SATHEE: Area Question 19

Area Question 19

Question 19

The area enclosed by the curves $y = \sin x + \cos x$ and $y = | \cos x - \sin x |$ over the interval $0, \frac{π}{2}$ is (2014 Adv.) (a) $4 (\sqrt{2} - 1)$ (c) $2 (\sqrt{2} + 1)$ (b) $2 \sqrt{2} (\sqrt{2} - 1)$ (d) $2 \sqrt{2} (\sqrt{2} + 1)$

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Solution:

PLAN To find the bounded area between $y = f (x)$ and $y = g (x)$ between $x =$ a to $x = b$ .

$∴$ Area bounded $= \int_{a}^{ε} [g (x) - f (x)] d x + \int_{a}^{b} [f (x) - g (x)] d x$

$= \int_{a}^{b} | f (x) - g (x) | d x$

Here,

$f (x) = y = \sin x + \cos x, when 0 \leq x \leq \frac{π}{2}$

and

$g (x) = y = | \cos x - \sin x |$

$= \begin{array}{ll} \cos x - \sin x, & 0 \leq x \leq \frac{π}{4} \sin x - \cos x, & \frac{π}{4} \leq x \leq \frac{π}{2} \end{array}$

could be shown as

$∴$ Area bounded $= \int_{0}^{π / 4} (\sin x + \cos x) - (\cos x - \sin x) d x$ $+ \int_{π / 4}^{π / 2} (\sin x + \cos x) - (\sin x - \cos x) d x$ $= \int_{0}^{π / 4} 2 \sin x d x + \int_{π / 4}^{π / 2} 2 \cos x d x$

$=-2[\cos x]{0}^{\pi / 4}+2[\sin x \cdot n]{\pi / 4}^{\pi / 2}=4-2 \sqrt{2}=2 \sqrt{2}(\sqrt{2}-1)$ sq units

Area Question 19

Question 19

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Area Question 19

Question 19

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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