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Mathematics Question on applications of integrals

The area bounded by the curves y=(x−1)2, y=(x+1)2\text{y} = \left(\text{x} - 1\right)^{2} \text{, } \text{y} = \left(\text{x} + 1\right)^{2} and y=14\text{y} = \frac{1}{4} is

A

13 sq unit\frac{1}{3} \text{ sq unit}

B

23 sq unit\frac{2}{3} \text{ sq unit}

C

14 sq unit\frac{1}{4} \text{ sq unit}

D

15 sq unit\frac{1}{5} \text{ sq unit}

Answer

13 sq unit\frac{1}{3} \text{ sq unit}

Explanation

Solution

The curves y=(x−1)2, y=(x+1)2\text{y} = \left(\text{x} - 1\right)^{2} \text{, } \text{y} = \left(\text{x} + 1\right)^{2} and y=14\text{y} = \frac{1}{4} are shown as where point of intersection are (x−1)2=14\left(\text{x} - 1\right)^{2} = \frac{1}{4} ⇒\Rightarrow x=12\text{x} = \frac{1}{2} and (x+1)2=14⇒x=−12\left(\text{x} + 1\right)^{2} = \frac{1}{4} \Rightarrow x = - \frac{1}{2} ∴\therefore Q(12, 14)\text{Q} \left(\frac{1}{2} \text{, } \frac{1}{4}\right) and R(−12, 14)\text{R} \left(- \frac{1}{2} \text{, } \frac{1}{4}\right) ∴\therefore Required area =2∫01/2[(x−1)2−14]dx= 2 \displaystyle \int _{0}^{1 / 2} \left[\left(\text{x} - 1\right)^{2} - \frac{1}{4}\right] \text{dx} =2([(x−1)33−14x])01/2= 2 \left(\left[\frac{\left(\text{x} - 1\right)^{3}}{3} - \frac{1}{4} \text{x}\right]\right)_{0}^{1 / 2} =2[−18⋅3−18−(−13−0)]=824=13 sq unit= 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}