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Y. Endah

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Pembahasan

$integral x squared open parentheses 2 minus x cubed close parentheses to the power of 1 half end exponent space d x A m b i l u equals 2 minus x cubed rightwards double arrow fraction numerator d u over denominator d x end fraction equals negative 3 x squared rightwards double arrow d x equals negative fraction numerator d u over denominator 3 x squared end fraction integral x squared u to the power of 1 half end exponent space open parentheses negative fraction numerator d u over denominator 3 x squared end fraction close parentheses minus 1 third integral u to the power of 1 half end exponent space d u minus 1 third open parentheses 2 over 3 close parentheses u to the power of 3 over 2 end exponent plus C minus 2 over 9 open parentheses 2 minus x cubed close parentheses to the power of 3 over 2 end exponent plus C$