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Absolutely Greatest (Posted on 2008-04-27) Difficulty: 2 of 5
Evaluate:

5
[|y-3| + [y]] dy
-1

Bonus Question:

Work out the following definite integral:

5
(|y-3| + [y]) dy
-1

Notes:

(i) [x] is defined as the greatest integer ≤ x

(ii) |x| is the absolute value of x.

  Submitted by K Sengupta    
Rating: 2.0000 (1 votes)
Solution: (Hide)
Let G(y) = [|y-3| + [y]]

Since, {y} = y – [y], we must have:

G(y) = [3 – {y}] = 2, whenever y < 3

Similarly, we have:

G(y) = |y-3 + [y]] = 2*[y] – 3, whenever y ≥ 3

Accordingly, it follows that:

Integral (-1 to 5) G(y) dy

= Integral (-1 to 3) G(y) dy + Integral (3 to 5) G(y) dy

= 4*2 + 3 + 5

= 16

Consequently, the required value of the definite integral is 16.

Solution To The Bonus Question:

Let H(y) = |y-3| + [y]

Then, we observe that:

H(y) = 3 – {y}, whenever y < 3, and:

H(y) = 2*[y] – 3 + {y}, whenever y ≥ 3

Accordingly, it follows that:

Integral (-1 to 5) H(y) dy

= Integral (-1 to 3) H(y) dy + Integral (3 to 5) H(y) dy

= 4(3 – Integral (0 to 1) y dy) + 2*Integral (0 to 1) y dy + 3 + 5

= 20 – 2*Integral (0 to 1)y dy

= 19

Consequently, the required value of the definite integral is 19.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionhoodat2008-04-28 16:37:29
SolutionsolutionCharlie2008-04-27 18:58:18
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