Q49 / 160 · AP EAMCET 2017 (26 Apr Shift 1) · Mathematics
The area (in sq. units) enclosed between the curves $y=x^2$ and $y=|x|$ is
- A$\frac{2}{3}$
- B$\frac{1}{6}$
- Correct answerC$\frac{1}{3}$
- D1
Solution
Given curves are


On solving Eqs. $(i)$ and $(i i)$, we get $$ x=0, x=1 $$ Thus, coordinates of point $B$ are $(1,1)$ $$ \begin{aligned} \therefore \quad \text { Total area }=2(\text { Area of region } O A B O) \\ \quad=2 \int_0^1\left(x-x^2\right) d x \\ \quad=2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=2\left[\frac{1}{2}-\frac{1}{3}\right] \\ \quad=2 \times \frac{1}{6}=\frac{1}{3} \text { sq unit } \end{aligned} $$


On solving Eqs. $(i)$ and $(i i)$, we get $$ x=0, x=1 $$ Thus, coordinates of point $B$ are $(1,1)$ $$ \begin{aligned} \therefore \quad \text { Total area }=2(\text { Area of region } O A B O) \\ \quad=2 \int_0^1\left(x-x^2\right) d x \\ \quad=2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=2\left[\frac{1}{2}-\frac{1}{3}\right] \\ \quad=2 \times \frac{1}{6}=\frac{1}{3} \text { sq unit } \end{aligned} $$
Hence, the correct answer is option: 'C'.