Correct Answer: Option B

Explanation

\(\int^1_0 x(x^2-2)^2 dx\)

\((x^2-2)^2=x^4-2x^2-2x^2+4\)

=\(x^4-4x^2+4\)

\(x(x^2-2)^2=x(x^4-4x^2+4)\)

=\(x^5-4x^3+4x\)

\(\int^1_0 x(x^2-2)^2 dx = \int^1_0 x^5 - 4x^3 + 4x dx\)

=\((\frac{x^6}{6} - x^4+2x^2)^1_0\)

= \((\frac{(1)^6}{6} - (1)^4 +2(1)^2)-(\frac{(0)^6}{6} - (0)^4+2(0)^2)\)

=\(\frac{7}{6} - 0 =\frac{7}{6}\)

\(\therefore 1\frac{1}{6}\)

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