Area Under Curves Question 10
Question 10 - 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola $(y-2)^{2}=x-1$, the line $x-2 y+4=0$ and the positive coordinate axes is
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Answer (5)
Solution
Solving the equations
$(y-2)^{2}=x-1$ and $x-2 y+4=0$
$$ x=2(y-2) $$
$\frac{x^{2}}{4}=x-1$
$x^{2}-4 x+4=0$
$(x-2)^{2}=0$
$x=2$
Exclose area (w.r.t. y-axis) $=\int_{0}^{3} \mathrm{x} d y-$ Area of $\Delta$.
$$ \begin{aligned} & =\int_{0}^{3}\left((y-2)^{2}+1\right) d y-\frac{1}{2} \times 1 \times 2 \ & =\int_{0}^{3}\left(y^{2}-4 y+5\right) d y-1 \ & =\left[\frac{y^{3}}{3}-2 y^{2}+5 y\right]_{0}^{3}-1 \text { hongo } \ & =9-18+15-1=5 \end{aligned} $$
