Find-0-n-x-1-3-dx-in-terms-of-n-n-N-

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

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})

\dots

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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