Definite Integration Question 140
Question: Area bounded by the curves $ y=[ \frac{x^{2}}{64}+2 ]([\cdot ] $ denotes the greatest integer function), $ v=x-1 $ and $ x=0 $ , above the x-axis is
Options:
A) 2 sq. unit
B) 3 sq. unit
C) 4 sq. unit
D) None of these
Show Answer
Answer:
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| x |<8 $
$ \Rightarrow y=[ \frac{x^{2}}{64}+2 ]=2, $ if $ | x |<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}[ {{(y+1)}^{2}} ]_0^{2}=\frac{9}{2}-\frac{1}{2}=4} $ sq. unit.
BETA
