Area of one loop formed by |y| = | sin x| is

Question

Area of one loop formed by |y| = | sin x| is (A)  0  (B)  2  (C)  4  (D)  2π

Tardigrade Admin · Accepted Answer

We have, |y|=| sin x| The graph of given curve <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ceacda7fcacc980426cd2893b7bf6fc3-.png" /> The area of one loop =2 ∫ limits0π sin x d x =2[- cos x]0π=2[- cos π+ cos 0] =2(1+1)=4

Q. Area of one loop formed by $∣ y ∣ = ∣ sin x ∣$ is

$0$

$2$

$4$

$2 π$

We have, $∣ y ∣ = ∣ sin x ∣$
The graph of given curve

The area of one loop
$= 2 0 \int π sin x d x$
$= 2 [- cos x]_{0}^{π} = 2 [- cos π + cos 0]$
$= 2 (1 + 1) = 4$

Q. Area of one loop formed by ∣y∣=∣sinx∣ is

0

2

4

2π

We have, ∣y∣=∣sinx∣ The graph of given curve The area of one loop =20∫π​sinxdx =2[−cosx]0π​=2[−cosπ+cos0] =2(1+1)=4

Q. Area of one loop formed by $∣ y ∣ = ∣ sin x ∣$ is

$0$

$2$

$4$

$2 π$

We have, $∣ y ∣ = ∣ sin x ∣$
The graph of given curve

The area of one loop
$= 2 0 \int π sin x d x$
$= 2 [- cos x]_{0}^{π} = 2 [- cos π + cos 0]$
$= 2 (1 + 1) = 4$