Math  /  Calculus

Questionx42x3+x2dx\int \sqrt{x^{4}-2 x^{3}+x^{2}} d x

Studdy Solution
Evaluate the simplified integral at the bounds.
(122133)(022033)=1213=3626=16 \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}
The final solution for the integral over 0x<10 \leq x < 1 is:
01x42x3+x2dx=16 \int_{0}^{1} \sqrt{x^4 - 2x^3 + x^2} \, dx = \frac{1}{6}

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