Question

The area bounded by the curves $\text{y} = \left(\text{x} - 1\right)^{2} \text{, } \text{y} = \left(\text{x} + 1\right)^{2}$ and $\text{y} = \frac{1}{4}$ is

• A. $\frac{1}{3} \text{ sq unit}$
• B. $\frac{2}{3} \text{ sq unit}$
• C. $\frac{1}{4} \text{ sq unit}$
• D. $\frac{1}{5} \text{ sq unit}$

Answer 1

The Correct Option is A

The curves $\text{y} = \left(\text{x} - 1\right)^{2} \text{, } \text{y} = \left(\text{x} + 1\right)^{2}$ and $\text{y} = \frac{1}{4}$ are shown as where point of intersection are $\left(\text{x} - 1\right)^{2} = \frac{1}{4}$ $\Rightarrow $ $\text{x} = \frac{1}{2}$ and $\left(\text{x} + 1\right)^{2} = \frac{1}{4} \Rightarrow x = - \frac{1}{2}$ $\therefore $ $\text{Q} \left(\frac{1}{2} \text{, } \frac{1}{4}\right)$ and $\text{R} \left(- \frac{1}{2} \text{, } \frac{1}{4}\right)$ $\therefore $ Required area $= 2 \displaystyle \int _{0}^{1 / 2} \left[\left(\text{x} - 1\right)^{2} - \frac{1}{4}\right] \text{dx}$ $= 2 \left(\left[\frac{\left(\text{x} - 1\right)^{3}}{3} - \frac{1}{4} \text{x}\right]\right)_{0}^{1 / 2}$ $= 2 \left[- \frac{1}{8 \cdot 3} - \frac{1}{8} - \left(- \frac{1}{3} - 0\right)\right] = \frac{8}{2 4} = \frac{1}{3} \text{ sq unit}$