The area bounded by the curves y=cos x and y=sin 2x,∀ x∈ [(π /6) , (π /2)] is equal to

Question

The area bounded by the curves y=cos x and y=sin 2x,∀ x∈ [(π /6) , (π /2)] is equal to (A) (π /2) sq. units (B) (π /3) sq. units (C) (π /6) sq. units (D) (1/4) sq. units

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-zdvpw96jdljzsn6b.jpg" /> A=∫(π/6)(π/2)( sin 2 x- cos x) d x=[-( cos 2 x/2)- sin x](π/6)(π/2) =[( cos 2 x/2)+ sin x](π/2)(π/6) =(1/2) × (1/2)+(1/2)+(1/2)-1=(1/4) sq. units

Q. The area bounded by the curves $y = cos x$ and $y = s in 2 x, \forall x \in [\frac{π}{6}, \frac{π}{2}]$ is equal to

$\frac{π}{2}$ sq. units

$\frac{π}{3}$ sq. units

$\frac{π}{6}$ sq. units

$\frac{1}{4}$ sq. units

$A = \int_{\frac{π}{6}}^{\frac{π}{2}} (sin 2 x - cos x) d x = [- \frac{c o s 2 x}{2} - sin x]_{\frac{π}{6}}^{\frac{π}{2}}$
$= [\frac{c o s 2 x}{2} + sin x]_{\frac{π}{2}}^{\frac{π}{6}}$
$= \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} + \frac{1}{2} - 1 = \frac{1}{4}$ sq. units

Q. The area bounded by the curves y=cosx and y=sin2x,∀x∈[6π​,2π​] is equal to

2π​ sq. units

3π​ sq. units

6π​ sq. units

41​ sq. units