Jika ∫⁰-₂ untuk nilai k bilangan bulat , maka k²-14=...a)140b)135c)130d)125e)120.... Question from @Anderson31

November 2019 1 62 Report

Jika ∫⁰-₂ $(cos(\pi + \frac{\pi \: kx}{2} ) + \frac{9x {}^{2} - 10x + 14}{k + 12} )dx=(k - 9)(k - 11)$
untuk nilai k bilangan bulat , maka k²-14=...
a)140
b)135
c)130
d)125
e)120

dwiafifah68

Verified answer

Untuk nilai k bilangan bulat, maka nilai k² - 14 = 130 dengan nilai k = 12, maka diperoleh

k² - 14 = 12² - 14

k² - 14 = 144 - 14

k² - 14 = 130

Simak pembahasan berikut.

Pembahasan

Diketahui:

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2} ) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = (k - 9)(k - 11)

untuk k bilangan bulat

Ditanya: nilai k² - 14

Jawab:

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$

Untuk mendapatkan nilai integral tersebut dihitung terlebih dahulu nilai $cos(\pi + \frac{\pi kx}{2})$

Ingat! cos (a + b) = cos a cos b - sin a sin b

$cos(\pi + \frac{\pi kx}{2})$ = cos π cos $\frac{\pi kx }{2}$ - sin π sin $\frac{\pi kx }{2}$

$cos(\pi + \frac{\pi kx}{2})$ = -1(cos $\frac{\pi kx }{2}$ ) - 0(sin $\frac{\pi kx }{2}$ )

$cos(\pi + \frac{\pi kx}{2})$ = -cos $\frac{\pi kx }{2}$

Subtitusikan -cos $\frac{\pi kx }{2}$ kedalam integral

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\int\limits^0_{-2} {(-cos \frac{\pi kx}{2} + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\int\limits^0_{-2} {-cos \frac{\pi kx}{2}} \, dx$ + $\int\limits^0_{-2} {(\frac{9x^{2} - 10x + 14}{k + 12})} \, dx$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\int\limits^0_{-2} {-cos \frac{\pi kx}{2}} \, dx$ + $\frac{1}{k+ 12}$ $\int\limits^0_{-2} {(9x^{2} - 10x + 14)} \, dx$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $-\frac{1}{\frac{k\pi}{2}} sin \frac{\pi kx}{2}$ $\left \| {{0} \atop {-2}} \right.$ + $\frac{1}{k+ 12}$ ( $\frac{9}{2 + 1}$ x²⁺¹ - $\frac{10}{1 + 1}$ x¹⁺¹ + 14x) $\left \| {{0} \atop {-2}} \right.$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $-\frac{2}{k\pi} sin \frac{\pi kx}{2}$ $\left \| {{0} \atop {-2}} \right.$ + $\frac{1}{k+ 12}$ ( $\frac{9}{3}$ x³ - $\frac{10}{2}$ x² + 14x) $\left \| {{0} \atop {-2}} \right.$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $-\frac{2}{k\pi} sin \frac{\pi kx}{2}$ $\left \| {{0} \atop {-2}} \right.$ + $\frac{1}{k+ 12}$ (3x³ - 5x² + 14x) $\left \| {{0} \atop {-2}} \right.$

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = ( $-\frac{2}{k\pi}) (sin \frac{\pi k(0)}{2} - sin \frac{\pi k(-2)}{2})$ + $\frac{1}{k+ 12}$ (3(0)³ - 5(0)² + 14(0) - (3(-2)³ - 5(-2)² + 14(-2)))

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = ( $-\frac{2}{k\pi})$ (0 - sin(-πk)) + $\frac{1}{k+ 12}$ (0 - 0 + 0 - (3(-8) - 5(4) + 14(-2)))

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $-\frac{2}{k\pi}$ (-sin(-πk)) + $\frac{1}{k+ 12}$ (0 - (-24 - 20 + (-28)))

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\frac{2}{k\pi}$ sin(-πk) + $\frac{1}{k+ 12}$ (0 - (-72)

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\frac{2}{k\pi}$ sin(-πk) + $\frac{1}{k+ 12}$ (0 + 72)

Ingat! sin kπ = 0, untuk k bilangan bulat, maka sin (-πk) = 0

$\int\limits^0_{-2} {(cos (\pi + \frac{\pi kx}{2}) + \frac{9x^{2} - 10x + 14}{k + 12})} \, dx$ = $\frac{2}{k\pi}$ (0) + $\frac{72}{k+ 12}$