Pertanyaan

Tentukanlah nilai integral di bawah ini: c. ∫ 0 2 ( x 3 − 6 x 2 + 8 x ) d x

Tentukanlah nilai integral di bawah ini:

c. $\int_{0}^{2} (x^{3} - 6 x^{2} + 8 x) d x$ $\;$

A. Armanda

Master Teacher

Mahasiswa/Alumni Universitas Indraprasta PGRI

Jawaban terverifikasi

Jawaban

.

integral subscript 0 superscript 2 open parentheses x cubed minus 6 x squared plus 8 x close parentheses space d x equals 4

Pembahasan

Ingat! Sehingga, Jadi, .

Ingat!

$table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript a superscript b f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses space d x end cell equals cell integral subscript a superscript b f open parentheses x close parentheses space d x plus-or-minus integral subscript a superscript b g open parentheses x close parentheses space d x end cell row cell integral subscript a superscript b k f left parenthesis x right parenthesis space d x end cell equals cell k integral subscript a superscript b f left parenthesis x right parenthesis space d x end cell row cell integral subscript a superscript b a x to the power of n space d x end cell equals cell open parentheses fraction numerator a over denominator n plus 1 end fraction x to the power of n plus 1 end exponent close parentheses subscript a superscript b space end cell end table$

Sehingga,

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral subscript 0 superscript 2 open parentheses x cubed minus 6 x squared plus 8 x close parentheses space d x end cell row blank equals cell integral subscript 0 superscript 2 x cubed space d x minus 6 integral subscript 0 superscript 2 x squared space d x plus 8 integral subscript 0 superscript 2 x space d x end cell row blank equals cell open parentheses fraction numerator 1 over denominator 3 plus 1 end fraction x to the power of 3 plus 1 end exponent close parentheses subscript 0 superscript 2 minus 6 open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent close parentheses subscript 0 superscript 2 plus 8 open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent close parentheses subscript 0 superscript 2 end cell row blank equals cell 1 fourth open parentheses x to the power of 4 close parentheses subscript 0 superscript 2 minus 6 times 1 third open parentheses x cubed close parentheses subscript 0 superscript 2 plus 8 times 1 half open parentheses x squared close parentheses subscript 0 superscript 2 end cell row blank equals cell 1 fourth open parentheses x to the power of 4 close parentheses subscript 0 superscript 2 minus 2 open parentheses x cubed close parentheses subscript 0 superscript 2 plus 4 open parentheses x squared close parentheses subscript 0 superscript 2 end cell row blank equals cell 1 fourth open parentheses 2 to the power of 4 minus 0 to the power of 4 close parentheses minus 2 open parentheses 2 cubed minus 0 cubed close parentheses plus 4 open parentheses 2 squared minus 0 squared close parentheses end cell row blank equals cell 16 over 4 minus 16 plus 16 end cell row blank equals 4 end table$

Jadi, integral subscript 0 superscript 2 open parentheses x cubed minus 6 x squared plus 8 x close parentheses space d x equals 4 .

Latihan Bab

Integral Fungsi Trigonometri

Integral Substitusi Trigonometri

Integral Tentu

Integral Parsial