2 integration - birkbeck, university of londonecon109.econ.bbk.ac.uk/brad/advcalc/notes 2...

Advanced Calculus Chapter 2 Integration 18

2 Integration

2.1 Methods of Integration

Recall that integration is the reverse process of differentiation,and is consequently sometimes referred to as anti-differentiation.More specifically, if f is a function of one variable x then an integralof f is a function F such that dF

dx= f(x). We write

F (x) =

∫f(x) dx.

For any constant c, since ddx

(c) = 0, we have ddx

(F (x) + c) = f(x).

Thus there are infinitely many possibilities for

∫f(x) dx. This does

not usually cause a problem as we always add a constant to anyintegral derived. An alternative approach is to include limits in theintegral and write

F (x) =

∫ x

a

f(x) dx.

The following table gives the integrals of some of the standard func-tions. Most of these follow directly from the derivatives of standardfunctions given in Chapter 1. Others can be checked by differenti-

ating

∫f(x) dx.

f(x)

∫f(x) dx f(x)

∫f(x) dx

xn xn+1

n+1+ c csc x − ln | csc x + cot x|+ c

(provided n 6= −1)1x

ln |x|+ c sec2 x tan x + c

sin x − cos x + c csc2 x − cot x + c

cos x sin x + c exp(x) or ex exp(x) + c

tan x ln | sec x|+ c 1√a2−x2 arcsin x

a+ c

cot x ln | sin x|+ c 1a2+x2

1aarctan x

a+ c

sec x ln | sec x + tan x|+ c

Integration by substitution

This method of integration is based on the chain rule for differen-

tiation. We change the variable of y =

∫f(x) dx using the substi-


tution x = g(u). Then, since integration is the reverse of differen-

tiation, we havedy

dx= f(x). Thus by the chain rule

dy

du=

dy

dx

dx

du

= f(x)dx

du

= f(g(u))dx

du

So integrating the final expression for dydu

, with respect to u, we get

y =

∫f(g(u))

dx

dudu. Comparing the two expressions for y we get

∫f(x) dx =

∫f(g(u))

dx

dudu.

When evaluating a definite integral, if we use a substitution tochange the variable from x to a function u of x, it is usually easiestto change the limits of integration from x values to u values. (Thissaves having to convert back to x again after integrating.)

Specifically, if u = u(x), then

∫ b

a

f(x) dx =

∫ u(b)

u(a)

f(u)dx

dudu.

Integration by Parts

If u and v are functions of x then the product rule for differentiationgives:

∫u

dv

dxdx = uv −

∫vdu

dxdx.

This is the formula for integration by parts.

Integration by parts is useful for integrating products where onefactor (u) will differentiate to give a simpler function, and where it

is possible to integrate the other factor

(dv

dx

).

Exercises 2.1

1. Integrate the following expressions with respect to x.


(a) 3x5 − 2x2 + 3x− 2; (b) 4x3 − 5

x5 ; (c) x3+3x2

x+1;

(d) (3x + 1)5; (e) sin(3x− 2); (f) 2x(x2 + 7)8;

(g) xe 2x2; (h) sin7 x cos x; (i) tan x;

(j) x+3x2+4

; (k) x cos x; (l) x(x + 1)10;

(m) x sin 2x; (n) xe−x; (o) x2ex;

(p) arctan x; (q) ln x; (q) arccos x.

2. Evaluate each of the following integrals using the given sub-stitution.

(a)

∫dx

x ln x, u = ln x.

(b)

∫4x

(3− 2x)2dx, u = 3− 2x.

(c)

∫sec5 x tan x dx, u = sec x.

(d)

∫2e2x

1 + e4xdx, u = e2x.

(e)

∫5 sin7 x dx, u = cos x.

(f)

∫14x2

√1 + x dx, x = u2 − 1.

(g)

∫ √1− x2 dx, x = sin u.

2.2 Integrals with infinite limits

It is sometimes possible to evaluate a definite integral that has one,or both, of its limits equal to infinity or negative infinity. We define

∫ ∞

a

f(x) dx = limX→∞

∫ X

a

f(x) dx.

If the limit exists then the definite integral exists and its value isthe limit; otherwise the definite integral does not exist.

When evaluating limits to infinity we make use of the following

results: (i) limX→∞

g(X) = limX→0

g

(1

X

); (ii) lim

X→∞A−X = 0, for all

real numbers A, with A > 1.


Example 1 Evaluate

∫ ∞

1

dx

x3.

∫ ∞

1

dx

x3= lim

X→∞

∫ X

1

dx

x3,

= limX→∞

∫ X

1

x−3 dx,

= limX→∞

[−1

2x−2

]X

1

,

= limX→∞

[− 1

2x2

]X

1

,

= limX→∞

(− 1

2X2+

1

2

),

= limX→0

(−1

2X2 +

1

2

),

=1

2.

Example 2 Let a ∈ R. Evaluate

∫ ∞

a

e−x dx.

∫ ∞

a

e−x dx = limX→∞

∫ X

a

e−x dx,

= limX→∞

[−e−x]X

a,

= limX→∞

(−e−X + e−a),

= e−a.

Exercises 2.2

1. Where possible, evaluate the following integrals:

(a)

∫ ∞

0

e−5x dx;

(b)

∫ ∞

1

dx√x

dx;

(c)

∫ ∞

0

dx

1 + x2dx;

(d)

∫ ∞

0

e−x cos x dx.

2. Determine all of the real numbers s for which

∫ ∞

1

xs dx is

defined.


3. Let f : (a, b] → R be a continuous function with

limx→a

f(x) = ∞. Suggest how you could define

∫ b

a

f(x) dx.

(Recall that (a, b] = {r ∈ R : a < r ≤ b}.)Use your suggestion to evaluate the following integrals:

(a)

∫ 1

0

dx√x

;

(b)

∫ 10

1

dx

(x− 1)23

.

2.3 Double integrals over rectangular regions

Let f : R2 → R. We want to find the volume enclosed betweenthe surface z = f(x, y), the xy-plane and the planes x = a, x = b,y = c and y = d.

Example 3 Find the volume enclosed between the surface of

z = 8x3y + 3x2y2

and the planes z = 0, x = 2, x = 3, y = 1 and y = 2.

The required volume is given by the double integral

∫ 2

1

∫ 3

2

(8x3y + 3x2y2) dx dy =

∫ 2

1

[2x4y + x3y2

]3

2dy,

=

∫ 2

1

(162y + 27y32− 32y − 8y2

)dy,

=

∫ 2

1

(130y + 19y2

)dy,

=

[65y2 +

19

3y3

]2

1

,

= 65× 4 +19

3× 8− 65− 19

3= 2391

3.

Exercises 2.3

1. Evaluate

∫ 3

1

∫ 2

0

x2y dx dy.

2. Evaluate

∫ π4

0

∫ π4

0

sin(x + y) dx dy.


2.4 Non-rectangular regions of integration

We now consider the problem of evaluating a double integral overa non-rectangular region D. We write this as

∫ ∫

D

f(x, y) dx dy.

As given, we are integrating first with respect to x. In this case thelimits of the inner integral are (possibly constant) functions of yand the limits of the outer integral are constants. When evaluatinga double integral the final answer represents a volume and so shouldnot involve either of the variables of integration; if your final answerdoes involve either variable then you should check your limits.

Example 4 Find the volume beneath the surface of z = xy overthe region of the (x, y)-plane enclosed by the y-axis, y = x andy = 2.

Here we require

∫ ∫

D

xy dx dy where D is the region of the (x, y)-

plane enclosed by the y-axis, y = x and y = 2.

We start by sketching the region of integration.

y

1.5

2

1

0

0.5

2.50 1.51 2

x

0.5

The region D

Since we are integrating with respect to x first we need to find thelimits of the inner integral as functions of y. To do this we draw aline parallel to the x-axis passing through D. Where the line crossesthe left hand side of D gives the lower limit (x = 0) and where itcrosses the right hand side of D gives the upper limit (x = y). Thelimits of the outer integral are then just the minimum (y = 0) and


maximum (y = 2) values of y over D. Thus we get:∫ ∫

D

xy dx dy =

∫ 2

0

∫ y

0

xy dx dy,

=

∫ 2

0

[1

2x2y

]y

0

dy,

=

∫ 2

0

1

2y3 dy,

=

[1

8y4

]2

0

= 2.

Example 5 Evaluate

∫ ∫

D

(xy +2) dx dy, where D is the bounded

region enclosed between y = 2x and y = x2.

First we sketch the region D.

y

3

4

2

01.50 21

x

0.5

1

The region D

From the line parallel to the x-axis we see that for the inner integralthe lower limit is x = 1

2y and the upper limit is x =

√y. For the

outer integral, the lower limit is y = 0 and the upper limit is y = 4.Thus∫ ∫

D

(xy + 2) dx dy =

∫ 4

0

∫ √y

12y

(xy + 2) dx dy,

=

∫ 4

0

[1

2x2y + 2x

]√y

12y

dy,

=

∫ 4

0

(1

2y2 + 2

√y − 1

8y3 − y

)dy,

=

[1

6y3 +

4

3y

32 − 1

32y4 − 1

2y2

]4

0

,

=1

6× 64 +

4

3× 8− 1

32× 256− 1

2× 16 =

16

3.


Splitting the integral

So far we have been able to determine the limits of the inner integralof a double integral a single functions of y. However, this is notalways possible. In such cases we have to split the integral intotwo, or more, integrals.

Example 6 Evaluate

∫ ∫

D

xy2 dx dy, where D is triangle with ver-

tices at (0, 0), (3, 1) and (0, 3).

First we sketch the region D.

y

2.5

0.5

3

2

1

2.500

10.5 2 31.5

x

1.5

The region D

From the lines parallel to the x-axis we see that the limits of theinner integral depend on whether y < 1 or y > 1. Thus we split theintegral into two integrals where the limits of the outer integralsare 0 and 1, and 1 and 3, respectively. Doing this, and working outthe limits of the inner integrals as before gives:∫ ∫

D

xy2 dx dy =

∫ 1

0

∫ 3y

0

xy2 dx dy +

∫ 3

1

∫ 32(3−y)

0

xy2 dx dy,

=

∫ 1

0

[1

2x2y2

]3y

0

dy +

∫ 3

1

[1

2x2y2

] 32(3−y)

0

dy,

=

∫ 1

0

9

2y4 dy +

∫ 3

1

9

8(3− y)2y2 dy,

=

∫ 1

0

9

2y4 dy +

∫ 3

1

9

8(y4 − 6y3 + 9y2) dy,

=9

2

[1

5y5

]1

0

+9

8

[1

5y5 − 3

2y4 + 3y3

]3

1

,

=9

10+

9

8

(1

5× 243− 3

2× 81 + 3× 27− 1

5+

3

2− 3

),

=9

10+

9

8× 62

5= 8 1

10.


Exercises 2.4

1. Evaluate

∫ ∫

D

(x + y) dx dy, where

D = {(x, y) : 0 ≤ y ≤ 1, y ≤ x ≤ 1 + 2y}.

2. Evaluate

∫ ∫

D

(x − y)2 dx dy, where D is the region enclosed

between x = 3, y = 1 and y = x2.

3. Evaluate

∫ ∫

D

xy dx dy, where D is the square with vertices

at (1, 0), (0, 1), (1, 2) and (2, 1).

4. Evaluate

∫ ∫

P

xy dx dy, where P is the pentagon with vertices

at (0, 0), (0, 2), (1, 2), (2, 1) and (2, 0).

5. Explain why

∫ ∫

D

dx dy gives the area of D. Use this result

to find the area of the region enclosed between the curvesx = 7y − xy2 and x = y2 + 3.

2.5 Changing the order of integration

In the examples considered so far, we have always integrated firstwith respect to x. This is not necessary, and there is no reasonwhy we should not choose to integrate first with respect to y. Intwo particular cases, it can be advantageous, or even necessary, tochange the order of integration. These cases are:

• to avoid having to split the integral into two, or more, regions;

• when it is difficult, or not possible, to integrate f(x, y) withrespect to x.


As an example of the first case, consider again Example 6, but thistime we integrate first with respect to y.

y

2.5

0.5

3

2

1

2.500

10.5 2 31.5

x

1.5

To find the limits of the inner integral we draw a line through theregion parallel to the y-axis; where the line crosses the region at thelowest point (y = 1

3x) gives the lower limit, and where it crosses

the highest point (y = 3 − 23x) gives the upper limit. The limits

of the outer integral are just the minimum (x = 0) and maximum(x = 3) values of x. Thus we get:∫ ∫

D

xy2 dx dy =

∫ ∫

D

xy2 dy dx,

=

∫ 3

0

∫ 3− 23x

13x

xy2 dy dx,

=

∫ 3

0

[1

3xy3

]3− 23x

13x

dx,

=

∫ 3

0

(1

3x(3− 2

3x)3 − 1

81x4

)dx,

=

∫ 3

0

(1

3x(27− 18x + 4x2 − 8

27x3)− 1

81x4

)dx,

=

∫ 3

0

(1

3x(27− 18x + 4x2 − 8

27x3)− 1

81x4

)dx,

=

∫ 3

0

(9x− 6x2 +

4

3x3 − 1

9x4

)dx,

=

[9

2x2 − 2x3 +

1

3x4 − 1

45x5

]3

0

,

=9

2× 9− 2× 27 +

1

3× 81− 1

45,

= 8 110

.

The following two examples illustrate the case when we change theorder of integration because it is difficult, or impossible, to evaluate


the integral in the order given.

Example 7 Evaluate I =

∫ 4

0

∫ 2

√x

x

1 + y5dy dx.

Here I is expressed so that first we have to integrate with respectto y. Since it is difficult to integrate 1

1+y5 with respect to y, wechange the order of integration. To do this we need to find the newlimits, and so we sketch the region of integration. Thus we sketchy =

√x and y = 2 (the limits of the inner integral of I) between

x = 0 and x = 4 (the limits of the outer integral of I). Thus theregion of integration is as follows.

y

2

0

2.5

1.5

0.5

2 430

x

1

1

The region D

From the line parallel to the x-axis, we see that the lower limit forx is x = 0, and the upper limit is x = y2. The corresponding limitsfor y are y = 0 and y = 2, respectively. Thus

I =

∫ 2

0

∫ y2

0

x

1 + y5dx dy,

=

∫ 2

0

[ 12x2

1 + y5

]y2

0

dy,

=

∫ 2

0

12y4

1 + y5dy,

=1

10

∫ 2

0

5y4

1 + y5dy,

=1

10

[ln(1 + y5)

]2

0,

=1

10ln 33.

Example 8 Evaluate I =

∫ 1

0

∫ 1

y

3y2e−x2

dx dy.


Here I is expressed so that first we have to integrate with respect to

x. Since if is not possible to express

∫e−x2

dx in terms of simple

functions, we change the order of integration. To find the newlimits, we sketch the region of integration. So we sketch x = y andx = 1 (the limits of the inner integral of I) between y = 0 andy = 1 (the limits of the outer integral of I). Thus the region ofintegration is as follows.

y

1

0.2

1.2

0.8

0.4

-0.2

0.80

1.20.4 0.60.2 10

x

0.6

-0.2

The region D

From the line parallel to the y-axis, we see that the lower limit fory is y = 0, and the upper limit is y = x. The corresponding limitsfor x are x = 0 and x = 1, respectively. Thus

I =

∫ 1

0

∫ x

0

3y2e−x2

dy dx,

=

∫ 1

0

[y3e−x2

]x

0dx,

=

∫ 1

0

x3e−x2

dx,

(Put w = x2, so dwdx

= 2x, when x = 0, w = 0,

and when x = 1, w = 1)

=

∫ 1

0

xwe−w dw

2x,

=1

2

∫ 1

0

we−w dw,

(Use integration by parts with u = w anddvdw

= e−w, so dudw

= 1 and v = −e−w.)


=1

2

[−we−w +

∫e−wdw

]1

0

,

=1

2

[−we−w − e−w]1

0,

= −1

2

[e−w(w + 1)

]1

0,

= −1

2(2e−1 − 1),

=1

2− e−1.

Exercises 2.5

1. For each of the following double integrals:

(i) evaluate the integral as given;

(ii) sketch the region of integration;

(iii) change the order of integration, and re-evaluate the in-tegral.

(a)

∫ 3

0

∫ y

0

(x2 + y2) dx dy.

(b)

∫ 2

0

∫ 4−2x

0

xy dy dx.

(c)

∫ 1

0

∫ √y

y

y2x dx dy.

2. Evaluate the following double integrals.

(a)

∫ 1

0

∫ 2

2y

6y√

1 + x3 dx dy;

(b)

∫ 2

1

∫ 1y

12

x3 cos(x2y) dx dy.

3. Evaluate

∫ 2

0

∫ 2y

y

sin(πy

x

)dx dy +

∫ 4

2

∫ 4

y

sin(πy

x

)dx dy.

2.6 Unbounded regions of integration

It is possible to perform some double integrals over an unboundedregion. In such cases, one or more of the limits is infinite.

Example 9 Evaluate

∫ ∫

D

e−(x+y) dx dy, where D = {(x, y) ∈ R2 :

0 ≤ y ≤ 1, y ≤ x}.


First we sketch the region of integration.

y

1

0.2

1.2

0.8

0.4

-0.20

0310.5 1.5 2.5

x

2

0.6

The region D

From the line parallel to the x-axis, we see that the lower limit forx is x = y, and the upper limit is x = ∞. The corresponding limitsfor y are y = 0 and y = 1, respectively. Thus

∫ ∫

D

e−(x+y) dx dy =

∫ 1

0

∫ ∞

y

e−(x+y) dx dy,

=

∫ 1

0

limX→∞

[−e−(x+y)]∞y

dy,

=

∫ 1

0

limX→∞

(−e−(X+y))

+ e−2y dy,

=

∫ 1

0

e−2y dy,

=

[−1

2e−2y

]1

0

,

= −1

2e−2 +

1

2,

=1

2(1− e−2).

Exercises 2.6

1. Evaluate

∫ ∫

D

dx dy

x2y2, where D = {(x, y) ∈ R2 : 1 ≤ y ≤

2, x ≥ y2}.

2. Evaluate

∫ ∫

D

xe−y2

dx dy, where D = {(x, y) ∈ R2 : x ≥0, y ≥ x2}.

2.7 Change of variables

Suppose we want to evaluate

∫ ∫

D

f(x, y) dx dy, and we change

the variables to u = h1(x, y), v = h2(x, y), where h1 and h2 are


invertible on D. Such a change of variables transforms the integralover D in the (x, y)-plane to an integral over ∆ in the (u, v)-plane.It remains to transform the function to a function in u and v, beingintegrated with respect to u and v. To do this, first note that sinceh1 and h2 are invertible on D, there are invertible functions g1 andg2 on ∆ such that x = g1(u, v) and y = g2(u, v). Now we define theJacobean by

∂(x, y)

∂(u, v)= det

(∂x∂u

∂x∂v

∂y∂u

∂y∂v

).

Then

Recall that if A =(

a bc d

)

then det A = ad− bc.

∫ ∫

D

f(x, y) dx dy =

∫ ∫

∆

f(g1(u, v), g2(u, v))

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ du dv.

If it is difficult to express x and y in terms of u and v then we canuse the fact that

∂(x, y)

∂(u, v)=

1∂(u,v)∂(x,y)

.

Example 10 Evaluate

∫ ∫

D

x2y2 dx dy, where D is the region en-

closed by the curves y =√

x, y =√

3x, y = 1x

and y = 2x.

The region D is illustrated in the following diagram.

The region D

y

2

0

2.5

1.5

x

20

0.5

1

0.5 1 2.51.5

The region D

Let u = y2

xand v = xy. This change of variables transforms the

curves y =√

cx and y = cx, for some constant c, to u = c and v = c,

respectively. Thus the region D is transformed into the region ∆that is enclosed between the lines u = 1, u = 3, v = 1 and v = 2.Now

∂(u, v)

∂(x, y)= det

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)= det

(− y2

x22yx

y x

)= −y2

x−2y2

x= −3y2

x.


Thus ∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ =1∣∣∣∂(u,v)

∂(x,y)

∣∣∣=

x

3y2.

Thus∫ ∫

D

x2y2 dx dy =

∫ ∫

∆

x2y2

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ du dv,

=

∫ 2

1

∫ 3

1

(x2y2)

(x

3y2

)du dv,

=

∫ 2

1

∫ 3

1

1

3

v2

udu dv,

=

∫ 2

1

1

3

[v2 ln u

]3

1dv,

=

∫ 2

1

1

3v2 ln 3 dv,

=

(1

3ln 3

)[1

3v3

]2

1

,

=1

9ln 3(8− 1),

=7

9ln 3.

Example 11 Evaluate

∫ ∫

D

y2 cos(y

x

)dx dy, where D is the re-

gion enclosed by the curves y = ax, y = bx, y = ax

and y = bx,

where a, b ∈ R with a > b > 0.

Let u = yx

and v = xy. This change of variables transforms thecurves y = cx and y = c

x, for some constant c, to u = c and v = c,

respectively. Thus the region D is transformed into the region ∆that is enclosed between the lines u = b, u = a, v = b and v = a.Now

∂(u, v)

∂(x, y)= det

(∂u∂x

∂u∂y

∂v∂x

∂v∂y

)= det

( − yx2

1x

y x

)= − y

x2x−y

x= −2

y

x.

Thus ∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ =1∣∣∣∂(u,v)

∂(x,y)

∣∣∣=

x

2y.


Thus∫ ∫

D

y2 cos(y

x

)dx dy =

∫ ∫

∆

y2 cos(y

x

) ∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ du dv,

=

∫ a

b

∫ a

b

y2 cos(y

x

) (x

2y

)du dv,

=

∫ a

b

∫ a

b

1

2xy cos

(y

x

)du dv,

=

∫ a

b

∫ a

b

1

2v cos u du dv,

=

∫ a

b

1

2v [sin u]ba dv,

=

∫ a

b

1

2(sin a− sin b)v dv,

=1

2(sin a− sin b)

[1

2v2

]a

b

,

=1

4(sin a− sin b)(a2 − b2).

Polar coordinates

Every non-zero point (x, y) ∈ R2 can be defined uniquely by itsdistance r from the origin and the angle θ between the positivex-axis and the line joining the the origin to (x, y). The pair (r, θ)are the polar coordinates of (x, y). (For completeness, the polarcoordinate of the origin are taken to be (0, 0).) From this, simpletrigonometry gives

x = r cos θ,

y = r sin θ.

It is often convenient to use polar coordinates instead of rectangularcoordinates. Here we consider only their use in a change of variablesin a double integral. This is usually advantageous when:

• the region of integration involves an area enclosed by a circle,or parts of a circle;

• f(x, y) contains terms involving x2 + y2.This is because

x2 + y2 = r2 cos2 x + r2 sin2 x,

= r2(cos2 x + sin2 x),= r2.

If we change to polar coordinates then

∂(x, y)

∂(r, θ)= det

(∂x∂r

∂x∂θ

∂y∂r

∂y∂θ

)

= det

(cos θ −r sin θsin θ r cos θ

)

= r(cos2 θ + sin2 θ) = r.


When changing variables from rectangular to polar coordinates you

may quote the fact that∂(x, y)

∂(r, θ)= r without going through the

above derivation.

Example 12 Find

∫ ∫

D

x2y dx dy, where

D = {(x, y) ∈ R2 : x ≥ 0, y ≥ 0 x2 + y2 ≤ a2}, for some a ∈ R.

Changing variables to polar coordinates we get:

∫ ∫

D

x2y dx dy =

∫ π2

0

∫ a

0

(r2 cos2 θ)(r sin θ)r dr dθ,

=

∫ π2

0

∫ a

0

r4 cos2 θ sin θ dr dθ,

=

∫ π2

0

[r5

5

]a

0

cos2 θ sin θ dr dθ,

=a5

5

∫ π2

0

cos2 θ sin θ dr dθ,

=a5

5

[−cos3 θ

3

]π2

0

=a5

5

1

3=

1

15a5.

Example 13 Evaluate

∫ ∫

D

xy dx dy, where D is the interior of

the circle of radius 1, centre (1, 1).

Here we first make the change of variable u = x− 1, v = y − 1. It

is easy to check that∂(u, v)

∂(x, y)= 1 and the region of integration ∆ is

the interior of the circle of radius 1, centre (0, 0). Thus

∫ ∫

D

xy dx dy =

∫ ∫

∆

(u + 1)(v + 1) du dv.

Now we change to polar coordinates with u = r cos θ and v = r sin θ.


This gives:

∫ ∫

∆

(u + 1)(v + 1) du dv =

∫ 2π

0

∫ 1

0

(r cos θ + 1)(r sin θ + 1)r dr dθ,

=

∫ 2π

0

∫ 1

0

(r3 cos θ sin θ + r2(cos θ + sin θ) + r

)dr dθ,

=

∫ 2π

0

[r4

4cos θ sin θ +

r3

3(cos θ + sin θ) +

r2

2

]1

0

dθ,

=

∫ 2π

0

(1

4cos θ sin θ +

1


1

2

)dθ,

=

∫ 2π

0

(1

8sin 2θ +

1


1

2

)dθ,

=

[− 1

16cos 2θ +

1

3(sin θ − cos θ) +

1

2θ

]2π

0

,

= − 1

16+

1

3(0− 1) + π +

1

16− 1

3(0− 1) + 0,

= π.

Note that in the process of evaluating the above double integral weused the trigonometric identity sin 2θ = 2 sin θ cos θ.

Exercises 2.7

1. Evaluate

∫ ∫

D

(x + y)(2x + 1) dx dy, where D is the region in

the upper right quadrant of the (x, y)-plane that is enclosedbetween the curves y = x2, y = x2+2, x+y = 4 and x+y = 6.

2. Evaluate

∫ ∫

D

dx dy

xy, where D is the region enclosed between

the curves y = 1x2 , y = 3

x2 , y =√

x3 and y =√

5x3.

3. Evaluate

∫ ∫

D

x2√

x2 + y2 dx dy, where

D = {(x, y) : x2 + y2 ≤ a2, y ≥ 0}, for some nonzero realnumber a.

4. Evaluate

∫ ∫

D

xy√

x2 + y2 dx dy, where

D = {(x, y) : 1 ≤ x2 + y2 ≤ 4, x ≥ 0, y ≥ 0}.5. Evaluate the integral given in Example 13 using the change

of variable x = 1 + r cos θ, y = 1 + r sin θ.

2 integration - birkbeck, university of londonecon109.econ.bbk.ac.uk/brad/advcalc/notes 2...

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