Evaluate : ∫_(−5) ^( 5) x^2 [x+(1/2)]dx = ? where [.]= greatest integer function


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Question Number 42823 by rahul 19 last updated on 03/Sep/18

$Evaluate :$ $\int_{- 5}^{5} x^{2} [x + \frac{1}{2}] d x = ?$ $w h e r e [.] = g r e a t e s t i n t e g e r f u n c t i o n$
Commented by prof Abdo imad last updated on 03/Sep/18
$changement x =−5+t give I = ∫_0 ^(10) (−5+t)^2 [−5+t +(1/2)]dt =∫_0 ^(10) (t^2 −10t +25){−5 +[t+(1/2)]}dt =−5 ∫_0 ^(10) (t^2 −10t +25)dt +∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt but ∫_0 ^(10) (t^2 −10t +25)dt =[(t^3 /3) −5t^2 +25t]_0 ^(10) =((10^3 )/3) −500 +250 =((1000)/3) −250 =((1000−750)/3) =((250)/3) also we have ∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt =Σ_(k=0) ^9 ∫_k ^(k+1) (t^2 −10t +25)[t+(1/2)]dt but we know that [x+y]=[x]+[y]+ξ withξ=0or ξ=1 ∫_0 ^(10) (t^2 −10t +25)[t+(1/2)]dt= =Σ_(k=0) ^9 ∫_k ^(k+1) (t^2 −10t +25)(k +ξ)dt =Σ_(k=0) ^9 (k+ξ) [(t^3 /3) −5t^2 +25t]_k ^(k+1) =Σ_(k=0) ^9 (k+ξ){ (((k+1)^3 )/3) −5(k+1)^2 +25(k+1) −(k^3 /3) +5k^2 −25k} =.....$
$c h a n g e m e n t x = - 5 + t g i v e$ $I = \int_{0}^{10} {(- 5 + t)}^{2} [- 5 + t + \frac{1}{2}] d t$ $= \int_{0}^{10} (t^{2} - 10 t + 25) {- 5 + [t + \frac{1}{2}]} d t$ $= - 5 \int_{0}^{10} (t^{2} - 10 t + 25) d t + \int_{0}^{10} (t^{2} - 10 t + 25) [t + \frac{1}{2}] d t$ $b u t \int_{0}^{10} (t^{2} - 10 t + 25) d t = {[\frac{t^{3}}{3} - 5 t^{2} + 25 t]}_{0}^{10}$ $= \frac{10^{3}}{3} - 500 + 250 = \frac{1000}{3} - 250 = \frac{1000 - 750}{3}$ $= \frac{250}{3} a l s o w e h a v e$ $\int_{0}^{10} (t^{2} - 10 t + 25) [t + \frac{1}{2}] d t$ $= \sum_{k = 0}^{9} \int_{k}^{k + 1} (t^{2} - 10 t + 25) [t + \frac{1}{2}] d t$ $b u t w e k n o w t h a t [x + y] = [x] + [y] + ξ w i t h ξ = 0 o r ξ = 1$ $\int_{0}^{10} (t^{2} - 10 t + 25) [t + \frac{1}{2}] d t =$ $= \sum_{k = 0}^{9} \int_{k}^{k + 1} (t^{2} - 10 t + 25) (k + ξ) d t$ $= \sum_{k = 0}^{9} (k + ξ) {[\frac{t^{3}}{3} - 5 t^{2} + 25 t]}_{k}^{k + 1}$ $= \sum_{k = 0}^{9} (k + ξ) {\frac{{(k + 1)}^{3}}{3} - 5 {(k + 1)}^{2} + 25 (k + 1)$ $- \frac{k^{3}}{3} + 5 k^{2} - 25 k} = . . . . .$
	Answered by MJS last updated on 03/Sep/18

	$in steps :$ $[x + \frac{1}{2}] = c$ $x \in [- 5; - 4.5 [\Rightarrow c = - 5$ $x \in [- 4.5; - 3.5 [\Rightarrow c = - 4$ $x \in [- 3.5; - 2.5 [\Rightarrow c = - 3$ $x \in [- 2.5; - 1.5 [\Rightarrow c = - 2$ $x \in [- 1.5; - . 5 [\Rightarrow c = - 1$ $x \in [- . 5; . 5 [\Rightarrow c = 0$ $x \in [. 5; 1.5 [\Rightarrow c = 1$ $x \in [1.5; 2.5 [\Rightarrow c = 2$ $x \in [2.5; 3.5 [\Rightarrow c = 3$ $x \in [3.5; 4.5 [\Rightarrow c = 4$ $x \in [4.5; 5] \Rightarrow c = 5$ $\underset{- 5}{\int^{5}} x^{2} [x + \frac{1}{2}] d x = Σ c \int x^{2} d x = Σ \frac{c}{3} {[x^{3}]}_{a}^{b} =$ $= - \frac{5}{3} ({(- 4.5)}^{3} - {(- 5)}^{3}) - \frac{4}{3} ({(- 3.5)}^{3} - {(- 4.5)}^{3}) - \frac{3}{3} ({(- 2.5)}^{3} - {(- 3.5)}^{3}) . . .$ $. . . + \frac{3}{3} (3 . 5^{3} - 2 . 5^{3}) + \frac{4}{3} (4 . 5^{3} - 3 . 5^{3}) + \frac{5}{3} (5^{3} - 4 . 5^{3}) =$ $= - \frac{5}{3} (5^{3} - 4 . 5^{3}) + \frac{5}{3} (5^{3} - 4 . 5^{3}) . . . = 0$
	Commented by rahul 19 last updated on 03/Sep/18
	Thank you sir ! ��
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