INTEGRAL (Lihat gambar sebelum menjawab) luas daerah yang diarsir dgn batas y=sin(pangkat 2)kx dan y.... Question from @Leviastrom

September 2018 1 48 Report

INTEGRAL (Lihat gambar sebelum menjawab) luas daerah yang diarsir dgn batas y=sin(pangkat 2)kx dan y=cos(pangkat 2)kx

hendrisyafa
y1= cos²kx
y2 = sin²kx
remember : cos²x + sin²x = 1

Cari batasan kurva :
                     y1       =   y2
cos²kx                      =   sin²kx
cos²kx - sin²kx          = 0
cos²kx - (1-cos²kx)    = 0
cos²kx - 1 + cos²kx   = 0
              2cos²kx -1  = 0
                  2 cos²kx = 1
                     cos²kx = 1/2
                 cos kx = √1/2     and cos kx = -1/2
                            = 1/2 √2   and             = -1/2√2

for :
cos kx = 1/2√2
           = cos 45
      kx = 45o or 1/4 π

and
cos kx = -1/2√2
           = cos  135o or 3/4 π

Area (luas) = $\int\limits^\frac{3\pi}{4}_\frac{\pi}{4} ({y1}^{2}-y2^{2} )dx$

$Area (luas) = \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} {(cos^{2}kx)}^{2}-(sin^{2}kx)^{2} dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} (cos^{2}kx+sin^{2}kx)(cos^{2}kx-sin^{2}kx) dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} 1.(cos^{2}kx-sin^{2}kx) dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} (cos^{2}kx-sin^{2}kx) dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} (cos^{2}kx-(1-cos^{2}kx)) dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} (2cos^{2}kx- 1) dx$

$= \int\limits^\frac{3\pi}{4}_\frac{\pi}{4} (cos2kx) dx$

$Area (Luas) = \frac{1}{2k}sin2kx \left \{ {{a=\frac{3\pi}{4} } \atop {b= \frac{\pi}{4}}} \right$

$A = \frac{1}{2k}[(sin(2\frac{3\pi}{4})k)-sin(2\frac{\pi}{4})k)$

dengan k = 1

$A = \frac{1}{2}[sin(\frac{3\pi}{2})-sin(\frac{\pi}{2})]$

$A= \frac{1}{2}(-1-1) = I-1I$

A= 1

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