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Integral Tentu
Kelas XII

[tex] \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 [/tex]
Misal u = 4x - 1
du/dx = 4
dx = ¼ du

Maka
[tex] \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} [/tex]

Jawaban : E