Question

Kerjakan Menggunakan Langkah-langkah Pengerjaan, Soal Nomor 4 dan 5, Kelas 12​

Answer 1

Verified answer

Penjelasan dengan langkah-langkah:

Nomor 4 :

[tex]\tt \int\limits {(2x-3)^\frac{1}{2}} \, dx\\\\= \int\limits{(2x-3)^\frac{1}{2} \times \frac{1}{2}} \, dx \\\\= \int\limits{\frac{(2x-3)^\frac{1}{2}}{2}} \, dx \to~sifat~integral~tentu:\int\limits^b_a {k.f(x)} \, dx =k.\int\limits^b_a {f(x)} \, dx\\\\= \frac{1}{2}.\int\limits {(2x-3)^\frac{1}{2}} \, dx \\\\=\frac{1}{2}.\frac{(2x-3)^{\frac{1}{2}+1}}{\frac{1}{2}+1} \\\\= \frac{1}{2}.\frac{(2x-3)^\frac{3}{2}}{\frac{3}{2}}\\\\= \frac{1}{2}.\frac{2\sqrt{(2x-3)^3}}{3}[/tex]

[tex]\tt = \frac{2}{2}.\frac{(2x-3)\sqrt{2x-3}}{3}\\ \\ = \frac{(2x-3)\sqrt{2x-3}}{3}+C[/tex]

Nomor 5 :

[tex]\tt \int\limits {(3x-8)^5} \, dx\\\\=\int\limits {(3x-8)^5\times \frac{1}{3}} \, dx \\\\=\int\limits {\frac{(3x-8)^5}{3}} \, dx\to~ ~sifat~integral~tentu:\int\limits^b_a {k.f(x)} \, dx =k.\int\limits^b_a {f(x)} \, dx\\\\= \frac{1}{3}.\int\limits {(3x-8)^5} \, dx \\\\=\frac{1}{3}.{\frac{(3x-8)^{5+1}}{5+1}} \, dx \\\\= \frac{1}{3}.{\frac{(3x-8)^6}{6} } \, dx \\\\= \frac{(3x-8)^6}{18}+C[/tex]