Pertanyaan

Diketahui ∫ 1 2 x + 2 1 d x = k .Jika k  = 0 ,maka nilai yang memenuhi ∫ 1 2 x + 2 4 x − 2 p d x = 4 k + 4 adalah ....

Diketahui $\int_{1}^{2} \frac{1}{x + 2} d x = k$ . Jika $k \neq = 0$ , maka nilai $p$ yang memenuhi $\int_{1}^{2} \frac{4 x - 2 p}{x + 2} d x = 4 k + 4$ adalah ....

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D. Natalia

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan perhitungan berikut! Jadi, jawaban yang tepat adalah E.

Perhatikan perhitungan berikut!

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript 1 superscript 2 fraction numerator 4 x minus 2 p over denominator x plus 2 end fraction space d x end cell equals cell integral subscript 1 superscript 2 fraction numerator 4 x plus 8 minus 8 minus 2 p over denominator x plus 2 end fraction space d x end cell row cell 4 k plus 4 end cell equals cell integral subscript 1 superscript 2 open square brackets open parentheses fraction numerator 4 x plus 8 over denominator x plus 2 end fraction close parentheses minus open parentheses fraction numerator 8 plus 2 p over denominator x plus 2 end fraction close parentheses close square brackets space d x end cell row cell 4 k plus 4 end cell equals cell integral subscript 1 superscript 2 fraction numerator 4 x plus 8 over denominator x plus 2 end fraction space d x minus integral subscript 1 superscript 2 fraction numerator 8 plus 2 p over denominator x plus 2 end fraction space d x end cell row cell 4 k plus 4 end cell equals cell integral subscript 1 superscript 2 fraction numerator 4 open parentheses x plus 2 close parentheses over denominator x plus 2 end fraction space d x minus open parentheses 8 plus 2 p close parentheses integral subscript 1 superscript 2 fraction numerator 1 over denominator x plus 2 end fraction space d x end cell row cell 4 k plus 4 end cell equals cell integral subscript 1 superscript 2 4 space d x minus open parentheses 8 plus 2 p close parentheses k end cell row cell 4 k plus 4 end cell equals cell open square brackets 4 x close square brackets subscript 1 superscript 2 minus open parentheses 8 plus 2 p close parentheses k end cell row cell 4 k plus 4 end cell equals cell 4 open parentheses 2 minus 1 close parentheses minus open parentheses 8 plus 2 p close parentheses k end cell row cell 4 k plus 4 end cell equals cell 4 minus open parentheses 8 plus 2 p close parentheses k end cell row cell 4 k end cell equals cell negative open parentheses 8 plus 2 p close parentheses k semicolon space k not equal to 0 end cell row 4 equals cell negative 8 minus 2 p end cell row cell 2 p end cell equals cell negative 12 end cell row p equals cell negative 6 end cell end table end style$

Jadi, jawaban yang tepat adalah E.