Can-you-evalate-A-a-b-f-x-dx-for-example-0-5-2-5-x-2-dx-

Question Number 2629 by Filup last updated on 24/Nov/15

Can you evalate :

A = \int_{a}^{b} ⌊ f (x) ⌋ d x

f o r e x a m p l e :

\int_{0.5}^{2.5} ⌊ x^{2} ⌋ d x

Commented by Filup last updated on 24/Nov/15

\int_{0.5}^{2.5} ⌊ x^{2} ⌋ d x = \int_{0.5}^{1} ⌊ x^{2} ⌋ d x + \int_{1}^{2} ⌊ x^{2} ⌋ d x + \int_{2}^{2.5} ⌊ x^{2} ⌋ d x

?

Commented by Yozzi last updated on 24/Nov/15

Y e s, b u t t h e r e i s a w a y t o s o l v e

i n t e g r a l s l i k e t h a t u s i n g t h e

g e o m e t r i c i n t e r p r e t a t i o n o f t h e

i n t e g r a l .

Answered by Yozzi last updated on 24/Nov/15

F o r 0 ⩽ x < 1 \Rightarrow ⌊ x^{2} ⌋ = 0

F o r 1 ⩽ x < \sqrt{2} \Rightarrow ⌊ x^{2} ⌋ = 1

F o r \sqrt{2} ⩽ x < \sqrt{3} \Rightarrow 2 ⩽ x^{2} < 3 \Rightarrow ⌊ x^{2} ⌋ = 2

F o r \sqrt{3} ⩽ x < 2 \Rightarrow 3 ⩽ x^{2} < 4 \Rightarrow ⌊ x^{2} ⌋ = 3

F o r 2 ⩽ x < 2.1 \Rightarrow 4 ⩽ x^{2} < 4.41 \Rightarrow ⌊ x^{2} ⌋ = 4

F o r 2.1 ⩽ x < 2.23 \Rightarrow 4.41 ⩽ x^{2} < 4.9729 \Rightarrow ⌊ x^{2} ⌋ = 4

F o r 2.23 ⩽ x < \sqrt{5} \Rightarrow ⌊ x^{2} ⌋ = 4

F o r \sqrt{5} ⩽ x < \sqrt{6} \Rightarrow ⌊ x^{2} ⌋ = 5

F o r \sqrt{6} ⩽ x ⩽ 2.5 \Rightarrow ⌊ x^{2} ⌋ = 6

∴ \int_{0}^{2.5} ⌊ x^{2} ⌋ d x = (1 - 0) \times 0 + (\sqrt{2} - 1) \times 1 + 2 (\sqrt{3} - \sqrt{2})

+ 3 (2 - \sqrt{3}) + 4 (2.1 - 2)

+ 4 (2.23 - 2.1) + 4 (\sqrt{5} - 2.23)

+ 5 (\sqrt{6} - \sqrt{5}) + 6 (2.5 - \sqrt{6})

\int_{0}^{2.5} ⌊ x^{2} ⌋ d x = (1) (\sqrt{2} - 1) + 2 (\sqrt{3} - \sqrt{2}) + 3 (\sqrt{4} - \sqrt{3}) + 4 (\sqrt{5} - \sqrt{4}) + 5 (\sqrt{6} - \sqrt{5}) + 6 (2.5 - \sqrt{6})

I = \int_{0}^{2.5} ⌊ x^{2} ⌋ d x = \underset{r = 1}{\sum^{5}} r (\sqrt{r + 1} - \sqrt{r}) + 6 (2.5 - \sqrt{6})

I = \int_{0.5}^{2.5} ⌊ x^{2} ⌋ d x a s w e l l s i n c e

\int_{0}^{2.5} ⌊ x^{2} ⌋ d x = \int_{0.5}^{2.5} ⌊ x^{2} ⌋ d x + \int_{0}^{0.5} ⌊ x^{2} ⌋ d x

a n d \int_{0}^{0.5} ⌊ x^{2} ⌋ d x = 0 (∵ f o r 0 ⩽ x < 1 \Rightarrow 0 ⩽ x^{2} < 1 \Rightarrow ⌊ x^{2} ⌋ = 0) .

\int_{0}^{\frac{n}{2}} ⌊ x^{2} ⌋ d x = \underset{r = 1}{\sum^{n}} r (\sqrt{r + 1} - \sqrt{r}) + (n + 1) (\frac{n}{2} - \sqrt{n + 1}) ? ? ? (n ⩾ 3, n o d d)

Commented by Filup last updated on 24/Nov/15

Wow thats amazing! So crazy!

Facebook Tweet Pin