perplexus.info :: Calculus : A Definite Integral Problem

A Definite Integral Problem (Posted on 2006-08-10)

Consider [x] as the greatest integer function of x and {x}=x–[x].

Evaluate ∫{√x} dx for x=1 to 484.

NOTE: The greatest integer function is defined as a function that produces the "greatest integer less than or equal to the number" operated upon, symbol [x] or sometimes [[x]]. If the number is an integer, use that integer. If the number is not an integer, use the next smaller integer.

Submitted by K Sengupta

Rating: 4.0000 (1 votes)

Solution: (Hide)

Integral({ sqrt(x)} dx; x= 1 to n^2
= Sum( (K=1 to K= n-1) (Integral (sqrt(x) – [sqrt(x)]) dx; x= K^2 to (K+1)^2))
Now, (Integral (sqrt(x) – [sqrt(x)]) dx; x= K^2 to (K+1)^2))
= Integral(sqrt(x) – K);dx; x= x= K^2 to (K+1)^2))
= (2/3*(x^ 3/2) – K*x ); lower limit(x)= K^2 and upper limit(x) (K+1)^2
= 2/3*(3*K^2 +3*K +1) – K(2K+1)
= K + 2/3

Consequently:

Integral({ sqrt(x)} dx; x= 1 to n^2
= Sum( K=1 to K= n-1) (K + 2/3)
= n(n-1)/2 + 2/3*(n-1)
=(n-1)(3n +4)/6

Substituting n = 22 , we have:

Integral ({sqrt(x)}) dx; x = 1 to 484

= 21*70/6 = 245

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: question	Richard	2006-08-10 20:03:22
question	Gatachiu	2006-08-10 18:04:04
solution	Gatachiu	2006-08-10 17:53:19
Confirmation	Richard	2006-08-10 14:31:28
Solution	Eric	2006-08-10 13:37:48

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