Area Question 24
Question 24
- The area bounded by the curves $y=(x-1)^{2}, y=(x+1)^{2}$ and $y=\frac{1}{4}$ is
(2005, 1M) (a) $\frac{1}{3}$ sq unit (b) $\frac{2}{3}$ sq unit (c) $\frac{1}{4}$ sq unit (d) $\frac{1}{5}$ sq unit
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Solution:
- The curves $y=(x-1)^{2}, y=(x+1)^{2}$ and $y=1 / 4$ are shown as
where, points of intersection are
$$ (x-1)^{2}=\frac{1}{4} \quad \Rightarrow \quad x=\frac{1}{2} \text { and }(x+1)^{2}=\frac{1}{4} \Rightarrow x=-\frac{1}{2} $$
i.e.
$$ Q \frac{1}{2}, \frac{1}{4} \quad \text { and } \quad R \quad-\frac{1}{2}, \frac{1}{4} $$
$\therefore \quad$ Required area $=2 \int_{0}^{1 / 2}(x-1)^{2}-\frac{1}{4} d x$
$$ \begin{aligned} & =2 \frac{(x-1)^{3}}{3}-\frac{1}{4} x^{1 / 2} \ & =2-\frac{1}{8 \cdot 3}-\frac{1}{8}–\frac{1}{3}-0 \quad=\frac{8}{24}=\frac{1}{3} \text { sq unit } \end{aligned} $$
