Evaluate the integral and interpret it as the area of a region. Sketch the region \( \int_{0}^{\pi/2}|sin x - cos 2x|\;dx \)

Solution:

Given,

\( \int_{0}^{\pi/2}|sin x - cos 2x|\;dx \)

We have,

sin x = cos 2x

sin x = 1 - 2 sin ²x

2 sin² x + sin x -1 = 0

2sin² x +2 sin x - sin x -1 = 0

2sin x(sin x + 1) - 1(sin x + 1) = 0

(2sin x - 1) (sin x + 1) = 0

The roots are:

2 sin x -1 = 0 ⇒ sin x = 1/2

⇒ x = 𝜋 /6, 𝜋 /6 belongs to 0 to 𝜋 /2

sin x +1 = 0 ⇒ sin x = -1

⇒ x = - 𝜋 /2, but this does not belong to 0 to 𝜋 /2

∴\( \int_{0}^{\pi/2}|sin x - cos 2x|\;dx = \int_{0}^{\pi /6}\left ( cos2x - sinx \right ) dx +\int_{\pi /6}^{\pi /2}\left ( sinx -cos 2x \right ) \)

\( \frac{sin2x}{2}\Biggr|_{0}^{\pi /6} + cos x\Biggr|_{0}^{\pi /6} - cos x\Biggr|_{\pi /6}^{\pi /2} - \frac{sin2x}{2}\Biggr|_{\pi /6}^{\pi /2} \)

= 1/2(√3/2 -0) + (√3/2 -1) - (0-√3/2) - 1/2 (0-√3/2)

= √3/4 + √3/2 -1 + √3/2 + √3/4

= (3√3)/2 - 1

Evaluate the integral and interpret it as the area of a region. Sketch the region [latex]\int_{0}^{\pi/2}|sin x - cos 2x|\;dx[/latex]

Summary:

The integral of \( \int_{0}^{\pi/2}|sin x - cos 2x|\;dx\) is (3√3)/2 - 1.