Area Question 16
Question 16
- The area (in sq units) of the region
$A={(x, y) \in R \times R \mid 0 \leq x \leq 3$,
$\left.0 \leq y \leq 4, y \leq x^{2}+3 x\right}$ is
(2019 Main, 8 April I) (a) $\frac{53}{6}$ (b) 8 (c) $\frac{59}{6}$ (d) $\frac{26}{3}$
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Solution:
- Here, $\left{(x, y) \in R^{2}: y \geq \sqrt{|x+3|}, 5 y \leq(x+9) \leq 15\right}$
$$ \begin{aligned} & \therefore \quad y \geq \sqrt{x+3} \end{aligned} $$
$$ \begin{aligned} & \text { or } \quad y^{2} \geq \begin{array}{c} x+3, \text { when } x \geq-3 \ -3-x, \text { when } x \leq-3 \end{array} \end{aligned} $$
Shown as
$$ \begin{array}{ll} \text { Also, } & 5 y \leq(x+9) \leq 15 \ \Rightarrow & (x+9) \geq 5 y \text { and } x \leq 6 \end{array} $$
Shown as
$\therefore \quad\left{(x, y) \in R^{2}: y \geq \sqrt{|x+3|}, 5 y \leq(x+9) \leq 15\right}$
$\therefore$ Required area $=$ Area of trapezium $A B C D$
- Area of $A B E$ under parabola - Area of $C D E$ under parabola
$=\frac{1}{2}(1+2)(5)-\int_{-4}^{-3} \sqrt{-(x+3)} d x-\int_{-3}^{1} \sqrt{(x+3)} d x$
$=\frac{15}{2}-\frac{(-3-x)^{3 / 2}}{-\frac{3}{2}}-\frac{(x+3)^{3 / 2}}{\frac{3}{2}}$
$=\frac{15}{2}+\frac{2}{3}[0-1]-\frac{2}{3}[8-0]=\frac{15}{2}-\frac{2}{3}-\frac{16}{3}=\frac{15}{2}-\frac{18}{3}=\frac{3}{2}$
