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Hasil dari ∫ 16 − 9 x 2 ​ d x = ....

Hasil dari ....

  1. begin mathsize 14px style 8 over 3 open parentheses 3 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end style 

  2. begin mathsize 14px style 4 over 3 open parentheses 3 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end style 

  3. begin mathsize 14px style 2 over 3 open parentheses 3 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end style 

  4. begin mathsize 14px style 2 over 3 open parentheses 4 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end style 

  5. begin mathsize 14px style 4 over 3 open parentheses 4 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end style 

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D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

jawaban yang tepat adalah C.

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Pembahasan

Misalkan: Sehingga diperoleh 4 adalah sisi miring sebuah segitiga siku-siku dengan kedua sisi tegaknya dan . Maka diperoleh dan Jika kita integralkan kedua ruas maka diperoleh Sehingga, Jadi, jawaban yang tepat adalah C.

Misalkan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of 16 minus 9 x squared end root end cell row cell y squared end cell equals cell 16 minus open parentheses 3 x close parentheses squared end cell row cell y squared plus open parentheses 3 x close parentheses squared end cell equals cell 4 squared end cell end table end style 

Sehingga diperoleh 4 adalah sisi miring sebuah segitiga siku-siku dengan kedua sisi tegaknya begin mathsize 14px style 3 x end style dan begin mathsize 14px style y end style.

Maka diperoleh

begin mathsize 14px style cos invisible function application theta equals fraction numerator square root of 16 minus 9 x squared end root over denominator 4 end fraction end style 

dan 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application theta end cell equals cell fraction numerator 3 x over denominator 4 end fraction rightwards arrow theta equals sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses end cell row x equals cell 4 over 3 sin invisible function application theta end cell row cell fraction numerator d x over denominator d theta end fraction end cell equals cell 4 over 3 cos invisible function application theta end cell end table end style 

Jika kita integralkan kedua ruas maka diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator d x over denominator d theta end fraction d theta end cell equals cell 4 over 3 integral cos invisible function application theta d theta end cell row cell integral d x end cell equals cell 4 over 3 integral cos invisible function application theta d theta end cell end table end style 

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral square root of 16 minus 9 x squared end root blank d x end cell equals cell 4 over 3 integral square root of 16 minus 9 open parentheses 4 over 3 sin invisible function application theta close parentheses squared end root times cos invisible function application theta d theta end cell row blank equals cell 4 over 3 integral square root of 16 minus 16 sin squared invisible function application theta end root times cos invisible function application theta d theta end cell row blank equals cell 4 over 3 integral square root of 16 open parentheses 1 minus sin squared invisible function application theta close parentheses end root times cos invisible function application theta d theta end cell row blank equals cell 4 over 3 integral square root of 16 cos squared invisible function application theta end root times cos invisible function application theta d theta end cell row blank equals cell 4 over 3 integral 4 cos squared invisible function application theta d theta end cell row blank equals cell 16 over 3 integral fraction numerator cos invisible function application 2 theta plus 1 over denominator 2 end fraction d theta end cell row blank equals cell 8 over 3 integral open parentheses cos invisible function application 2 theta plus 1 close parentheses d theta end cell row blank equals cell 8 over 3 open parentheses 1 half sin invisible function application 2 theta plus theta close parentheses plus C end cell row blank equals cell 8 over 3 open parentheses 1 half times 2 sin invisible function application theta cos invisible function application theta plus theta close parentheses plus C end cell row blank equals cell 8 over 3 open parentheses fraction numerator 3 x over denominator 4 end fraction times fraction numerator square root of 16 minus 9 x squared end root over denominator 4 end fraction plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end cell row blank equals cell 2 over 3 open parentheses 3 x square root of 16 minus 9 x squared end root plus sin to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 x over denominator 4 end fraction close parentheses close parentheses plus C end cell end table end style 

Jadi, jawaban yang tepat adalah C.

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