Area enclosed between the curves y= sin 2 x and y= cos 2 x in the interval 0 ≤ x ≤ π, is equal to

Question

Area enclosed between the curves y= sin 2 x and y= cos 2 x in the interval 0 ≤ x ≤ π, is equal to (A) 2 (B) (1/2) (C) 1 (D) (π/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0d8573689d424eb568ce9509bdf0ff99-.png" /> Area =∫ limitsπ / 43 π / 4( sin 2 x- cos 2 x) d x .=-∫ limitsπ / 43 π / 4 cos 2 x d x=(-1/2)( sin 2 x)]π / 43 π / 4 =(-1/2)(-1-1)=1

Q. Area enclosed between the curves $y = sin^{2} x$ and $y = cos^{2} x$ in the interval $0 \leq x \leq π$ , is equal to

2

$\frac{1}{2}$

1

$\frac{π}{2}$

Area $= π /4 \int 3 π /4 (sin^{2} x - cos^{2} x) d x$
$= - π /4 \int 3 π /4 cos 2 x d x = \frac{- 1}{2} (sin 2 x)]_{π /4}^{3 π /4}$
$= \frac{- 1}{2} (- 1 - 1) = 1$

Q. Area enclosed between the curves y=sin2x and y=cos2x in the interval 0≤x≤π, is equal to

2

21​

1

2π​

Area =π/4∫3π/4​(sin2x−cos2x)dx =−π/4∫3π/4​cos2xdx=2−1​(sin2x)]π/43π/4​ =2−1​(−1−1)=1

Q. Area enclosed between the curves $y = sin^{2} x$ and $y = cos^{2} x$ in the interval $0 \leq x \leq π$ , is equal to

$\frac{1}{2}$

$\frac{π}{2}$

Area $= π /4 \int 3 π /4 (sin^{2} x - cos^{2} x) d x$
$= - π /4 \int 3 π /4 cos 2 x d x = \frac{- 1}{2} (sin 2 x)]_{π /4}^{3 π /4}$
$= \frac{- 1}{2} (- 1 - 1) = 1$