Find ∫_0 ^( n) [(x)^(1/3) ]dx=? in terms of n. (n∈N)


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Question Number 79946 by mr W last updated on 29/Jan/20

$F i n d$ $\int_{0}^{n} [\sqrt[3]{x}] d x = ?$ $i n t e r m s o f n . (n \in N)$
	Answered by key of knowledge last updated on 29/Jan/20

	$⌊^{3} \sqrt{n} ⌋ = a$ $\int_{m^{3}}^{{(m + 1)}^{3}} [^{3} \sqrt{x^{3}}] dx = m \int_{m^{3}}^{{(m + 1)}^{3}} dx = {3 m}^{3} + {3 m}^{2} + m$ $\int_{0}^{n} [\sqrt[3]{x}] d x = \underset{0}{\sum^{a - 1}} ({3 m}^{3} + {3 m}^{2} + m) + \int_{a^{3}}^{n} [\sqrt{x^{3}}] =$ $\underset{0}{\sum^{a - 1}} ({3 m}^{3} + {3 m}^{2} + m) + a (n - a^{3})$
	Commented by mr W last updated on 29/Jan/20

	$t h a n k s a l o t s i r!$
	Commented by mr W last updated on 29/Jan/20

	$i t h i n k y o u m e a n t ⌊ \sqrt[3]{n} ⌋ = a i n t h e f i r s t$ $l i n e, r i g h t ?$
	Commented by key of knowledge last updated on 29/Jan/20

	$yes . Mr W . thank$
	Answered by mr W last updated on 29/Jan/20
	$let h=⌊(n)^(1/3) ⌋ x=[0,1^3 ): [(x)^(1/3) ]=0 ⇒∫_0 ^1^3 [(x)^(1/3) ]dx=0×(1^3 −0) x=[1^3 ,2^3 ): [(x)^(1/3) ]=1⇒∫_0 ^1^3 [(x)^(1/3) ]dx=1×(2^3 −1^3 ) x=[2^3 ,3^3 ): [(x)^(1/3) ]=2⇒∫_1^3 ^2^3 [(x)^(1/3) ]dx=2×(3^3 −2^3 ) x=[3^3 ,4^3 ): [(x)^(1/3) ]=3⇒∫_2^3 ^3^3 [(x)^(1/3) ]dx=3×(4^3 −3^3 ) ... x=[(h−1)^3 ,h^3 ): [(x)^(1/3) ]=h−1⇒∫_((h−1)^3 ) ^h^3 [(x)^(1/3) ]dx=(h−1)×(h^3 −(h−1)^3 ) x=[h^3 ,n]: [(x)^(1/3) ]=h⇒∫_h^3 ^n [(x)^(1/3) ]dx=h×(n−h^3 ) ∫_0 ^( n) [(x)^(1/3) ]dx=Σ_(k=0) ^(h−1) ∫_k^3 ^((k+1)^3 ) [(x)^(1/3) ]dx+∫_h^3 ^n [(x)^(1/3) ]dx =Σ_(k=0) ^(h−1) {k[(k+1)^3 −k^3 ]}+h(n−h^3 ) =Σ_(k=1) ^(h−1) (3k^3 +3k^2 +k)+h(n−h^3 ) =3(((h−1)^2 h^2 )/4)+3(((h−1)h(2h−1))/6)+(((h−1)h)/2)+h(n−h^3 ) =((h^2 (h−1)(3h+1)+4h(n−h^3 ))/4) ⇒∫_0 ^( n) [(x)^(1/3) ]dx=nh−((h^2 (h+1)^2 )/4) example n=100: h=⌊((100))^(1/3) ⌋=4 ∫_0 ^( 100) [(x)^(1/3) ]dx=100×4−((4^2 ×5^2 )/4)=300$
	$l e t h = ⌊ \sqrt[3]{n} ⌋$ $x = [0, 1^{3}) : [\sqrt[3]{x}] = 0 \Rightarrow \int_{0}^{1^{3}} [\sqrt[3]{x}] d x = 0 \times (1^{3} - 0)$ $x = [1^{3}, 2^{3}) : [\sqrt[3]{x}] = 1 \Rightarrow \int_{0}^{1^{3}} [\sqrt[3]{x}] d x = 1 \times (2^{3} - 1^{3})$ $x = [2^{3}, 3^{3}) : [\sqrt[3]{x}] = 2 \Rightarrow \int_{1^{3}}^{2^{3}} [\sqrt[3]{x}] d x = 2 \times (3^{3} - 2^{3})$ $x = [3^{3}, 4^{3}) : [\sqrt[3]{x}] = 3 \Rightarrow \int_{2^{3}}^{3^{3}} [\sqrt[3]{x}] d x = 3 \times (4^{3} - 3^{3})$ $. . .$ $x = [{(h - 1)}^{3}, h^{3}) : [\sqrt[3]{x}] = h - 1 \Rightarrow \int_{{(h - 1)}^{3}}^{h^{3}} [\sqrt[3]{x}] d x = (h - 1) \times (h^{3} - {(h - 1)}^{3})$ $x = [h^{3}, n] : [\sqrt[3]{x}] = h \Rightarrow \int_{h^{3}}^{n} [\sqrt[3]{x}] d x = h \times (n - h^{3})$ $\int_{0}^{n} [\sqrt[3]{x}] d x = \underset{k = 0}{\sum^{h - 1}} \int_{k^{3}}^{{(k + 1)}^{3}} [\sqrt[3]{x}] d x + \int_{h^{3}}^{n} [\sqrt[3]{x}] d x$ $= \underset{k = 0}{\sum^{h - 1}} {k [{(k + 1)}^{3} - k^{3}]} + h (n - h^{3})$ $= \underset{k = 1}{\sum^{h - 1}} (3 k^{3} + 3 k^{2} + k) + h (n - h^{3})$ $= 3 \frac{{(h - 1)}^{2} h^{2}}{4} + 3 \frac{(h - 1) h (2 h - 1)}{6} + \frac{(h - 1) h}{2} + h (n - h^{3})$ $= \frac{h^{2} (h - 1) (3 h + 1) + 4 h (n - h^{3})}{4}$ $\Rightarrow \int_{0}^{n} [\sqrt[3]{x}] d x = n h - \frac{h^{2} {(h + 1)}^{2}}{4}$ $e x a m p l e n = 100 :$ $h = ⌊ \sqrt[3]{100} ⌋ = 4$ $\int_{0}^{100} [\sqrt[3]{x}] d x = 100 \times 4 - \frac{4^{2} \times 5^{2}}{4} = 300$
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