Determine-and-prove-if-true-0-n-x-2-dx-lt-k-1-n-k-2-

Question Number 9533 by FilupSmith last updated on 14/Dec/16

Determine and prove if true :

\int_{0}^{n} x^{2} d x < \underset{k = 1}{\sum^{n}} k^{2}

Answered by geovane10math last updated on 14/Dec/16

\frac{n^{3}}{3} + C < 1 + 4 + 9 + \dots + n^{2}

\frac{n^{3}}{3} + C < \frac{n (n + 1) (2 n + 1)}{6}

C < \frac{(n^{2} + n) (2 n + 1)}{6} - \frac{n^{3}}{3}

C < \frac{2 n^{3} + n^{2} + 2 n^{2} + n - 2 n^{3}}{6}

C < \frac{3 n^{2} + n}{6}

It depends of the {integration}^{'} s constant . .

0 < \frac{3 n^{2} + n}{6}

3 n^{2} + n > 0

n (3 n + 1) = 0

n_{1} = 0 n_{2} = - \frac{1}{3}

S = {n \in R ∣ n < - \frac{1}{3} or n > 0}

Commented by geovane10math last updated on 14/Dec/16

D O N O T C O N S I D E R T H E C,

I W R O N G S O R R Y

Answered by sou1618 last updated on 14/Dec/16

o t h e r s o l u t i o n \dots

k = 0, 1, 2, 3, \dots .

w h e n x : k ⩽ x ⩽ k + 1

\Rightarrow k^{2} ⩽ x^{2} ⩽ {(k + 1)}^{2}

\Rightarrow \int_{k}^{k + 1} k^{2} d x < \int_{k}^{k + 1} x^{2} d x < \int_{k}^{k + 1} {(k + 1)}^{2} d x

\Leftrightarrow k^{2} (k + 1 - k) < \int_{k}^{k + 1} x^{2} d x < {(k + 1)}^{2} (k + 1 - k)

\Leftrightarrow k^{2} < \int_{k}^{k + 1} x^{2} d x < {(k + 1)}^{2}

\Rightarrow \underset{k = 0}{\sum^{n - 1}} k^{2} < \underset{k = 0}{\sum^{n - 1}} \int_{k}^{k + 1} x^{2} d x < \underset{k = 0}{\sum^{n - 1}} {(k + 1)}^{2}

\Leftrightarrow \underset{k = 0}{\sum^{n - 1}} k^{2} < \int_{0}^{n} x^{2} d x < \underset{k = 0}{\sum^{n - 1}} {(k + 1)}^{2} = \underset{k = 1}{\sum^{n}} k^{2}

* \underset{k = 0}{\sum^{n - 1}} {(k + 1)}^{2} = 1^{2} + 2^{2} + \dots + n^{2} = \underset{k = 1}{\sum^{n}} k^{2}

∴ \underset{k = 0}{\sum^{n - 1}} k^{2} < \int_{0}^{n} x^{2} d x < \underset{k = 1}{\sum^{n}} k^{2}

Answered by mrW last updated on 14/Dec/16

\int_{0}^{n} x^{2} d x = \frac{1}{3} {[x^{3}]}_{0}^{n} = \frac{n^{3}}{3}

\underset{k = 1}{\sum^{n}} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}

= \frac{{2 n}^{3} + {3 n}^{2} + n}{6} = \frac{n^{3}}{3} + (\frac{n^{2}}{2} + \frac{n}{6})

> \frac{n^{3}}{3} = \int_{0}^{n} x^{2} d x

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