The area enclosed by the curves y= cos x, y=1+ sin 2 x and x=(3 π/2) as x varies from 0 to (3 π/2), is

Question

The area enclosed by the curves y= cos x, y=1+ sin 2 x and x=(3 π/2) as x varies from 0 to (3 π/2), is (A) (3 π/2)-2 (B) (3 π/2) (C) 2+(3 π/2) (D) 1+(3 π/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ead5ae47fbe14b48adb2b015200c2adc-.png" /> A=∫ limits03 π / 2(1+ sin 2 x- cos x) d x =(x-(1/2) cos 2 x- sin x)0(3 π/2) ((3 π/2)+(1/2)+1)-(0-(1/2)) =2+(3 π/2)

Q. The area enclosed by the curves $y = cos x, y = 1 + sin 2 x$ and $x = \frac{3 π}{2}$ as $x$ varies from 0 to $\frac{3 π}{2}$ , is

$\frac{3 π}{2} - 2$

$\frac{3 π}{2}$

$2 + \frac{3 π}{2}$

$1 + \frac{3 π}{2}$

$A = 0 \int 3 π /2 (1 + sin 2 x - cos x) d x$
$= (x - \frac{1}{2} cos 2 x - sin x)_{0}^{\frac{3 π}{2}}$
$(\frac{3 π}{2} + \frac{1}{2} + 1) - (0 - \frac{1}{2})$
$= 2 + \frac{3 π}{2}$

Q. The area enclosed by the curves y=cosx,y=1+sin2x and x=23π​ as x varies from 0 to 23π​, is

23π​−2

23π​

2+23π​

1+23π​