Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 pdf

Hint, Solution Exercise 4.3 mathematica/calculus/integral/fundamental.nbEvaluateR x√x2+ 3 dx.. Hint, Solution Exercise 4.4 mathematica/calculus/integral/fundamental.nbEvaluateR cos xsin

4.6 Exercises

Exercise 4.1 (mathematica/calculus/integral/fundamental.nb)EvaluateR (2x + 3)10dx

Hint, Solution

Exercise 4.2 (mathematica/calculus/integral/fundamental.nb)EvaluateR (ln x) 2

x dx

Hint, Solution

Exercise 4.3 (mathematica/calculus/integral/fundamental.nb)EvaluateR x√x2+ 3 dx

Hint, Solution

Exercise 4.4 (mathematica/calculus/integral/fundamental.nb)EvaluateR cos xsin xdx

Hint, Solution

Exercise 4.5 (mathematica/calculus/integral/fundamental.nb)EvaluateR x3x−52 dx

where ∆x = b−a

N and xn = a + n∆x, to show that

Z 1 0

Z f (x) g(x)

f0(x − ξ) dξ

b From the above identity show that

f (x) = f (0) + xf0(0) +

Z x 0

ξf00(x − ξ) dξ

c Using induction, show that

1n!ξ

Exercise 4.16 (mathematica/calculus/integral/partial.nb)EvaluateR x3 +xx+12 −6xdx.

Hint, Solution

Exercise 4.17 (mathematica/calculus/integral/improper.nb)EvaluateR04 (x−1)1 2 dx

Hint, Solution

Exercise 4.18 (mathematica/calculus/integral/improper.nb)EvaluateR1

Hint, Solution

Hint 4.9

CONTINUE

Hint 4.10

a Evaluate the integral

b Use integration by parts to evaluate the integral

c Use integration by parts with u = f(n+1)(x − ξ) and dv = n!1ξn

Hint 4.11

The arc length from 0 to x is

Z x 0

a(x − 2) +

b(x + 2)

1 = a(x + 2) + b(x − 2)

Set x = 2 and x = −2 to solve for a and b.

Hint 4.17

Z 4 0

1(x − 1)2dx = lim

δ→0 +

Z 1−δ 0

1(x − 1)2 dx + lim

→0 +

Z 4 1+

1(x − 1)2 dxHint 4.18

Z 1 0

Z1

x2 + a2 dx = 1

aarctan

xa



4.8 Solutions

Solution 4.1

Z(2x + 3)10dxLet u = 2x + 3, g(u) = x = u−3

2 , g0(u) = 1

2.Z(2x + 3)10dx =

= (2x + 3)

11

22Solution 4.2

Z(ln x)2

x dx =

Z(ln x)2d(ln x)

dx dx

= (ln x)

3

3Solution 4.3

d(x2)

dx dx

= 12

(x2+ 3)3/23/2

= (x

2+ 3)3/2

3

Solution 4.4

Zcos xsin x dx =

Z1sin x

x3− 5

13

= 12

Solution 4.7

Let u = sin x and dv = sin x dx Then du = cos x dx and v = − cos x

Z π 0

sin2x dx = − sin x cos xπ0 +

Z π 0

cos2x dx

=

Z π 0

cos2x dx

=

Z π 0

(1 − sin2x) dx

= π −

Z π 0

sin2x dx

2

Z π 0

sin2x dx = π

Z π 0

sin2x dx = π

2Solution 4.8

Let H0(x) = h(x)

ddx

Z f (x) g(x)

h(ξ) dξ = d

dx(H(f (x)) − H(g(x)))

= H0(f (x))f0(x) − H0(g(x))g0(x)

= h(f (x))f0(x) − h(g(x))g0(x)Solution 4.9

First we compute the area for positive integer n

An =

Z 1 0

= 1

2 − 1

n + 1

Then we consider the area in the limit as n → ∞.

In Figure4.3 we plot the functions x1, x2, x4, x8, , x1024 In the limit as n → ∞, xn on the interval [0 1] tends tothe function

(

0 0 ≤ x < 1

1 x = 1Thus the area tends to the area of the right triangle with unit base and height

0.2 0.4 0.6 0.8 1 0.2

0.4 0.6 0.8 1

Figure 4.3: Plots of x1, x2, x4, x8, , x1024

Solution 4.10

1

f (0) +

Z x 0

f0(x − ξ) dξ = f (0) + [−f (x − ξ)]x0

= f (0) − f (0) + f (x)

= f (x)2

f (0) + xf0(0) +

Z x 0

ξf00(x − ξ) dξ = f (0) + xf0(0) + [−ξf0(x − ξ)]x0 −

Z x 0

1n!ξ

n+1f(n+1)(x − ξ)

x 0

−

Z x 0

Z x 0

1(n + 1)!ξ

n+1f(n+2)(x − ξ) dξThis shows that the hypothesis holds for n = m + 1 By induction, the hypothesis hold for all n ≥ 0

Solution 4.11

First note that the arc length from a to b is 2(b − a)

Z b a

p

1 + (f0(x))2dx =

Z b 0

p

1 + (f0(x))2dx −

Z a 0

f (x) is a continuous, piecewise differentiable function which satisfies f0(x) = ±√

3 at the points where it is differentiable.One example is

f (x) =√

3xSolution 4.12

= −x cos x + sin x + CSolution 4.14

Let u = x3 and dv = e2x dx Then du = 3x2dx and v = 12e2x

x2e2x dx

Let u = x2 and dv = e2x dx Then du = 2x dx and v = 1

2 e2x.Z

 1

2x e

2x−12

A(x − 2) +

B(x + 2)

1 = A(x + 2) + B(x − 2)Setting x = 2 yields A = 14 Setting x = −2 yields B = −14 Now we can do the integral

x2− 4dx =

Z 

14(x − 2) − 1

4(x + 2)

dx

x − 2

x + 2

+ C

310(x − 2) − 2

15(x + 3)

dx

Z 4 0

1(x − 1)2dx = lim

δ→0 +

Z 1−δ 0

1(x − 1)2 dx + lim

→0 +

Z 4 1+

1(x − 1)2 dx

Solution 4.18

Z 1 0

Z ∞ 0

1

x2+ 4dx = limα→∞

Z α 0

α 0

= 12

π

2 − 0

= π4

4.10 Quiz Solutions

Solution 4.1

Let a = x0 < x1 < · · · < xn−1 < xn = b be a partition of the interval (a b) We define ∆xi = xi+1− xi and

∆x = maxi∆xi and choose ξi ∈ [xi xi+1]

Z b a

√

x dx

=

Z 1 0

√

x dx +

Z 2 0

+ 2

3x

3/2

2 0

ddx

Solution 4.4

First we expand the integrand in partial fractions

1 + x + x2(x + 1)3 = a

 d

dx(1 + x + x

2)

 ... 0 .5 1. 5< /small>

2 -1< /small> -0 .5< /sup>

0 0 .5 1< /small>

0 0 .5 1. 5 2

Figure 5 .10 : Paraboloid, Tangent Plane and Line Connecting (1, ... point (1, 1, 2) this is

∇f (1, 1, 2) = 2i + 2j − k

We know a point on the tangent plane, (1, 1, 2), and the normal, ∇f (1, 1, 2) The equation of the plane is

∇f (1, 1, 2) ·... −

16

(1) (1 − 2) (1 − 3) =

12

(2)(2 − 1) (2 − 3) = −

12

(3)(3 − 1) (3 − 2) =

16 1

x(x − 1) (x − 2)(x −