SATHEE: Area Question 12

Area Question 12

Question 12

The area of the region $A = (x, y); 0 \leq y \leq x | x | + 1 $ a n d $ - 1 \leq x \leq 1$ in sq. units, is

(2019 Main, 9 Jan II) (a) 2 (b) $\frac{4}{3}$ (c) $\frac{1}{3}$ (d) $\frac{2}{3}$

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Solution:

We have,

$A = (x, y) : 0 \leq y \leq x | x | + 1 $ a n d $ - 1 \leq x \leq 1$

When $x \geq 0$ , then $0 \leq y \leq x^{2} + 1$

and when $x < 0$ , then $0 \leq y \leq - x^{2} + 1$

Now, the required region is the shaded region.

$[∵ y = x^{2} + 1 \Rightarrow x^{2} = (y - 1)$ , parabola with vertex $(0, 1)$ and $y = - x^{2} + 1 \Rightarrow x^{2} = - (y - 1)$ ,

parabola with vertex $(0, 1)$ but open downward] We need to calculate the shaded area, which is equal to

$\begin{aligned} \int_{- 1}^{0} (- x^{2} + 1) & d x + \int_{0}^{1} (x^{2} + 1) d x & = - \frac{x^{3}}{3} + x_{- 1}^{0} + \frac{x^{3}}{3} + x_{0}^{1} & = 0 - \frac{(- 1)^{3}}{3} + (- 1) + \frac{1}{3} + 1 - 0 & = - \frac{1}{3} - 1 + \frac{4}{3} & = \frac{2}{3} + \frac{4}{3} = 2 \end{aligned}$

Area Question 12

Question 12

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Area Question 12

Question 12

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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