Q115 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics
The area bounded by the curve $y=x^2$ and $y-6=-|x|$ is
- A$\frac{37}{4}$
- B$\frac{22}{3}$
- Correct answerC$\frac{44}{3}$
- D$\frac{38}{3}$
Solution
Let us first draw the curves $y=x^2 \& y-6=-|x|$ :

For point $\mathrm{A}$, solving $y=x^2$ ang $y-6=-x$, we get: $x=2 \& y=4 \Rightarrow \mathrm{A} \equiv(2,4)$ The required area $=$ Area $(O A B C)=2 \times$ Area $(O A B)$ $\begin{aligned} & =2 \times(\text { Area }(\mathrm{OAD})+\text { Area }(\mathrm{DAB})) \\ & =2\left(\int_0^4 \sqrt{x} d x+\int_4^6(-x+6) d x\right)=2\left[\frac{16}{3}+2\right]=\frac{44}{3}\end{aligned}$

For point $\mathrm{A}$, solving $y=x^2$ ang $y-6=-x$, we get: $x=2 \& y=4 \Rightarrow \mathrm{A} \equiv(2,4)$ The required area $=$ Area $(O A B C)=2 \times$ Area $(O A B)$ $\begin{aligned} & =2 \times(\text { Area }(\mathrm{OAD})+\text { Area }(\mathrm{DAB})) \\ & =2\left(\int_0^4 \sqrt{x} d x+\int_4^6(-x+6) d x\right)=2\left[\frac{16}{3}+2\right]=\frac{44}{3}\end{aligned}$
Hence, the correct answer is option: 'C'.