Question

Sketch the region enclosed by the curves and find its area.

9. J = ਸੰਢ, 3 = z| 10. y = 3 = 2

Answer 1

$\small {\color{Red} (9)}\\ \text{Consider the curves}\\ y=\frac{2}{1+x^2} \: \: \: y=\left | x \right |\\ \text{The absolute function is as follows}\\ y=\left | x \right |=\begin{cases} x & \text{ if } x\geq 0 \\ -x & \text{ if } x < 0 \end{cases}\\ \text{The point of intersection of the two curves is as follows}\\ \frac{2}{1+x^2}=x\Rightarrow x=1\\ \frac{2}{1+x^2}=-x\Rightarrow x=-1\\ \text{Therefore, the point of intersection are (-1,1) and (1,1)} \\$

The area between the two curves is as follows Area = " ir (ar) – g(x)] de - L P -1-)] dr + L [1722-(-2) de = (zaretan (a) + + (2 aretan (a) – = -1 = 2.14

(-1,1)) (11) (0.0)

$\small {\color{Red} (10)}\\ \text{Consider the curves}\\ y=\frac{1}{\sqrt{1-x^2}} \: \: \: y=2\\ \text{The point of intersection of the two curves is as follows}\\ \frac{1}{\sqrt{1-x^2}}=2\\ \sqrt{1-x^2}=\frac{1}{2}\\ 1-x^2=\frac{1}{4} \Rightarrow x=\frac{\sqrt{3}}{2},\:x=-\frac{\sqrt{3}}{2}\\ \text{Therefore, the point of intersection are }\left (-\frac{\sqrt{3}}{2},2 \right ) ,\; \left (\frac{\sqrt{3}}{2},2 \right )\\ \text{The area between the two curves is as follows}\\ \mathrm{Area}=\int_{a}^{b}\left [ f(x)-g(x) \right ]dx\\ =\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}(2-\frac{1}{\sqrt{1-x^2}})dx\\ =\left[2x-\arcsin \left(x\right)\right]^{\frac{\sqrt{3}}{2}}_{-\frac{\sqrt{3}}{2}}\\ =2\sqrt{3}-2\cdot \frac{\pi }{3}\\ =1.3697\\$

(-0.866, 2) (0.866, 2)