Please bantuin, dikumpulin hari selasa ini . yah yah yah :D hehhe.... Question from @dewie19970730

dewie19970730 @dewie19970730

September 2018 1 44 Report

Please bantuin, dikumpulin hari selasa ini . yah yah yah :D hehhe

acim $1) \int\limits^ \frac{ \pi }{6} _0 {(cos^2x - sin^2x)} \, dx \\ \\ = \int\limits^ \frac{ \pi }{6} _0 {cos(2x)} \, dx \\ \\ = [\frac{1}{2}sin(2x)] [0,\frac{\pi}{6}] \\ \\ = \frac{1}{2}sin(2(\frac{\pi }{6})) - \frac{1}{2}sin(2(0)) \\ \\ = \frac{1}{2}sin(\frac{\pi }{3}) - \frac{1}{2}sin(0) \\ \\ = \frac{1}{2}(\frac{1}{2} \sqrt{3}) - 0 \\ \\ = \frac{1}{4}\sqrt{3} \\ \\ = \frac{1}{4} \sqrt{3}$

2) $\int\limits^7_4 { \frac{dx}{ \sqrt{x-3}}} \, \\ \\ misal u = x - 3 \\ du = dx \\ \\ = \int\limits^7_4 { \frac{du}{ \sqrt{u}}} \\ \\ = \int\limits^7_4 {\frac{du}{u^{ \frac{1}{2}}} \\ \\ = \int\limits^7_4 {u^{ -\frac{1}{2}}{du}} \\ \\ = [2u^{ \frac{1}{2}}] [4,7] \\ \\ = [2(x-3)^{ \frac{1}{2}}] [4,7] \\ \\ = [2 \sqrt{x-3}] [4,7]$

$= (2 \sqrt{7-3}) - (2 \sqrt{4-3}) \\ \\ = 4 - 2 \\ \\ = 2$

1 votes Thanks 1

dewie19970730 maap tapi nmr satu yg kamu kerjain jawaban'a ga ada di pg :D hehhe

acim berarti optionnya salah , hehe ... cek no 1 di link berikut ini : (wolfram gak bisa bohong) http://www.wolframalpha.com/input/?i=integral&a=*C.integral-_*Calculator.dflt-&f2=%28cos%28x%29%29^2+-+%28sin%28x%29%29^2&f=Integral.integrand\u005f%28cos%28x%29%29^2+-+%28sin%28x%29%29^2&f3=0&f=Integral.rangestart\u005f0&f4=pi%2F6&f=Integral.rangeend_pi%2F6&a=*FVarOpt.1-_**-.***Integral.variable---.**Integral.rangestart-.*Integral.rangeend---

dewie19970730 gabisa dicopy klo lewat hape :D

acim yg sdh, sy kasih tau aja... nomer 1 dijamin bener 1000 %

dewie19970730 okey deh, thanks so much yaaa :)