Evaluate \(\int^1_0 x(x^2-2)^2 dx\)
- \(\frac{6}{7}\)
-
\(1\frac{1}{6}\)
- \(\frac{1}{7}\)
- \(3\frac{1}{6}\)
The correct answer is: B
Explanation
\(\int^1_0 x(x^2-2)^2 dx\)
\((x^2-2)^2=x^4-2x^2-2x^2+4\)
=\(x^4-4x^2+4\)
\(x(x^2-2)^2=x(x^4-4x^2+4)\)
=\(x^5-4x^3+4x\)
\(\int^1_0 x(x^2-2)^2 dx = \int^1_0 x^5 - 4x^3 + 4x dx\)
=\((\frac{x^6}{6} - x^4+2x^2)^1_0\)
= \((\frac{(1)^6}{6} - (1)^4 +2(1)^2)-(\frac{(0)^6}{6} - (0)^4+2(0)^2)\)
=\(\frac{7}{6} - 0 =\frac{7}{6}\)
\(\therefore 1\frac{1}{6}\)