The area under the curve y = | cos x - sin x |, 0 le x le (π/2), and above x-axis is :

Question

The area under the curve y = | cos x - sin x |, 0 le x le (π/2), and above x-axis is : (A) 2 √2 (B) 2 √2 - 2  (C) 2 √2 + 2 (D) 0

Tardigrade Admin · Accepted Answer

y = | cos x - sin x | <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/f46fce8263ec069f2d614c5e35f9364e-.jpeg" /> Required area = 2 ∫ limitsπ /40 ( cos x - sin x)dx = 2[ sin x + cos x]0π/4 = 2[(2/√2) - 1 ] = (2√2 -2) sq. units

Q. The area under the curve $y = ∣ cos x - sin x ∣, 0 \leq x \leq \frac{π}{2},$ and above x-axis is :

$22$

$22 - 2$

$22 + 2$

0

$y = ∣ cos x - sin x ∣$

Required area $= 2 0 \int π /4 (cos x - sin x) d x$
$= 2 [sin x + cos x]_{0}^{π /4}$
$= 2 [\frac{2}{2} - 1] = (22 - 2)$ sq. units

Q. The area under the curve y=∣cosx−sinx∣,0≤x≤2π​, and above x-axis is :

22​

22​−2

22​+2

0

y=∣cosx−sinx∣ Required area =20∫π/4​(cosx−sinx)dx =2[sinx+cosx]0π/4​ =2[2​2​−1]=(22​−2) sq. units

Q. The area under the curve $y = ∣ cos x - sin x ∣, 0 \leq x \leq \frac{π}{2},$ and above x-axis is :

$22$

$22 - 2$

$22 + 2$

$y = ∣ cos x - sin x ∣$

Required area $= 2 0 \int π /4 (cos x - sin x) d x$
$= 2 [sin x + cos x]_{0}^{π /4}$
$= 2 [\frac{2}{2} - 1] = (22 - 2)$ sq. units