Area Question 32
Question 32
- Find the area bounded by the curves $x^{2}=y, x^{2}=-y$ and $y^{2}=4 x-3$.
$(2005,4$ M)
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Solution:
- The region bounded by the curves $y=x^{2}, y=-x^{2}$ and $y^{2}=4 x-3$ is symmetrical about $X$-axis, where $y=4 x-3$ meets at $(1,1)$.
$\therefore$ Area of curve $(O A B C O)$
$$ =2 \quad \int_{0}^{1} x^{2} d x-\int_{3 / 4}^{1}(\sqrt{4 x-3}) d x $$
$$ \begin{aligned} & =2{\frac{x^{3}}{3}}{0}^{1}-{\frac{(4 x-3)^{3 / 2}}{3 \cdot 4 / 2}}{3 / 4}^{1} \ & =2 \frac{1}{3}-\frac{1}{6} \ & =1 \cdot \frac{1}{6}=\frac{1}{3} \mathrm{sq} \text { unit } \end{aligned} $$
