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$(1)$ Calculation of value of $\displaystyle \int_{0}^{n}\cos\left(\lfloor x \rfloor\cdot \{x\}\right)dx\;,$ Where $\lfloor x \rfloor$ is floor function of $x$ and $\{x\} = x-\lfloor x \rfloor.$ and $n$ is a positive integer.

$(2)\; $Calculation of least positive integer $n$ for which $\displaystyle \int_{1}^{n}\lfloor x \rfloor \cdot \lfloor \sqrt{x} \rfloor dx>60\;,$ Where $\lfloor x \rfloor$ is floor

function of $x$

$\bf{My\; Try::}$ For $(1)$ one:: We can write $\displaystyle \int_{0}^{n}\cos\left(\lfloor x \rfloor\cdot \{x\}\right)dx = \displaystyle \int_{0}^{n}\cos\left(\lfloor x \rfloor\cdot \left(x-\lfloor x \rfloor\right)\right)dx$

So $\displaystyle \int_{0}^{n}\cos\left(\lfloor x \rfloor\cdot \left(x-\lfloor x \rfloor\right)\right)dx = \int_{0}^{1}\cos(0)dx+\int_{1}^{2}\cos(1\cdot (x-1))dx+\int_{2}^{3}\cos(2\cdot(x-2))dx+..........+\int_{n-1}^{n}\cos((n-1)(x-(n-1)))dx$

Now How can I solve after that, Help me, Thanks

$\bf{My\; Try}::$ For $(2)$ one:: We can write

$\displaystyle \int_{1}^{n}\lfloor x \rfloor \cdot \lfloor \sqrt{x} \rfloor dx = \int_{0}^{1}0\cdot 0dx+\int_{1}^{2}1\cdot 1dx+\int_{2}^{3}2\cdot 1dx++\int_{3}^{4}3\cdot 1dx+......+\int_{n-1}^{n}(n-1)\lfloor \sqrt{n-1}\rfloor x$

Now How can i solve after that, Help me, Thanks

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  • $\begingroup$ Have you tried substituting $y=k(x-k)$ and evaluating the integrals? $\endgroup$ Nov 2, 2014 at 16:53

1 Answer 1

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Hint: For $k \geq 1$,

$$ \int_k^{k+1} \cos(k(x-k)) \ dx = \int_0^1 \cos(kx) \ dx = \frac{1}{k} \sin(k)$$

Now try adding them up.

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