Pembahasan Integral Simak UI 2018 Matematika IPA kode 421

Soal yang Akan Dibahas
Jika $ \int \limits_{-2}^0 \left( \cos \left(\pi + \frac{\pi kx}{2} \right) + \frac{9x^2 - 10x + 14}{k+12} \right) dx = (k-9)(k-11) $ untuk nilai $ k $ bilangan bulat, maka $ k^2 - 14 = .... $
A). $ 140 \, $ B). $ 135 \, $ C). $ 130 \, $ D). $ 125 \, $ E). $ 120 \, $

$\spadesuit $ Konsep Dasar
*). Rumus integral :
$ \int ax^n \, dx = \frac{a}{n+1}x^{n+1} + c $
$ \int \cos ax \, dx = \frac{1}{a} \sin ax + c $
*). Rumus trigonometri :
$ \cos ( \pi + A ) = -\cos A $
*). Sifat integral tentu :
$ \int \limits_a^b ( f(x) + g(x)) dx = \int \limits_a^b f(x) dx + \int \limits_a^b g(x) dx $
*). untuk $ k $ bilangan bulat, maka $ \sin ( 2\pi k ) = 0 $

$\clubsuit $ Pembahasan
*). Menyusun persamaan :
$\begin{align} \int \limits_{-2}^0 \left( \cos \left(\pi + \frac{\pi kx}{2} \right) + \frac{9x^2 - 10x + 14}{k+12} \right) dx& = (k-9)(k-11) \\ \int \limits_{-2}^0 \left( - \cos \left( \frac{\pi kx}{2} \right) + \frac{9x^2 - 10x + 14}{k+12} \right) dx &= (k-9)(k-11) \\ \int \limits_{-2}^0 - \cos \left( \frac{\pi kx}{2} \right) dx + \int \limits_{-2}^0 \, \frac{9x^2 - 10x + 14}{k+12} dx & = (k-9)(k-11) \\ \int \limits_{-2}^0 - \cos \left( \frac{\pi kx}{2} \right) dx + \frac{1}{k+12} \int \limits_{-2}^0 \, 9x^2 - 10x + 14 dx & = (k-9)(k-11) \\ -\frac{1}{\frac{\pi k}{2} } [\sin \left( \frac{\pi kx}{2} \right) ]_{-2}^0 + \frac{1}{k+12} [ 3x^3 - 5x^2 + 14x ]_{-2}^0 & = (k-9)(k-11) \\ -\frac{1}{\frac{\pi k}{2} } [0 - 0 ] + \frac{1}{k+12} [72] & = (k-9)(k-11) \\ 0 + \frac{72}{k+12} & = (k-9)(k-11) \\ \frac{72}{k+12} & = (k-9)(k-11) \\ (k+12)(k-9)(k-11) & = 72 \\ (k+12)(k-9)(k-11) & = 24.3.2 \\ \end{align} $
terpenuhi untuk $ k = 12 $
Sehingga nilai $ k^2 - 14 = 12^2 - 14 = 144 - 14 = 130 $
Jadi, nilai $ k^2 - 14 = 130. \, \heartsuit $

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