Area Ques 61
Question
- The area (in $\mathrm{sq}$ units) of the region $A=\left\{(x, y): x^{2} \leq y \leq x+2\right\}$ is $\quad$
(2019 Main, 9 April I)
(a) $\frac{13}{6}$
(b) $\frac{9}{2}$
(c) $\frac{31}{6}$
(d) $\frac{10}{3}$
Show Answer
Answer:
Correct Answer: 61.(b)
Solution:
Formula:
- Given region is $A=\left\{(x, y): x^{2} \leq y \leq x+2\right\}$
Now, the region is shown in the following graph
For intersecting points $A$ and $B$
$ \begin{array}{ll} \text { Taking, } & x^{2}=x+2 \Rightarrow x^{2}-x-2=0 \\ \Rightarrow & x^{2}-2 x+x-2=0 \\ \Rightarrow & x(x-2)+1(x-2)=0 \\ \Rightarrow & x=-1,2 \Rightarrow y=1,4 \end{array} $
So, $A(-1,1)$ and $B(2,4)$.
Now, shaded area $=\int _{-1}^{2}\left[(x+2)-x^{2}\right] d x$
$=(\frac{x^{2}}{2}+2 x-{\frac{x^{3}}{3}})_{-1}^{2}=(\frac{4}{2}+4-\frac{8}{3})-(\frac{1}{2}-2+\frac{1}{3})$
$=8-\frac{1}{2}-\frac{9}{3}=8-\frac{1}{2}-3=5-\frac{1}{2}=\frac{9}{2}$ sq units
BETA
