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Nilai dari ∫ − 1 1 ( x 3 + 2 x − 1 ) 2 d x adalah...

Nilai dari $\int_{- 1}^{1} (x^{3} + 2 x - 1)^{2} d x$ adalah...

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

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Pembahasan

Jadi, tidak ada jawaban yang tepat.

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral subscript negative 1 end subscript superscript 1 left parenthesis x cubed plus 2 x minus 1 right parenthesis squared d x end cell row blank equals cell integral subscript negative 1 end subscript superscript 1 left parenthesis x to the power of 6 plus 2 x to the power of 4 minus x cubed plus 2 x to the power of 4 plus 4 x squared minus 2 x minus x cubed minus 2 x plus 1 right parenthesis d x end cell row blank equals cell integral subscript negative 1 end subscript superscript 1 left parenthesis x to the power of 6 plus 2 x to the power of 4 plus 2 x to the power of 4 minus x cubed minus x cubed plus 4 x squared minus 2 x minus 2 x plus 1 right parenthesis d x end cell row blank equals cell integral subscript negative 1 end subscript superscript 1 left parenthesis x to the power of 6 plus 4 x to the power of 4 minus 2 x cubed plus 4 x squared minus 4 x plus 1 right parenthesis d x end cell row blank equals cell open square brackets 1 over 7 x to the power of 7 plus 4 over 5 x to the power of 5 minus 2 over 4 x to the power of 4 plus 4 over 3 x cubed minus 4 over 2 x squared plus x close square brackets subscript negative 1 end subscript superscript 1 end cell row blank equals cell open parentheses 1 over 7 1 to the power of 7 plus 4 over 5 1 to the power of 5 minus 2 over 4 1 to the power of 4 plus 4 over 3 1 cubed minus 4 over 2 1 squared plus 1 close parentheses minus end cell row blank blank cell open parentheses 1 over 7 left parenthesis negative 1 right parenthesis to the power of 7 plus 4 over 5 left parenthesis negative 1 right parenthesis to the power of 5 minus 2 over 4 left parenthesis negative 1 right parenthesis to the power of 4 plus 4 over 3 left parenthesis negative 1 right parenthesis cubed minus 4 over 2 left parenthesis negative 1 right parenthesis squared plus left parenthesis negative 1 right parenthesis close parentheses end cell row blank equals cell open parentheses 1 over 7 plus 4 over 5 minus 2 over 4 plus 4 over 3 minus 4 over 2 plus 1 close parentheses minus open parentheses negative 1 over 7 minus 4 over 5 minus 2 over 4 minus 4 over 3 minus 4 over 2 minus 1 close parentheses end cell row blank equals cell 2 over 7 plus 8 over 5 plus 8 over 3 plus 2 end cell row blank equals cell fraction numerator 30 plus 168 plus 280 plus 210 over denominator 105 end fraction end cell row blank equals cell 688 over 105 end cell end table$

Jadi, tidak ada jawaban yang tepat.