question_answer

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

A) 2 sq. unit

B) 3 sq. unit

C) 4 sq. unit

D) None of these

Correct Answer: C

Solution :

[c] We have, \[0\le \frac{{{x}^{2}}}{64}<1,\] if \[-8<x<8\] \[\Rightarrow 2\le \frac{{{x}^{2}}}{64}+2<3,\,\,if\,\,\left| x \right|<8\] \[\Rightarrow y=\left[ \frac{{{x}^{2}}}{64}+2 \right]=2,\] if \[\left| x \right|<8\] The graphs of the given curves is as shown in figure. Req. area = area of the shaded region \[=\int_{0}^{2}{xdy}\] \[=\int_{0}^{2}{(y+1)dy=\frac{1}{2}\left[ {{(y+1)}^{2}} \right]_{0}^{2}=\frac{9}{2}-\frac{1}{2}=4}\] sq. unit.