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Question: The value of ![](https://cdn.pureessence.tech/canvas_274.png?top_left_x=540&top_left_y=791&width=300...

The value of dx, where [x] is the greatest integer less than or equal to x is -

A

2

B

8/3

C

4

D

None of these

Answer

None of these

Explanation

Solution

For x Ī [0, 2], x2 + 1 Ī [1, 5], we must break

[0, 2] = [0, 1] Č [1, ∫02af(x)dx={0, if f(2a−x)=−f(x)2∫0af(x)dx, if f(2a−x)=f(x)\int _ { 0 } ^ { 2 a } f ( x ) d x = \left\{ \begin{array} { l l } 0 & , \text { if } f ( 2 a - x ) = - f ( x ) \\ 2 \int _ { 0 } ^ { a } f ( x ) d x , & \text { if } f ( 2 a - x ) = f ( x ) \end{array} \right. ] Č [I=∫0πecos⁡2(π−x)⋅cos⁡3(2n+1)(π−x)dxI = \int _ { 0 } ^ { \pi } e ^ { \cos ^ { 2 } ( \pi - x ) } \cdot \cos ^ { 3 } ( 2 n + 1 ) ( \pi - x ) d x,I=−∫0πecos⁡2x⋅cos⁡3(2n+1)xdxI = - \int _ { 0 } ^ { \pi } e ^ { \cos ^ { 2 } x } \cdot \cos ^ { 3 } ( 2 n + 1 ) x d x] Č [ I=−II = - I , 2]. Hence ∫02x[x2+1]\int _ { 0 } ^ { 2 } x ^ { \left[ x ^ { 2 } + 1 \right] } dx

= ∫01x[x2+1]\int _ { 0 } ^ { 1 } \mathrm { x } ^ { \left[ \mathrm { x } ^ { 2 } + 1 \right] } dx + ∫12x[x2+1]\int _ { 1 } ^ { \sqrt { 2 } } x ^ { \left[ x ^ { 2 } + 1 \right] } dx + dx + ∫32x[x2+1]\int _ { \sqrt { 3 } } ^ { 2 } x ^ { \left[ x ^ { 2 } + 1 \right] } dx

= 12\frac { 1 } { 2 } + 13\frac { 1 } { 3 } [23/2 – 1] + 14\frac { 1 } { 4 }[9 – 4] + 15\frac { 1 } { 5 } [32 – 35/2]

= 13\frac { 1 } { 3 }23/2 – 1535/2\frac { 1 } { 5 } 3 ^ { 5 / 2 }