solids. We therefore givetype of limit a special name and notation.

If f is a continuous function defined for a ≤ x ≤ b, we divide the interval [a, b]into n subintervals of equal width ∆x = (b − a)/n. We let x0(= a), x1, x2, . . .,xn(= b) be the endpoints of these subintervals and we choose simple points

x∗1, x

∗2, . . . , x

∗n in these subintervals, so x∗

i lies in the ith subinterval [xi−1, xi].Then the definite integral of f from a to b is

∫ b

a

f(x) dx = limn→∞

n∑

i=1

f(x∗i )∆x (5.7)

Definition 5.3.

Note:

1. In the notation

∫ b

a

f(x) dx,

• the symbol

∫

was introduced by Leibniz and is called an integral sign,

• f(x) is called integrand, and

• a and b are called the limit of integration ;

– a is the lower limit and

– b is the upper limit,

• the symbol dx has no official meaning by itself.

The procedure of calculating an integral is called integration.

2. The definite integral

∫ b

a

f(x) dx is a number; it does not depend on x. In fact, we

could use any letter in place of x without changing the value of the integral:

∫ b

a

f(x) dx =

∫ b

a

f(t) dt =

∫ b

a

f(r) dr

3. Because we have assumed that f is continuous, it can be proved that the limit inDefinition 5.3 always exists and gives the same value no matter how we choose thesample points x∗

i .


4. The sumn

∑

i=1

f(x∗i )∆x

that occurs in Definition 5.3 is calledRiemann sum after the German mathemati-cian Bernhard Riemann (1826 - 1866). We know that if f happens to be positive,then the Riemann sum can be interpreted as a sum of areas of approximating rect-angles

(

See Figure 5.1 (a))

. By comparing Definition 5.3 with the definition of

area, we see that the definite integral

∫ b

a

f(x) dx can be interpreted as the area

under the curve y = f(x) from a to b(

see Figure 5.1 (b))

.

a b0 x

y

x∗i

∆x

(a)

a b0 x

y

y = f(x)

(b)Figure 5.1:

If f take on both positive an negative values, then the Riemann sum is the sumof areas of the rectangles that lie above the x-axis and the negative of the areas ofthe rectangles that lie below the x-axis. When we take the limit of such Riemannsums, we get the situation illustrated in the following Figure.

a b0 x

y

y = f(x)+

−

+

A definite integral can be interpreted as a net area, that is, a difference of areas:

∫ b

a

f(x) dx = A1 −A2

where A1 is the area of the region above the x-axis and below the graph of f andA2 is the area of the region below the x-axis and above the graph of f .

5. Although we have defined

∫ b

a

f(x) dx by dividing [a, b] into subintervals of equal

width, there are situations in which it is advantageous to work with subintervalsof unequal width.

When we defined the definite integral, we implicitly assumed that a < b. But thedefinition as a limit of Riemann sums makes sense even if a > b. Notice that if we reverse


a and b, than ∆x changes fromb− a

nto

a− b

n. Therefore

∫ a

b

f(x) dx = −∫ b

a

f(x) dx (5.8)

If a = b, then ∆x = 0 and so∫ a

a

f(x) dx = 0

We now develop some basic properties of integrals. We assume that f and g arecontinuous on an interval [a, b].

1.

∫ b

a

c dx = c(b− a), where c is any constant

2.

∫ b

a

cf(x) dx = c

∫ b

a

f(x) dx, where c is any constant

3.

∫ b

a

[

f(x) + g(x)]

dx =

∫ b

a

f(x) dx+

∫ b

a

g(x) dx

4.

∫ b

a

[

f(x)− g(x)]

dx =

∫ b

a

f(x) dx−∫ b

a

g(x) dx

Theorem 5.3 (Properties of the Integral).

If f is continuous on an interval [a, b] and c is any constant, then

∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx.

Theorem 5.4.

For the case where f(x) ≥ 0 and a < c < b, this theorem can be seen from the geometricinterpretation in the following Figure.

a c b x

y

0

y = f(x)


The area under y = f(x) from a to c plus the area from c to b is equal to the total areafrom a to b.

Let f and g be continuous on an interval [a, b] and g(x) ≤ f(x) for all x in [a, b].Then

∫ b

a

g(x) dx ≤∫ b

a

f(x) dx

Theorem 5.5.

If it is known that

∫ 8

0

f(x) dx = 3,

∫ 5

0

f(x) dx = −4,

∫ 6

5

g(x) dx = 1, and∫ 8

6

g(x) dx = −2, find

∫ 8

5

(

2f(x) + g(x))

dx.

Example 5.11.

Solution

Exercise 5.3

1. Express the limit as a definite integral on the given interval.

(a) limn→∞

n∑

i=1

xi sin xi△x, [0, π]

(b) limn→∞

n∑

i=1

exi

1 + xi△x, [1, 5]

(c) limn→∞

n∑

i=1

[2(x∗i )

2 − 5x∗i ]△x, [0, 1]

(d) limn→∞

n∑

i=1

√

x∗i △x, [1, 4]


2. Use the form of the definition of the integral given in (4.7) to evaluate the integral.

(a)

∫ 1

0

2x dx (b)

∫ 5

−1

(1 + 3x) dx

(c)

∫ 5

1

(2 + 3x− x2) dx (d)

∫ 2

0

(2− x2) dx

(e)

∫ 5

0

(1 + 2x3) dx (f)

∫ 2

1

x3 dx

3. Prove that

∫ b

a

x dx =b2 − a2

2.

4. Prove that

∫ b

a

x2 dx =b3 − a3

3.

5. Write the given sum or difference as a single integral in the form∫ b

af(x) dx.

(a)

∫ 2

0

f(x) dx+

∫ 3

2

f(x) dx (b)

∫ 3

0

f(x) dx−∫ 3

2

f(x) dx

(c)

∫ 2

0

f(x) dx+

∫ 1

2

f(x) dx (d)

∫ 2

−1

f(x) dx+

∫ 3

2

f(x) dx

(e)

∫ 3

1

f(x) dx+

∫ 6

3

f(x) dx+

∫ 12

6

f(x) dx

6. If

∫ 8

2

f(x) dx = 1.8 and

∫ 8

5

f(x) dx = 3.2, find

∫ 5

2

f(x) dx.

7. If

∫ 1

0

f(x) dx = 3,,

∫ 4

0

f(x) dx = −7, and

∫ 4

3

f(x) dx = 2, find

∫ 3

1

f(x) dx.


1. (a)

∫ π

0

x sin xdx (b)

∫ 5

1

ex

1 + xdx (c)

∫ 1

0

(2x2 − 5x)dx (d)

∫ 4

1

√x dx

2. (a) 1 (b) 42 (c) 2 (d) 43

(e) 317.5 (f) 3.75

5. (a)

∫ 3

0

f(x) dx (b)

∫ 2

0

f(x) dx (c)

∫ 1

0

f(x) dx (d)

∫ 3

−1

f(x) dx (e)

∫ 12

1

f(x) dx

6. −1.4 7. −12

5.4 The Fundamental Theorem of Calculus

In the previous section we defined the concept of the definite integral but did not giveany general methods for evaluating them. In this section we shall give a method forusing antiderivatives to evaluate definite integrals.


If f is continuous on [a, b], then the function F defined by

F (x) =

∫ x

a

f(t) dt, a ≤ x ≤ b

is continuous on [a, b] and differentiable on (a, b), and F ′(x) = f(x).

Theorem 5.6 (The First Fundamental Theorem of Calculus).

Using Leibniz notation for derivative, the result in Theorem 5.6 can be expressed bythe formula

d

dx

∫ x

a

f(t) dt = f(x)

Find the derivative of the function g(x) =

∫ x

0

√1 + t2 dt.

Example 5.12.

Solution

Find the derivative of the function g(x) =

∫ x4

1

sec t dt.

Example 5.13.

Solution


Let F (x) =

∫ x3

x2

(

et2

+ 1)

dt. Find F ′(x).

Example 5.14.

Solution

If f is continuous on [a, b], then

∫ b

a

f(x) dx = F (b)− F (a) = F (x)]b

a

where F is any antiderivative of f , that is, a function such that F ′ = f .

Theorem 5.7 (The Second Fundamental Theorem of Calculus).

Evaluate

∫ 8

1

x− 13√x2

dx.

Example 5.15.

Solution


Evaluate

∫ π/4

1

1 + cos2 θ

cos2 θdθ.

Example 5.16.

Solution

Exercise 5.4

1. Use the First Fundamental Theorem of Calculus to find the derivative.

(a) f(x) =

∫ x

0

(t2 − 3t+ 2) dt (b) g(x) =

∫ x

0

√1− 2t dt

(c) g(y) =

∫ y

2

t2 sin t dt (d) F (x) =

∫ 2

x

cos(t2) dt

(e) f(x) =

∫ x2

0

(e−t2 + 1) dt (f) f(x) =

∫ −1

x

ln(t2 + 1) dt

(g) h(x) =

∫ 1/x

2

arctan t dt (h) y =

∫

√x

3

cos t

tdt

(i) y =

∫ 1

1−3x

u3

1 + u2du (j) y =

∫ x3

√x

√t sin t dt


2. Evaluate the definite integrals using the Second Fundamental Theorem of Calculus.

(a)

∫ 3

−1

x5 dx (b)

∫ 2

0

(2x− 3) dx

(c)

∫ 4

0

√x dx (d)

∫ 4

0

(√x+ 3x) dx

(e)

∫ 2

1

3

x4dx (f)

∫ 1

0

(x√x+ x−1/2) dx

(g)

∫ 3

3

√x5 + 2 dx (h)

∫ π/2

0

2 sin x dx

(i)

∫ π

π/2

sec x tan x dx (j)

∫ π

π/2

(2 sin x− cosx) dx

(k)

∫ 9

1

1

2xdx (l)

∫ 1

0

(ex − e−x) dx

(m)

∫ 9

8

2x dx (n)

∫ 3

0

(3ex − x2) dx

(o)

∫ π/2

π/6

(

x+2

sin2 x

)

dx (p)

∫

√3

1

6

1 + x2dx

(q)

∫ 2

0

|2x− 3| dx (r)

∫ 3π/4

0

| cosx| dx

(s)

∫ 3

−2

f(x) dx, where f(x) =

{

−x, x ≥ 0x2, x < 0

(t)

∫ 2

0

f(x) dx, where f(x) =

{

x4, 0 ≤ x < 1x5, 1 ≤ x ≤ 2


1. (a) f ′(x) = x2 − 3x+ 2 (b) g′(x) =√1 + 2x (c) g′(y) = y2 sin y

(d) F ′(x) = − cos(x2) (e) f ′(x) = (e−x4

+ 1)2x (f) f ′(x) = − ln(x2 + 1)

(g) h′(x) = − arctan(1/x)/x2 (h) y′ =cos

√x

2x(i) y′ =

3(1− 3x)3

1 + (1− 3x)2

(j) y′ = 3x7/2 sin(x3)− (sin√x)/(2 4

√x)

2. (a) 3643

(b) −2 (c) 163

(d) 883

(e) 78

(f) 125

(g) 0 (h) 2 (i) Does not exist

(j) 3 (k) ln 3 (l) e+ e−1 − 2 (m) 28

ln 2(n) 3e3 − 12 (o) π2

9+ 2

√3 (p) π

2

(q) 52

(r) 2−√22

(s) −116

(t) 10.7


5.5 Evaluating Definite Integrals by Substitution

There is only one slight difference in using substitution for evaluating a definite integral:If you change variables, you must also change the limits of integration to correspond

to the new variable.That is, when you introduce the new variable u = g(x), you must also change the

limits of integration from x = a and x = b to the corresponding limits for u : u = g(a)and u = g(b). We have

∫ b

a

f(g(x))g′(x) dx =

∫ g(b)

g(a)

f(u) du.

Evaluate

∫ π/8

0

sin5 2x cos 2x dx.

Example 5.17.

Solution

Evaluate

∫ e

1

√1 + ln x

xdx.

Example 5.18.

Solution


Exercise 5.5

Evaluate the definite integral, if it exists.

1.

∫ 2

0

(x− 1)25 dx 2.

∫ 2

0

x√x2 + 1 dx

3.

∫ 1

0

x2(1 + 2x3)5 dx 4.

∫ 1

−1

x

(x2 + 1)2dx

5.

∫ 1

0

cosπx dx 6.

∫ π

π/2

4 cosx

(sin x+ 1)2dx

7.

∫ 4

1

1

x2

√

1 +1

xdx 8.

∫ π/2

π/4

cotx dx

9.

∫ 3

0

dx

2x+ 310.

∫ 4

1

x− 1√x

dx

11.

∫ π/3

0

sin x

cos2 xdx 12.

∫ 13

0

dx3

√

(1 + 2x)2

13.

∫ 2

1

x√x− 1 dx 14.

∫ e4

e

dx

x√lnx

15.

∫ a

0

x√x2 + a2 dx (a > 0)


1. 0 2. 53

√5− 1

33. 81

44. 0 5. 0 6. −2 7. 4

√2

3− 5

√5

128. 1

2ln 2 9. 1

2ln 3

10. 83

11. 1 12. 3 13. 1615

14. 2 15. 13(2√2− 1)a3

Chapter 5 Integrationmathstat.sci.tu.ac.th/~archara/MA111/MA111-118/... · Deﬁnition 5.1. For...

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Transcript of Chapter 5 Integrationmathstat.sci.tu.ac.th/~archara/MA111/MA111-118/... · Deﬁnition 5.1. For...