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

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Welcome to March Calculus Madness!

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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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