Sunday, March 20, 2022

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

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Welcome to March Calculus Madness!


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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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