Area bounded by the curves y-x264-2- - - denotes the greatest integer function y-x-1 and x-0 above the x-axis is-

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Geometric Interpretation of Def.Int as Limit of Sum

Area bounded ...

Question

Area bounded by the curves $y = [\frac{x^{2}}{64} + 2]$ ( $[\cdot]$ denotes the greatest integer function) $y = x - 1$ and $x = 0$ above the x-axis is?

2

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Solution

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Similar questions

y = [\frac{x^{2}}{64} + 2], y = x - 1

x = 0

x -

[.]

y = x^{2}, y = [x + 1], 0 \leq x < 1

y -

[.]

y = \sqrt{(sin [x] + [sin x])},

[.]

x = 1

x = \frac{π}{2}

x -

f (x) = \frac{1}{x^{2} + [x]^{2} + 2 {x} + 1 - 2 x [x]}

x = \frac{- 3}{2}

x = \frac{5}{2}

The area bounded by $y = x^{2}, y = [x + 1], x \leq 1$ and the $y -$ axis is
$([\cdot] denotes the greatest integer function.)$

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