hello floor function ∫_a ^b ⌊x⌋ dx a,b∈z and b>a =∫_0 ^b ⌊x⌋ dx −∫_0 ^a ⌊x⌋ dx =((b^2 −b)/2)−((a^2 −a)/2) ....(1) now ∫_m ^k ⌊x⌋ dx when m,k∉z when m<a<b<k b=[k] and a=[m] ∫_m ^a ⌊x⌋dx +∫_a ^b ⌊x⌋dx +∫_b ^k ⌊x⌋dx =(a−m)⌊m⌋+((b^2 −b)/2)−((a^2 −a)/2)+(k−b)⌊k⌋ =(⌊m⌋−m)⌊m⌋+((⌊k⌋^2 −⌊k⌋)/2)−((⌊m⌋^2 −⌊m⌋)/2)+(k−⌊k⌋)⌊k⌋ ⌊m⌋^2 −m⌊m⌋ +(1/2)⌊k⌋^2 −(1/2)⌊k⌋−(1/2)⌊m⌋^2 −(1/2)⌊m⌋+k⌊k⌋−⌊k⌋^2 =k⌊k⌋−m⌊m⌋+(1/2)⌊m⌋^2 −(1/2)⌊k⌋^2 +(1/2)⌊m⌋−(1/2)⌊k⌋ ∴∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 example ∫_(−1.5) ^(3.7) ⌊x⌋dx=(3.7)3−((−1.5)(−2))+(1/2)(−2−3)(−2+3+1) =11.1−3−5=3.1


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Question Number 88955 by M±th+et£s last updated on 14/Apr/20

$h e l l o$ $f l o o r f u n c t i o n$ $\int_{a}^{b} ⌊ x ⌋ d x a, b \in z a n d b > a$ $= \int_{0}^{b} ⌊ x ⌋ d x - \int_{0}^{a} ⌊ x ⌋ d x = \frac{b^{2} - b}{2} - \frac{a^{2} - a}{2} . . . . (1)$ $n o w$ $\int_{m}^{k} ⌊ x ⌋ d x w h e n m, k \notin z w h e n m < a < b < k$ $b = [k] a n d a = [m]$ $\int_{m}^{a} ⌊ x ⌋ d x + \int_{a}^{b} ⌊ x ⌋ d x + \int_{b}^{k} ⌊ x ⌋ d x$ $= (a - m) ⌊ m ⌋ + \frac{b^{2} - b}{2} - \frac{a^{2} - a}{2} + (k - b) ⌊ k ⌋$ $= (⌊ m ⌋ - m) ⌊ m ⌋ + \frac{⌊ k ⌋^{2} - ⌊ k ⌋}{2} - \frac{⌊ m ⌋^{2} - ⌊ m ⌋}{2} + (k - ⌊ k ⌋) ⌊ k ⌋$ $⌊ m ⌋^{2} - m ⌊ m ⌋ + \frac{1}{2} ⌊ k ⌋^{2} - \frac{1}{2} ⌊ k ⌋ - \frac{1}{2} ⌊ m ⌋^{2} - \frac{1}{2} ⌊ m ⌋ + k ⌊ k ⌋ - ⌊ k ⌋^{2}$ $= k ⌊ k ⌋ - m ⌊ m ⌋ + \frac{1}{2} ⌊ m ⌋^{2} - \frac{1}{2} ⌊ k ⌋^{2} + \frac{1}{2} ⌊ m ⌋ - \frac{1}{2} ⌊ k ⌋$ $∴ \int_{m}^{k} ⌊ x ⌋ d x = k ⌊ k ⌋ - m ⌊ m ⌋ + \frac{1}{2} (⌊ m ⌋ - ⌊ k ⌋) (⌊ m ⌋ + ⌊ k ⌋) + 1$ $e x a m p l e$ $\int_{- 1.5}^{3.7} ⌊ x ⌋ d x = (3.7) 3 - ((- 1.5) (- 2)) + \frac{1}{2} (- 2 - 3) (- 2 + 3 + 1)$ $= 11.1 - 3 - 5 = 3.1$
Commented byM±th+et£s last updated on 14/Apr/20

$n i c e w o r k s i r$
Commented bymathmax by abdo last updated on 14/Apr/20
$let A =∫_n ^m [x]dx with m≥n ⇒∃q∈N /m=n+q ⇒ A =∫_n ^(n+q) [x]dx =_(x=n+t) ∫_0 ^q [n+t]dt =∫_0 ^q ndt +∫_0 ^q [t]dt =nq +Σ_(p=0) ^(q−1) ∫_p ^(p+1) p dt =nq +Σ_(p=0) ^(q−1) p =nq +(((q−1)q)/2) but q =m−n ⇒ A =n(m−n)+(((m−n−1)(m−n))/2) ⇒ A =((2n(m−n))/2) +((m−n)/2)(m−n−1) =((m−n)/2){ 2n+m−n−1} =((m−n)/2)(n+m−1) example ∫_2 ^7 [x]dx =((7−2)/2)(2+7−1) =(5/2)×8 =20 let verify ∫_2 ^7 [x]dx =∫_2 ^3 [x]dx +∫_3 ^4 [x]dx +∫_4 ^5 [x]dx +∫_5 ^6 [x]dx +∫_6 ^7 [x]dx =2+3 +4 +5 +6 =20 (the relation is correct) ∫_(−2) ^5 [x]dx =((5+2)/2)(−2+5−1) =(7/2)×2 =7 let verify ∫_(−2) ^5 [x]dx =∫_(−2) ^(−1) [x]dx +∫_(−1) ^0 [x]dx +∫_0 ^1 [x]dx +∫_1 ^2 [x]dx +∫_2 ^3 [x]dx +∫_3 ^4 x]dx +∫_4 ^5 [x]dx =−2 +(−1)+0 +1 +2+3+4 =−3+3+3+4=7 (correct!)$
$l e t A = \int_{n}^{m} [x] d x w i t h m ⩾ n \Rightarrow \exists q \in N / m = n + q \Rightarrow$ $A = \int_{n}^{n + q} [x] d x =_{x = n + t} \int_{0}^{q} [n + t] d t = \int_{0}^{q} n d t + \int_{0}^{q} [t] d t$ $= n q + \sum_{p = 0}^{q - 1} \int_{p}^{p + 1} p d t = n q + \sum_{p = 0}^{q - 1} p = n q + \frac{(q - 1) q}{2}$ $b u t q = m - n \Rightarrow A = n (m - n) + \frac{(m - n - 1) (m - n)}{2} \Rightarrow$ $A = \frac{2 n (m - n)}{2} + \frac{m - n}{2} (m - n - 1)$ $= \frac{m - n}{2} {2 n + m - n - 1} = \frac{m - n}{2} (n + m - 1)$ $e x a m p l e \int_{2}^{7} [x] d x = \frac{7 - 2}{2} (2 + 7 - 1) = \frac{5}{2} \times 8 = 20 l e t v e r i f y$ $\int_{2}^{7} [x] d x = \int_{2}^{3} [x] d x + \int_{3}^{4} [x] d x + \int_{4}^{5} [x] d x + \int_{5}^{6} [x] d x + \int_{6}^{7} [x] d x$ $= 2 + 3 + 4 + 5 + 6 = 20 (t h e r e l a t i o n i s c o r r e c t)$ $\int_{- 2}^{5} [x] d x = \frac{5 + 2}{2} (- 2 + 5 - 1) = \frac{7}{2} \times 2 = 7 l e t v e r i f y$ $\int_{- 2}^{5} [x] d x = \int_{- 2}^{- 1} [x] d x + \int_{- 1}^{0} [x] d x + \int_{0}^{1} [x] d x + \int_{1}^{2} [x] d x + \int_{2}^{3} [x] d x$ $+ \int_{3}^{4} x] d x + \int_{4}^{5} [x] d x = - 2 + (- 1) + 0 + 1 + 2 + 3 + 4$ $= - 3 + 3 + 3 + 4 = 7 (c o r r e c t!)$
Commented bymr W last updated on 14/Apr/20

$p l e a s e c h e c k w h a t y o u g e t :$ $\int_{- 2}^{4} ⌊ x ⌋ d x = ?$ $y o u g e t 7, b u t i t s h o u l d b e 3 .$
Commented byM±th+et£s last updated on 14/Apr/20

$w h e r e y o u s e e a m i s t a k e i n t h e s o l u t i o n$
Commented byM±th+et£s last updated on 14/Apr/20

$i f a, b \in z$ $\frac{b^{2} - b}{2} - \frac{a^{2} - a}{2} = \frac{16 - 4}{2} - \frac{4 + 2}{2} = 6 - 3 = 3 ? ?$ $i t s r i g h t$
Commented bymr W last updated on 14/Apr/20

$w i t h k = 4, m = - 2 y o u g e t f r o m$ $\int_{m}^{k} ⌊ x ⌋ d x = k ⌊ k ⌋ - m ⌊ m ⌋ + \frac{1}{2} (⌊ m ⌋ - ⌊ k ⌋) (⌊ m ⌋ + ⌊ k ⌋) + 1$ $t h e r e s u l t 7 .$
Commented byM±th+et£s last updated on 14/Apr/20

$t h i s i s f o r a, b \notin z$
Commented bymr W last updated on 14/Apr/20

$i s e e . b u t w e s h o u l d f i n d a f o r m u l a$ $f o r a n y v a l u e s o f a a n d b . m y f o r m u l a$ $w o r k s f o r a n y v a l u e s o f a a n d b .$
Commented byM±th+et£s last updated on 14/Apr/20

$y o u a r e r i g h t s i r t h a n k y o u$
Commented bymathmax by abdo last updated on 15/Apr/20

$t h a n k x$
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