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Practice Gateway. The structure of the practice exam is identical to the in-class exam. Refer to email instructions on how to take practice exams. Current setting — you may take the practice exam one time every 45 minutes. Num 1 2 3 4 5 6 7 8

Topic u -Substitution

Integration-By-Parts Other Integration Techniques Double Integrals: Reverse Order Double Integrals: Set up Bounds Triple Integrals 1 Triple Integrals 2 Change of Variable

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Exam Randomization. 1. There are 8 problems on the exam. 2. One question is randomly selected from each of the topic listed on the left. 3. In addition, each exam problem has randomly generated coefficients. In short, no two exams will have the same problems and/or coefficients. Webwork Tips. Click HERE for full listing. • You do not need to simplify. For: Enter: Not:

2 + 212 − 216

2 + 1/2ˆ2 - 1/2ˆ6 2.234375

• Mathematical Constants: Symbol π

Usage pi e

e

• Common Functions: Function p

x

ln x ex 3! −1

tan

(x)

Usage sqrt(x) or xˆ(1/2) ln(x) eˆ(x) 3! atan(x) or arctan(x)

Contents

Half-Angle Formulas 1 − cos(2x) 2 1 + cos(2x) cos2 (x) = 2 sin2 (x) =

u-Substitution

2

u-Sub. with Definite Integrals

4

Integration by Parts

4

Integrals

Solve for Integral

5

R

Trigonometric Integral

5

Trigonometric Substitution

6

Area of 2D Region

7

d 2 d x [tan(x)] = sec (x)

Reverse Order of Integration

7

d d x [sec(x)] = sec(x) tan(x)

Change to Polar Coord.

9

R

³ ´ 1 1 −1 x d x = tan +C x2 + a2 a a sec3 (x) d x =

Derivatives

Contains 2

3D Rectangular Coordinates

10

Integral in Spherical

11

Transformation

11

Jacobian

11

Transformation Examples

12

2

a −x a2 + x2 x2 − a2

9

Integral in Cylindrical

1 1 sec(x) tan(x) + ln | sec(x) + tan(x)| +C 2 2

Try Substitution x = a sin(θ) x = a tan(θ) x = a sec(θ)

Related Identity sin2 (θ) + cos2 (θ) = 1 1 + tan2 (θ) = sec2 (θ) 1 + tan2 (θ) = sec2 (θ)

Idea (u -Substitution). The goal of a substitution is to simplify the integrand into a familiar form with a known integral. Suppose we have the composite function f (g (x)). We may differentiate by applying the chain rule: d [ f (g (x))] = f 0 (g (x))g 0 (x). dx

Reversing this process, we may apply the fundamental theorem of calculus to undo the differentiation: Z

0

0

f (g (x))g (x)d x =

Z

d [ f (g (x))]d x = f (g (x)) +C . dx

Now, when we need to find an integral, it is often a matter of recognition: what is f and what is g ? To assist in the simplification and recognition of f , we may make a u -substitution. Notice: if we make the following substitution, u = g (x),

d u = g 0 (x)d x,

the integrand simplifies and the integral is easier to find, Z

f 0 (g (x))g 0 (x)d x =

Z

f 0 (u)d u = f (u) +C = f (g (x)) +C .

2

(1)

Example 1 (Substitution Simplifies to

f 0 (u)d u ). Integrate Z (x 3 + 3x + 1)9 (3x 2 + 3)d x.

R

♠

Solution. We choose u to be the function raised to the ninth power, u = x 3 + 3x + 1,

d u = (3x 2 + 3)d x,

then the integrand simplifies and we may compute Z

(x 3 + 3x + 1)9 (3x 2 + 3)d x =

Z

u9d u =

u 10 (x 3 + 3x + 1)10 +C = +C . 10 10

■

Example 2 (Constant Multiple of d u ). Integrate Z

cos5 (2x) sin(2x)d x.

♠

Solution. We may choose u to be u = cos(2x),

1 d u = −2 sin(2x)d x =⇒ − d u = sin(2x)d x, 2

which the integrand may be rewritten and integrated as Z

5

cos (2x) sin(2x)d x =

Z

µ ¶ Z 1 1 1 u6 1 u − du = − u5d u = − +C = − cos6 (x) +C . 2 2 2 6 12 5

■

Example 3 (Substitution Does Not Yield Constant Multiple of d u ). Integrate Z

(x + 3)9 (x + 1)d x.

♠

Solution. Let u = x + 3. Clearly, d u = d x . However, the presence of x in (x + 1) causes the previous two examples to not directly apply. We need to rewrite x + 1 in terms of u . We may observe that u − 2 = x + 1. We may now use the substitution to find the integral: Z

(x + 3)9 (x + 1)d x =

Z

u 9 (u − 2)d u =

Z

u 10 d u − 2

Z

u9d u =

µ ¶ u 11 u 10 1 −2 +C = (x + 3)10 (x + 3) − +C . 11 10 5

■

Example 4 (Two-Stage Substitution). Let a, b, k be positive constants. Integrate Z

(a + ln(x))4 (b − ln(x)) dx kx

♠

Solution. First, observe that if we choose u = ln(x) with u = 1/xd x , then we can take care of the x in the denominator: Z

(a + ln(x))4 (b − ln(x)) dx = kx

3

Z

(a + u)4 (b − u) d u. k

Now, this problem is a variation of the previous example. Choose w = a + u with d w = d u . This choice yields w 4 (b − (w − a)) dw k Z Z 1 b+a =− w 5d w + w 4d w k k 1 w6 b + a w5 + +C =− k 6 k 5 µ ¶ (a + ln(x))5 a + ln(x) b + a =− − +C . k 6 5

(a + u)4 (b − u) du = k

Z

Z

■

Theorem. Let u = g (x). Suppose f and g 0 (x) are continuous functions. Then, b

Z

a

f 0 (g (x))g 0 (x)d x =

u=g (b)

Z

u=g (a)

f 0 (u)d u = f (g (b)) − f (g (a)).

p

Example 5. Let k = b + 1 > 1. Integrate bq

Z 0

p k − xd x.

♠

p

Solution. If we choose u = k − x , we have

p u=k− x 1 d u = − p d x =⇒ −2(k − u)d u = d x 2 x

We may use this differential relationship to write bq

Z 0

p

k − xd x = = =

p k− b p

Z Z

p k− 0 1

p p −2k u + 2u ud u

k 1

Z k

u(−2)(k − u)d u

−2ku 1/2 + 2u 3/2 d u

· ¸1 u 3/2 u 5/2 = −2k +2 3/2 5/2 k µ ¶ µ ¶ 1 1 k 3/2 k 5/2 = −2k +2 − −2k +2 . 3/2 5/2 3/2 5/2

■

Theorem (Integration by Parts). Let u and v be differentiable functions. Then, Z

u d v = uv −

Z

v du

Example 6. Integrate Z

ln(x) d x.

4

♠

Solution. Choose u = ln(x) and d v = 1 d x . Then, Z

ln(x) d x = x ln(x) −

Z

1 d x = x ln(x) − x +C .

■

Example 7 (Solve for Integral). Integrate Z

e x sin(x) d x.

♠

Solution. We need to integrate by parts twice, and then solve for the integral. We choose u = ex du = exd x

d v = sin(x)d x v = − cos(x)

Table 1: Choice for first integration by parts.

Z

e x sin(x) d x = −e x cos(x) − (−1)

Z

e x cos(x)d x = −e x cos(x) +

Z

e x cos(x)d x.

(2)

Be consistent: choose u in the same manner as in the first integration by parts, u = ex du = exd x

d v = cos(x)d x v = sin(x)

Table 2: Choice for second integration by parts.

Z

e x cos(x)d x = e x sin(x) −

Z

e x sin(x)d x

(3)

We now substitute (3) into (2), and solve for the integral:

2

Z

e x sin(x) d x = −e x cos(x) + e x sin(x) −

Z

e x sin(x) d x = −e x cos(x) + e x sin(x)

Z

e x sin(x) d x =

Z

e x sin(x)d x

ex (sin(x) − cos(x)) 2

■

Example 8 (Trigonometric Integral). Integrate Z

sin6 (x) d x.

♠

Solution. The plan is to utilize half-angle formulas to reduce the power of the trigonometric term: Z

sin6 (x) d x =

Z

¡ 2 ¢3 sin (x) d x =

Z µ

1 − cos(2x) 2

5

¶3

dx =

1 8

Z

(1 − cos(2x))3 d x.

After we expand, we utilize both the half-angle formula (again) and a u -substitution: 1 8

Z

1 (1 − cos(2x)) d x = 8 3

= = = = =

Z

à ! à ! à ! 3 3 3 2 cos3 (2x) d x cos (2x) − cos(2x) + 1− 3 2 1

Z 1 1 − 3 cos(2x) + 3 cos2 (2x) − cos3 (2x) d x 8 ¶ µ Z ¡ ¢ 1 1 + cos(4x) − 1 − sin2 (2x) cos(2x) d x 1 − 3 cos(2x) + 3 8 2 µ ¶ Z 1 1 + cos(4x) 1 − 3 cos(2x) + 3 − cos(2x) + sin2 (2x) cos(2x) d x 8 2 Z 1 5 3 − 4 cos(2x) + cos(4x) + sin2 (2x) cos(2x) d x 8 2 2 · ¸ 3 1 1 5 3 x − 2 sin(2x) + sin(4x) + sin (2x) +C 8 2 8 6

where we have used the u -substitution u = sin(2x): Z

sin2 (2x) cos(2x) d x =

Z

1 1 u3 1 u2 d u = +C = sin3 (2x) +C 2 2 3 6

■

If the integral contains a term with a form a 2 ± x 2 or x 2 − a 2 , a trigonometric substitution may be effective. See the table below. Contains

Try Substitution

Related Identity

Related Derivative

a2 − x2 a2 + x2 x2 − a2

x = a sin(θ) x = a tan(θ) x = a sec(θ)

sin2 (θ) + cos2 (θ) = 1 1 + tan2 (θ) = sec2 (θ) 1 + tan2 (θ) = sec2 (θ)

d 2 d x [tan(θ)] = sec (θ) d d x [sec(θ)] = sec(θ) tan(θ)

Table 3: Trigonometric substitution information. Example 9 (Trigonometric Substitution). Integrate Z p

a 2 − x 2 d x.

♠

Solution. Below, we utilize the trig. substitution x = a sin(θ): Z p

a2 − x2 d x

=

Z q

a 2 − a 2 sin2 (θ)a cos(θ) d θ = a 2

Z q Z 2 2 1 − sin (θ) cos(θ) d θ = a cos(θ) cos(θ) d θ.

We may use the half-angle formula for cos2 (x): a

2

Z

2

cos (θ) d θ = a

2

Z µ

¶ µ ¶ 1 + cos(2θ) sin(2θ) 2 θ dθ = a + +C . 2 2 4

If we wish to rewrite the integral in terms of x , we will have to manipulate sin(2θ). We know the identity, sin(2θ) = 2 sin(θ) cos(θ),

and we can express cos(θ) in terms of x as follows: cos(θ) =

q

2

1 − sin (θ) =

6

r 1−

³ x ´2 a

.

Thus, Z p

¶ θ sin(2θ) +C + 2 4   q ¡ x ¢2 ¡ ¢ x −1 x 2a 1− a   sin a = a2  +  +C 2 4 µ

a2 − x2 d x = a2

à ! ³ ´ xr ³ x ´2 a2 −1 x sin + +C = 1− 2 a a a

■

Example 10 (Completing the Square). Integrate 1

Z

x 2 − 6x + 34

dx

♠

Solution. Complete the square of the denominator, 1 dx = x 2 − 6x + 34

Z

1 dx = (x 2 − 6x + 9) − 9 + 34

Z

Z

1 d x, (x − 3)2 + 25

and then perform a u -substitution, u = x − 3, and subsequently a trig. substitution, u = 5 tan(θ), Z

1 du = 2 u + 25

1 1 5 sec2 (θ) d θ = 2 25 tan (θ) + 25 5

Z

Z

1 1 sec2 (θ) d θ = 2 tan (θ) + 1 5

Z

1 1 d θ = θ +C . 5

We now unravel the substitutions to write the answer in terms of the original variable x : Z

¶ µ 1 1 −1 x − 3 +C . d x = tan x 2 − 6x + 34 5 5

■

Theorem. The area of a bounded region R in the plane is found as the double integral with integrand 1: A(R) =

Ï

1 d A.

R

Example 11. Find the area of the region R given as © ª R = (x, y) | 3y 2 ≤ x ≤ 16 − y 2 , −2 ≤ y ≤ 2 .

♠

Solution. For this problem, we naturally integrate in x first, and then y (left diagram): Ï R

1d A =

2

Z

Z

16−y 2

−2 3y 2

1d xd y =

Z

2 −2

16−y 2

[x]3y 2

dy =

Z

2 −2

(16 − y 2 ) − 3y 2 d y = 4

Z

2 −2

4 − y 2d y

We exploit the even integrand to write 4

Z

2 −2

2

4− y dy = 8

2

Z 0

· ¸2 · ¸ y3 128 8 4 − y d y = 8 4y − = 8 8− = 3 0 3 3 2

■

Example 12 (Reverse Order of Integration). Integrate Z 1Z 0

1 y

2

e x d xd y.

7

♠

y

y

x

R

x R1

(a) R = (x, y) | 3y 2 ≤ x ≤ 16 − y 2 , −2 ≤ y ≤ 2 ©

R2

(b) R = R 1 ∪ R 2

ª

Figure 1: Integration in x and then y is more natural Î for this Î region.ÎIntegration in y and then x would require the computation of two double integrals: R 1d A = R1 1d A + R2 1d A . y

y

R

R

x

(a) R = (x, y) | y ≤ x ≤ 1, 0 ≤ y ≤ 1 ©

x

(b) R = (x, y) | 0 ≤ y ≤ x, 0 ≤ x ≤ 1

ª

©

ª

Solution. Notice: If the substitution u = x 2 was chosen, there is no corresponding 2x in d u = 2xd x . We must reverse the order of integration to proceed. Sketch the region in the x y -plane and write R in set notation, Using the diagram on the right, we see that y is bounded between 0 and x . Thus, the integral of interest

8

may be computed as Z 1Z 0

1 y

2

e x d xd y = =

Z 1Z 0

Z

0 u=1

u=0

x

2

e x d yd x =

1

Z 0

ex

2

x

·Z 0

¸ Z 1 2 dy dx = e x xd x 0

µ ¶ 1 1 £ ¤1 1 eu d u = e u 0 = [e − 1] 2 2 2

■

Example 13 (Change to Polar Coordinates). Find the volume beneath the surface z = 16 − x 2 − y 2 and above the x y -plane. Solution. The volume beneath the surface and above the x y -plane is computed via the double integral Ï

2

R

2

2π Z 4

Z

Z

2

2π Z 4

3

(16 − r )r d r d θ = 16r − r d r d θ = 2π 0 0 ¸ ¤ £ 1 = 2π 8(4)2 − (4)4 = 2π 2(4)3 − (4)3 = 2π43 = 128π 4

16 − x − y d A =

0

4

Z

0

0

¸ · 1 4 4 2 16r − r d r = 2π 8r − r 4 0 3

·

■ y 4 3 2

R

1

10

−4 −3 −2 −1 −1 0

−2

0

2

4

2

3

x

4

−2

2

0

1

−3

−2

−4

(a) The portion of the surface z = 16 − x 2 − y 2 lying above the x y -plane.

(b) Region in the x y -plane.

Example 14 (Modification to Example 13). Find the volume beneath the surface in the first octant. Solution. The region of integration lies in the first quadrant. Because of symmetry, the volume is 1/4 the previous amount:

♠

■

y 4 3

Ï R

2

2

16 − x − y d A =

π/2 Z

Z 0

4 0

1 (16 − r )r d r d θ = 4 2

1 = 128π = 32π 4

2π Z

Z 0

4 0

3

16r − r d r d θ

2

R

1 −4 −3 −2 −1 −1

1

2

3

4

x

−2 −3 −4

Example 15. Find the volume of the the solid bounded by x = 0, x = 1 − z 2 , y = 0, z = 0, and z = 1 − y .

9

♠

Solution. Examine the solid by studying the intersection of the surfaces. We integrate in x first, so we describe the bounds on x T = {(x, y, z) | f (y, z) ≤ x ≤ g (y, z); (y, z) ∈ R} = {(x, y, z) | 0 ≤ x ≤ 1 − z 2 ; (y, z) ∈ R}

Now, we examine the y z -plane to fully describe the bounds of the solid: T = {(x, y, z) | 0 ≤ x ≤ 1 − z 2 , 0 ≤ z ≤ 1 − y, 0 ≤ y ≤ 1}.

The bounds are determined, so integrate: V=

Z 1Z 0

1−y

1−z 2

Z

0

0

1 d xd zd y =

Z 1Z 0

1−y

1 − z 2 d zd y =

0

Z 1· 0

1 z − z3 3

¸1−y 0

dy =

1

Z 0

1 (1 − y) − (1 − y)3 d y. 3

A u = 1 − y substitution simplifies the computation: V=

u=0

Z

u=1

1 (u − u 3 )(−1) d u = 3

1

Z 0

· 2 ¸1 1 u u4 1 1 5 u − u3 d u = − = − = 3 2 12 0 2 12 12

■

Idea. Cylindrical and spherical coordinate systems simplify integration on solids that are roughly cylindrical and spherical, respectively. The mathematics convention is to label the angle in the x y -plane measured in a CCW manner relative to the positive x axis as θ . In spherical, the angle measured relative to the positive z -axis is labeled φ (this is not the case in physics). Example 16 (Change to Cylindrical Coordinates). Evaluate the following integral by changing to cylindrical coordinates

I=

Z

1

Z p1−y 2 Z p

−1 −

1−y 2

1 −1

(x 2 + y 2 )3/2 d zd xd y

♠

Solution. The solid of integration is ¾ ½ q q D = (x, y, z) | − 1 ≤ z ≤ 1, − 1 − y 2 ≤ x ≤ 1 − y 2 , −1 ≤ y ≤ 1 .

This is a cylinder that opens up along the z -axis. In the x y -plane a cross sectional slice, which is constant with respect to z , is a unit circle. ½ ¾ q q © ª 2 2 R = (x, y) | − 1 − y ≤ x ≤ 1 − y , −1 ≤ y ≤ 1 = (x, y) | x 2 + y 2 ≤ 1

It is advantageous to switch to cylindrical coordinates. The solid of integration in cylindrical coordinates is D˜ = {(r, θ, z) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, −1 ≤ z ≤ 1}

The integral is I=

Z

1

Z

−1 0

2π Z 1 0

(r 2 )3/2 r d r d θd z = 2 · 2π

1

Z 0

10

r 3 r d r = 2 · 2π

1

Z 0

1 4π r 4 d r = 2 · 2π = 5 5

■

Example 17 (Change to Spherical Coordinates). For a sphere centered at the origin with radius equal to 4, find the volume of sphere that lies above the plane z = 2. ♠ Solution. Recall, the horizontal plane z = a > 0 in spherical coordinates is ρ cos(φ) = a =⇒ ρ =

a = a sec(φ) cos(φ)

The angle φ∗ at which the plane intersects the sphere is found by solving the following equation: 4 cos(φ∗ ) = 2 =⇒ cos(φ∗ ) =

π 1 =⇒ φ∗ = 2 3

Thus, the solid of integration is © ª D = (x, y, z) | 2 sec(φ) ≤ ρ ≤ 4, 0 ≤ φ ≤, 0 ≤ θ ≤ 2π .

The volume V is given by the integral

V=

2π Z π/3 Z 4

Z 0

= 2π =

0

2 sec(φ)

π/3

Z 0

2π 3 4 3 ·

sin(φ) π/3

Z 0

·

ρ 2 sin(φ)d ρd φd θ = 2π

ρ3 3

¸4 2 sec(φ)

sin(φ)d φ − 8

dφ = π/3

Z 0

2π 3

π/3

Z 0

π/3

Z 0

sin(φ)

4

Z

2 sec(φ)

ρ 2 d ρd φ

£ ¤ sin(φ) 43 − (2 sec(φ))3 d φ

¸ ¤ 2π £ 3 sin(φ) sec3 (φ)d φ =: 4 I 1 + 8I 2 3

We compute both of the remaining integrals, I1 =

π/3

Z 0

sin(φ)d φ = [− cos(φ)]π/3 0 = [− cos(π/3) + cos(0)] = [−1/2 + 1] = 1/2

and I2 =

π/3

Z 0

3

sin(φ) sec (φ)d φ =

π/3

Z 0

tan2 (φ) tan(φ) sec (φ)d φ = 2 ·

2

¸π/3 0

£ ¤ p = tan2 (π/3) − 0 = ( 3)2 − 0 = 3

Hence, V=

· ¸ 2π 3 1 2π 112π 4 · +8·3 = [32 + 24] = 3 2 3 3

■

Idea (Transformation). A transformation T relates two sets of variables (u, v) ∈ S and (x, y) ∈ R , ( T (u, v) = (x, y),

where

x = g (u, v)

y = h(u, v)

In the context of integration, a change of variable usually simplifies the region of integration from R to the simpler S . Now, the area of a region in x y -space is likely scaled when it is mapped to uv -space. To ensure that the change of variable does not alter the value of the integral, a scaling factor needs to be incorporated. The absolute value of the Jacobian is this scaling factor.

11

Definition (Jacobian). For a transformation T : x = g (u, v), y = h(u, v), the Jacobian of T is ¯ ¯ ∂x ¯ ∂(x, y) ¯¯ ∂u J (u, v) = =¯ (u, v) ¯ ∂y ¯ ∂u

∂x ∂v ∂y ∂v

¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯

Example 18 (Change of Variable). Evaluate the following integral using the indicated transformation Ï

x y d A;

R

R is the square: (0, 0), (1, 1), (2, 0), (1, −1),

using the transformation (

x =u+v

y =u−v

♠

Solution. The corners of the square map to the following points in uv -space T −1 (0, 0) = (0, 0) T −1 (1, 1) = (1, 0) T −1 (2, 0) = (1, 1)

T −1 (1, −1) = (0, 1)

We also know that linear transformations take lines to lines, so there is also a square in uv -space. It follows that the region of integration in uv -space is S = {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}

The Jacobian is ¯ ∂(x, y) ¯¯ 1 =¯ 1 (u, v)

¯ 1 ¯¯ = −1 − 1 = −2. −1 ¯

We now compute the integral: Ï R

xy d A =

Ï S

=2

(u + v)(u − v)| − 2| d ud v

Z 1Z 0

1 0

u 2 − v 2 d ud v

¸1 u3 − uv 2 d v 3 0 0 Z 1 1 =2 − v2 d v 0 3 · ¸1 1 v3 =2 v− =0 3 3 0 =2

Z 1·

■

Example 19 (Change of Variable). Let R be the region bounded by the ellipse x2 y 2 + = 1, a2 b2

Evaluate

Î R

|x y|d A .

a, b > 0. ♠

12

Solution. Let T be the transformation (

x = au y = bv

This transformation takes the ellipse to the unit circle in uv -space: (au)2 (bv)2 + = 1 =⇒ u 2 + v 2 = 1. a2 b2

The Jacobian is ¯ ∂(x, y) ¯¯ a = (u, v) ¯ 0

¯ 0 ¯¯ = ab. b ¯

We now evaluate the integral below, I=

Ï R

|x y|d A =

Ï

|au · bv| · ab d ud v = (ab)2

S

Ï S

|uv| d ud v = 4(ab)2

Ï

uv d ud v.

S1

Above, symmetry of f (u, v) = |uv| about both the x and y axes has been exploited, where S 1 is the sector of the circle lying in the first quadrant. Below, we change to polar coordinates, u = r cos(θ), v = r sin(θ): I = 4(ab)

2

= 4(ab)2

Ï S1

π/4

Z

= 4(ab)2 ·

uv d ud v = 4(ab)

0

cos(θ) sin(θ)d θ

2

π/4 Z 1

Z 0 1

Z 0

1 1 (ab)2 · = 2 4 2

0

r 2 cos(θ) sin(θ) r d r d θ

r 3 d r = 4(ab)2

1

Z 0

w dw ·

1 4 ■

13