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Nilai dari ∫02​(x2−2x+1)dx=....

Pertanyaan

Nilai space dari space integral subscript 0 superscript 2 left parenthesis x squared minus 2 x plus 1 right parenthesis d x equals....

  1. 2 over 3

  2. 1

  3. 4 over 3

  4. 5 over 3

  5. 2

Y. Laksmi

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

integral subscript 0 superscript 2 left parenthesis x squared minus 2 x plus 1 right parenthesis d x equals right enclose 1 third x cubed minus x squared plus x end enclose subscript 0 superscript 2  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals open square brackets 1 third left parenthesis 2 right parenthesis cubed minus left parenthesis 2 right parenthesis squared plus 2 close square brackets minus 0  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 8 over 3 minus 4 plus 2  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2 over 3

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Nilai dari ∫−11​(2x2+6x−1)dx=....

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