March Calculus Madness Sweet Sixteen - Day 16: A Parametric Integral Example

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Let:

x(t) = a * cos t

y(t) = b * sin t

Then:

∫ y(x) dx = ∫ y(t) * x'(t) dt

x'(t) = -a * sin t

And:

∫ y(t) * x'(t) dt

= ∫ -a * sin t * b * sin t dt

= -a * b * ∫ sin^2 t dt

= -a * b * ∫1/2 - 1/2 * cos(2*t) dt

= -a * b * (t/2 - sin(2*t)/4) + C

That wraps up March Calculus Madness 2022.  

Next Post:  April 9, 2022

Have a great day and hope you enjoyed the series!  

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 15: r = α * (1 - cos Θ)

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r = α * (1 - cos Θ)

r = α * (1 - cos Θ)

r^2 = α^2 * (1 - 2 * cos Θ + cos^2 Θ)

dr/dΘ = α * sin Θ

(dr/dΘ)^2 = α^2 * sin^2 Θ

Area from 0 ≤ Θ ≤ 2*π

1/2 * ∫ α^2 * (1 - cos Θ)^2  dΘ from Θ = 0 to Θ = 2*π

= α^2/2 * ∫ 1 - 2 * cos Θ + cos^2 Θ dΘ from Θ = 0 to Θ = 2*π

= α^2/2 * ∫ 1 - 2 * cos Θ + 1/2 * cos(2*Θ) + 1/2 dΘ from Θ = 0 to Θ = 2*π

= α^2/2 * (3/2 * Θ - 2 * sin Θ + 1/4 * sin(2*Θ) from Θ = 0 to Θ = 2*π)

= 3/2 * π * α^2

Arc Length from 0 ≤ Θ ≤ 2*π

r^2 + (dr/dΘ)^2

= α^2 * (1 - 2 * cos Θ + cos^2 Θ) + α^2 * sin^2 Θ

= α^2 - 2  * α^2 * cos Θ + α^2 * (cos^2 Θ + sin^2 Θ)

= α^2 - 2  * α^2 * cos Θ + α^2 

= 2 * α^2 - 2 * α^2 * cos Θ

= 2 * α^2 *(1  - cos Θ)

Arc Length:

∫ 2 * α^2 *(1  - cos Θ) dΘ from Θ = 0 to Θ = 2*π

= α * √2 * ∫ √(1  cos Θ) dΘ from Θ = 0 to Θ = 2*π

= α * √2 * 4 * √2

= 8 * α

Eddie 

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March Calculus Madness Sweet Sixteen - Day 14: The Arc Length of a Spiral

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The Length of a Spiral from 0 ≤ Θ ≤ m

The equation of a spiral:  r = α * Θ

The arc length of a polar equation r(Θ):  ∫ √(r(Θ)^2 + (dr/dΘ)^2) dΘ

r = α * Θ

r^2 = α^2 * Θ^2

dr = α dΘ

(dr/dΘ)^2 = α^2

∫ √(α^2 * Θ^2 + α^2) dΘ from Θ = 0 to Θ = m

= α * ∫ √(Θ^2 + 1) dΘ from Θ = 0 to Θ = m

= α/2 * ( ln|Θ + √(1 + Θ^2)| + Θ * √(1 + Θ^2) for Θ = 0 to Θ = m)

(see below)

= α/2 * ( ln|m + √(1 + m^2)| + m * √(1 + m^2) )

Aside:

∫ √(1 + Θ^2) dΘ

Let Θ = tan x

dΘ = sec^2 x dx

∫ √(1 + Θ^2) dΘ

= ∫ √(1 + tan^2 x) * sec^2 x  dx

= ∫ √(sec^2 x) * sec^2 x dx

= ∫ sec^3 x dx

= 1/2 * ∫ sec x dx + (sec x * tan x)/2 + C

(per reduction integration rule for sec x)

= 1/2 * ln|tan x + sec x| + 1/2 * sec x * tan x + C

with: 

Θ = tan x

arctan Θ = x

sec(arctan Θ) = sec x

√(1 + x^2) = sec x

= 1/2 * ln|Θ + √(1 + Θ^2)| + 1/2 * Θ * √(1 + Θ^2) + C

Eddie 

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March Calculus Madness Sweet Sixteen - Day 13: Some Double Integrals

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Double Integration Time

(I) 

∫ ∫ sin(a*x + b*y) dx dy

= ∫ -1/a * cos(a*x + b*y) + C1 dy

= -1/(a*b) * sin(a*x + b*y) + C1*y + C2

(II) 

∫ ∫ sin(a*x + b*y) dy dx

= ∫ -1/b * cos(a*x + b*y)+ C1 dx

= -1/(a*b) * sin(a*x + b*y) + C1 * x + C2

For (I) and (II) to equal,  x = y

(III)

∫ ∫ e^(a*x + b*y) dx dy

= ∫ 1/a * e^(a*x + b*y)+ C1 dy

= 1/(a*b) * e^(a*x + b*y) + C1 * y + C2

(IV)

∫ ∫ e^(a*x + b*y) dy dx

= ∫ 1/b * e^(a*x + b*y) + C1  dx

= 1/(a*b) * e^(a*x + b*y) + C1 * x + C2

For (III) and (IV) to be equal, x = y

Eddie

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March Calculus Madness Sweet Sixteen - Day 12: Arc Length of Sine and Cosine

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What is the arc length of y = sin x and y = cos x from x = 0 to x = 2 * π

The arc length of y(x) is calculated by:

∫ √(1 + (dy/dx)^2) dx for x = a to x = b

For y(x) = sin x, dy/dx = cos x, (dy/dx)^2 = cos^2 x

Arc length of y = sin x from x = 0 to x = 2*π

∫ √(1 + cos^2 x) dx for x = 0 to x = 2*π ≈ 7.64039557806

Likewise, for y(x) = cos x, dy/dx = -sin x, (dy/dx)^2 = sin^2 x

∫ √(1 + sin^2 x) dx for x = 0 to x = 2*π ≈ 7.64039557806

Yes, the arc lengths are the approximately the same.  

Eddie

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March Calculus Madness Sweet Sixteen - Day 11: ∫ int(x) dx

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f(x) = IP(x), integer function, HP Prime

int(x):  integer part function

Domain:

0 ≤ x < 1;  int(x) = 0

1 ≤ x < 2;  int(x) = 1

2 ≤ x < 3;  int(x) = 2

3 ≤ x < 4;  int(x) = 3

Hence:

∫ int(x) dx for x = 1 to x = 2 

= lim a→2- ∫ 1 dx for x = 1 to x = a

= lim a→2-  a - 1

= 2 - 1

= 1

∫ int(x) dx for x = 2 to x = 3 

= lim a→3- ∫ 1 dx for x = 1 to x = a

= lim a→3-  2 * a - 2 * 2

= 2 * 3 - 4 

= 2

∫ int(x) dx for x = 3 to 4 

= lim a→4- ∫ 1 dx for x = 1 to a

= lim a→4-  3 * a - 3 * 3 

= 4 * 3 - 9

= 3

and so on...

∫ int(x) dx for x = 1 to x =3 

= (∫ int(x) dx for x =1 to x=2 )+ (∫ int(x) dx for x =2 to x=3) + (∫ int(x) dx for x=3 to x=4 )

= 1 + 2 + 3

= 6

The General Integral ∫ int(x) dx for x = 1 to x = t

∫ int(x) dx for x = 1 to x = t

= ∫ int(x) dx for x = 1 to x = int(t) + ∫ int(x) dx for x = int(t) to x = t

= lim a→int(t)- ∫ int(x) dx for x = 1 to x = a + ∫ int(x) dx for x = int(t) to x = t

= (1 + 2 + 3 + 4 + .... + t-1) + t * int(t) - int(t) * int(t)

= t * (t-1)/2 + t * int(t) - int^2(t)

Example:

∫ int(x) dx for x = 1 to x = 8.3

= (7 * 8)2 + (8.3 * 8 - 8^2)

= 30.4

Eddie 

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March Calculus Madness Sweet Sixteen - Day 10: ∫ frac(x) dx

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frac(x) - HP Prime

frac(x):  fractional part function

Domain:  0 ≤ frac(x) < 1

∫ frac(x) dx for x = 0 to 1

According to the graph above, the area between resembles as a right triangle.

When 0≤x<1, frac(x) = x

Hence:

∫ frac(x) dx for x = 0 to x = 1

Note:  

lim a→1-  (∫ frac(x) dx for x = 0 to x = a)

= lim a→1-  (∫ x dx for x = 0 to x = a)

= lim a→1-  (∫ x dx for x = 0 to x = a)

= lim a→1-  (a^2/2 - 0)

= 1/2

What if the upper limit is less than 1? 

Let b where, 0≤x≤b<1  (b<1):

∫ frac(x) dx for x = 0 to x = b

= ∫ x dx for x = 0 to x = b

= b^2/2

Eddie 

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 9: ∫ e^x/(e^x + 1) dx and ∫ e^x/(e^x - 1) dx

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∫ e^x/(e^x + 1) dx

Let z = e^x

Then: 

ln z = x

1/z dz = dx

∫ e^x/(e^x + 1) dx

= ∫ z/(z + 1) * 1/z dz

= ∫ 1/(z + 1) dz

= ln |z + 1| + C

= ln |e^x + 1| + C

∫ e^x/(e^x - 1) dx

Again, let z = e^x

∫ e^x/(e^x - 1) dx

= ∫ z/(z - 1) * 1/z dz

= ∫ 1/(z - 1) dz

= ln |z - 1| + C

= ln |e^x - 1| + C

Eddie 

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March Calculus Madness Sweet Sixteen - Day 8: ∫ ln^2 x dx and ∫ ln^3 x dx

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Integration by parts to the rescue!

∫ ln^2 x dx

u = ln^2 x,   dv = dx

du = 2 * ln * 1/x dx,   v = x

∫ ln^2 x dx

= x * ln^2 x - ∫ 2 * ln x * 1/x * x dx

= x * ln^2 x - ∫ 2 * ln x dx

u = ln x, dv = 2 dx

du = 1/x dx, v = 2 * x

= x * ln^2 x - (2 * x * ln x - ∫ 1/x * 2 * x dx)

= x * ln^2 x - (2 * x * ln x - ∫ 2 dx)

= x * ln^2 x - 2 * x * ln x + ∫ 2 dx

= x * ln^2 x - 2 * x * ln x + 2 * x + C

∫ ln^3 x dx

u = ln^3 x dx, dv = dx

du = 3 * ln^2 x dx, v = x

= x * ln^3 x - ∫3 * ln^2 x * 1/x * x dx

= x * ln^3 x - 3 * ∫ ln^2 x dx

= x * ln^3 x - 3 * (x * ln^2 x - 2 * x * ln x + 2 * x + C)

(see the above)

=  x * ln^3 x - 3 * x * ln^2 x + 6 * x * ln x - 6 * x + D   

(where D = 3 * C, C and D are constants)

Eddie

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March Calculus Madness Sweet Sixteen - Day 7: ∫ 1/(1 + cos x) dx and ∫ 1/(1 + sin x) dx

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∫ 1/(1 + cos x) dx and ∫ 1/(1 + sin x) dx

To tackle these integrals, we will make use of the following derivatives and trig identities:

d/dx tan x = sec^2 x dx

d/dx cot x = -csc^2 x dx

d/dx csc x = -cot x * csc x

d/dx sec x = tan x * sec x

sin^2 x + cos^2 x = 1

cot x = 1/tan x,  csc x = 1/sin x,  sec x = 1/cos x

∫ 1/(1 + cos x) dx

= ∫ 1/(1 + cos x) * (1 - cos x)/(1 - cos x) dx

= ∫ (1 - cos x)/(1 - cos^2 x) dx

= ∫ (1 - cos x)/sin^2 x dx

= ∫ csc^2 x - cot x * csc x dx

= -cot x + csc x + C

∫ 1/(1 + sin x) dx

= ∫ 1/(1 + sin x) * (1 - sin x)/(1 - sin^2 x) dx

= ∫ (1 - sin x)/(1 - sin^2 x) dx

= ∫ (1 - sin x)/(cos^2 x) dx

= ∫ sec^2 x - tan x * sec x dx

= tan x - sec x + C

Note:  CAS systems in graphing calculators will default to using sin x, cos x, and tan x.  

Eddie

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March Calculus Madness Sweet Sixteen - Day 5: x^n ∙ √(1 + x)

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d/dx x^n ∙ √(1 + x)

Here we can make use the multiplication rule:

d/dx f(x) ∙ g(x) = f(x) ∙ g'(x) + f'(x) ∙ g(x)

In this case:

f(x) = x^n

g(x) = √(1 + x) = (1 + x)^(1/2)

Then:

f'(x) = n ∙ x^(n-1)

g'(x) = 1/2 ∙ (1 + x)^(-1/2)

And: 

d/dx x^n ∙ √(1 + x) 

= x^n ∙ 1/2 ∙ (1 + x)^(-1/2) + n ∙ x^(n-1) ∙ (1 + x)^(1/2)

For indefinite integrals, I will do two specific cases of n.

∫ x ∙ √(1 + x) dx   (n = 1)

Using integration by parts:

u = x,  dv = (1 + x)^(1/2) dx

du = dx,  v = 2/3 ∙ (1+x)^(3/2)

∫ x ∙ √(1 + x) dx

= 2/3 ∙ (1+x)^(3/2) ∙ x - ∫ (1 + x)^(1/2) dx

= 2/3 ∙ (1+x)^(3/2) ∙ x - 2/3 ∙ (1 + x)^(3/2) + C

= 2/3 ∙ (1 + x)^(3/2) ∙ (x - 1) + C

∫ x^2 ∙ √(1 + x) dx   (n = 2)

Let z = (1 + x)^(1/2)

dz = 1/2 ∙ (1+ x)^(-1/2) dx

2 ∙ (1+x)^(1/2) dz = dx

2 ∙ z  dz = dx

z^2 = 1 + x

z^2 - 1 = x

z^4 - 2 ∙ z^2 + 1 = x^2

∫ x^2 ∙ √(1 + x) dx   

= ∫ (z^4 - 2 ∙ z^2 + 1) ∙ z ∙ 2 ∙ z dz

= ∫ (z^4 - 2 ∙ z^2 + 1) ∙ 2 ∙ z^2 dz

= ∫ 2 ∙ z^6 - 4 ∙ z^4 + 2 ∙ z^2 dz

= 2/7 ∙ z^7 - 4/5 ∙ z^5 + 2/3 ∙ z^3 + C

= 2/7 ∙ (1 + x)^(7/2) - 4/5 ∙ (1 + x)^(5/2) + 2/3 ∙ (1 + x)^(3/2) + C

(z = (1 + x)^(1/2))

Eddie

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March Calculus Madness Sweet Sixteen - Day 4: ∫x^2 / √(1 - x^2) dx

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∫x^2 / √(1 - x^2)  dx

Substitute:  z = arcsin x

sin z = x

sin^2 z = x^2

and dz = 1/√(1 - x^2) dx

Then:

∫x^2 / √(1 - x^2)  dx

= ∫ sin^2 z dz

= ∫  1/2 - 1/2 ∙ cos(2∙z) dz    

(by trigonometric identity of sin^2 z = 1/2 - 1/2 ∙ cos(2∙z))

= z/2 - 1/4 ∙ sin(2∙z) + C

= z/2 - 1/2 ∙ sin z ∙ cos z  + C

(by trigonometric identity of sin(2∙z) = 2 ∙ cos z ∙ sin z)

= 1/2 ∙ arcsin x - 1/2 ∙ sin(arcsin x) ∙ cos(arcsin x) + C

= 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - x^2) + C

(sin(arcsin x) = x, cos(arcsin x) = √(1 - x^2)

Summary:

∫x^2 / √(1 - x^2)  dx  = 1/2 ∙ arcsin x - 1/2 ∙ x ∙ √(1 - x^2) + C

Eddie

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 3: Derivative and Integral of x^n∙(1+x)^2

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x^n ∙ (1 + x)^2

= x^n ∙ (1 + 2 ∙ x + x^2)

= x^n + 2 ∙x^(n+1) + x^(n+2)

d/dx x^n ∙ (1 + x)^2 

= d/dx x^n + 2 ∙x^(n+1) + x^(n+2)

= n ∙ x^(n-1) + 2 ∙ (n+1) ∙ x^n + (n+2) ∙ x^(n+1)

∫ x^n ∙ (1 + x)^2 dx

= ∫ x^n + 2 ∙x^(n+1) + x^(n+2) dx

= x^(n+1)/(n+1) + (2 ∙ x^(n+2))/(n+2) + x^(n+3)/(n+3) + C

Eddie

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 2: Derivative and Integral of the Absolute Value Function

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What is the derivative and the indefinite integral of the absolute value?

By defintion:

| x | = x when x ≥ 0, -x when x < 0

Hence:

d/dx | x | =   1 when x ≥ 0, and -1 when x < 0

and 

∫ | x | dx = x^/2 + C when x ≥ 0, abnd -x^2/2 + C when x < 0

What about |a∙x + b|?

The function |a∙x + b| hits the x-axis when:

a∙x + b = 0

a∙x = -b

x = -b/a

|a∙x + b| = 

(a∙x + b) when x ≥ (-b/a), 

and -(a∙x + b) when < (-b/a)

d/dx |a∙x + b| = 

a when x ≥ (-b/a),

and -A when < (-b/a)

∫ |a∙x + b| dx = 

A ∙ x^2/2 + C  when x ≥ (-b/a), 

and -A ∙ x^2/2 + C  when < (-b/a)

Eddie  

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

March Calculus Madness Sweet Sixteen - Day 1: Double Integration

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For the two-variable function f(x,y), can we assume that ∫ ∫ f(x,y) dx dy = ∫∫ f(x,y) dy dx?  

Two simple examples:

Equation 1:

∫ ∫ x^2 + y^2 dx dy

= ∫ x^3/3 + y^2 ∙ x + C1 dy

x^3 ∙ y/3 + y^3 ∙ x/3 + C1 ∙ y + C2

Equation 2: 

∫ ∫ x^2 + y^2 dy dx

∫ x^2 ∙ y + y^3/3 + C1 dx

x^3 ∙ y/3 + y^3 ∙ x/3 + C1∙ x + C2

However, for both Equation 1 and Equation 2 to be equal:

x^3 ∙ y/3 + y^3 ∙ x/3 + C1 ∙ y + C2 = x^3 ∙ y/3 + y^3 ∙ x/3 + C1∙ x + C2

C1 ∙ y = C1 ∙ x

y = x

By this example alone, we cannot assume that ∫ ∫ f(x,y) dx dy = ∫∫ f(x,y) dy dx.

Eddie

All original content copyright, © 2011-2022.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.