Tolong dibantu kak soalnya.... Question from @Jillaja

May 2019 1 52 Report

Tolong dibantu kak soalnya

Verified answer

Integral Tentu
Kelas XII

$\int_{\frac12}^{1} (\cos \pi x - \sqrt{4x-1} ) dx \\ = \int_{\frac12}^{1} \cos \pi x \ dx - \int_{\frac12}^{1} \sqrt{4x-1} \ dx \\ \\ Kita \ tinjau \ \int_{\frac12}^{1} \sqrt{4x-1} \ dx$
Misal u = 4x - 1
du/dx = 4
dx = ¼ du

Maka
$\int_{\frac12}^{1} \sqrt{4x-1} \ dx \\ \int_{\frac12}^{1} \frac14 \sqrt{u} \ du \\ = | \frac14 \times \frac23 \ u \sqrt{u} |_{\frac12}^{1} \\ = | \frac{(4x-1) \sqrt{4x-1}}{6} |_{\frac12}^{1} \\ = ( \frac{(4(1) - 1) \sqrt{4(1) - 1)}}{6} ) - (\frac{(4(\frac12) - 1) \sqrt{4(\frac12) - 1}}{6}) \\ = \frac{ \sqrt{3}}{3} - \frac16 \\ \\ Tinjau \ \int_{\frac12}^{1} \cos \pi x dx \\ \int_{\frac12}^{1} \cos \pi x dx \\ = | \frac{\sin \pi x}{ \pi} |_{\frac12}^{1} \\ = \frac{\sin \pi }{ \pi} - \frac{\sin \frac12 \pi }{ \pi} \\ = - \frac{1}{ \pi } \\ Maka \\ \int_{\frac12}^{1} (\cos \pi x - \sqrt{4x-1} ) dx \\ = - \frac{1}{ \pi} - ( \frac{ \sqrt{3}}{3} - \frac16 ) \\ = \frac{\pi - 3 \sqrt3 \pi - 6}{6 \pi}$

Jawaban : E

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