soal integra tolong Bantu jawab

Jawaban:

Integral :

Rumus dasar integral tak tentu

[tex] {\boxed{\ \tt \int x {}^{n}dx = \frac{1}{n + 1} x {}^{n+ 1} + C }} [/tex]

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PENYELESAIAN

# Soal 1

[tex] \int \rm ( {3x}^{2} + 2x + 4)dx[/tex]

[tex] = \rm \frac{3}{2 + 1} x {}^{2 + 1} + \frac{2}{1 + 1} x {}^{1 + 1} + 4x + C [/tex]

[tex] = \rm \frac{3}{3}x^{3} + \frac{2}{2}x^{2} + 4x + C [/tex]

[tex] = \rm \: x^{3} + x^{2} + 4x + C [/tex]

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# Soal 2

[tex] \int \rm (x . \sqrt{x}) dx [/tex]

[tex] \int \rm (x . x^{\frac{1}{2}})dx [/tex]

[tex] \int \rm \: x^{\frac{3}{2}} dx [/tex]

[tex] = \rm \frac{1}{\frac{3}{2} + 1} x^{\frac{3}{2} + 1} + C [/tex]

[tex] = \rm \frac{1}{\frac{5}{2}} x^{\frac{5}{2}} + C [/tex]

[tex] = \rm \frac{2}{5}x^{\frac{5}{2}} + C [/tex]

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# Soal 3

[tex] \int \rm (x + 1)^{2} dx [/tex]

[tex] \int \rm ( x^{2} + 2x + 1) dx [/tex]

[tex] = \rm \frac{1}{2 + 1}x^{2 + 1} + \frac{2}{1 + 1}x^{1 + 1} + x + C [/tex]

[tex] = \rm \frac{1}{3}x^{3} + \frac{2}{2}x^{2} + x + C [/tex]

[tex] = \rm \frac{1}{3}x^{3} + x^{2} + x + C [/tex]

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# Soal 4

[tex] \int \rm (x^{\frac{1}{3}} - 3) dx [/tex]

[tex] = \rm \frac{1}{\frac{1}{3} + 1}x^{\frac{1}{3} + 1} - 3x + C [/tex]

[tex] = \rm \frac{1}{\frac{4}{3}}x^{\frac{4}{3}} - 3x + C [/tex]

[tex] = \rm \frac{3}{4}x^{\frac{4}{3}} - 3x + C [/tex]

••••••••••••••••••••••♪♪♪♪•••••••••••••••••••••

Detail jawaban

♬ Mapel: Matematika

♬ Kelas: XI

♬ Materi: Bab 10 - Integral tak tentu

♬ Kata kunci: integral

♬ Kode Soal: 2

♬ Kode Kategorisasi: 11.2.10

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