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

Correct answer is (c) 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 -

$\frac{3 π}{2} - 2$

$\frac{3 π}{2}$

$2 + \frac{3 π}{2}$

$1 + \frac{3 π}{2}$

Correct answer is (c) $2 + \frac{3 π}{2}$

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​

Correct answer is (c) 2+23π​

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 -

$\frac{3 π}{2} - 2$

$\frac{3 π}{2}$

$2 + \frac{3 π}{2}$

$1 + \frac{3 π}{2}$

Correct answer is (c) $2 + \frac{3 π}{2}$