SATHEE: Area Question 43

Area Question 43

Question 43

Sketch the region bounded by the curves $y = x^{2}$ and $y = 2 / (1 + x^{2})$ . Find its area.

$(1992, 4$ M)

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Solution:

The curve $y = x^{2}$ is a parabola. It is symmetric about $Y$ -axis and has its vertex at $(0, 0)$ and the curve $y = \frac{2}{1 + x^{2}}$ is a bell shaped curve. $X$ -axis is its asymptote and it is symmetric about $Y$ -axis and its vertex is $(0, 2)$ .

$\begin{aligned} Since, y = x^{2} & and y = \frac{2}{1 + x^{2}} & \Rightarrow y = \frac{2}{1 + y} & \Rightarrow y^{2} + y - 2 = 0 & \Rightarrow (y - 1) (y + 2) = 0 \Rightarrow y = - 2, 1 & But y \geq 0, so y = 1 \Rightarrow x = \pm 1 \end{aligned}$

Therefore, coordinates of $C$ are $(- 1, 1)$ and coordinates of $B$ are $(1, 1)$ .

$∴$ Required area $O B A C O = 2 \times$ Area of curve $O B A O$

$= 2 \int_{0}^{1} \frac{2}{1 + x^{2}} d x - \int_{0}^{1} x^{2} d x$

$=2\left[2 \tan ^{-1} x\right]{0}^{1}-{\frac{x^{3}}{3}}{0}^{1}=2 \frac{2 \pi}{4}-\frac{1}{3}=\pi-\frac{2}{3}$ sq unit

Area Question 43

Question 43

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Area Question 43

Question 43

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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