Question

How can we find the integral of (x-x^2) / (2x^(1/3)) from -8 to -1?

Answer 1

\[\int\limits_{-8}^{-1} \frac{ x-x ^{2} }{ 2\sqrt[3]{x} }\]

Answer 2

Try writing the bottom cube root as an exponent, and then distribute the denominator between the two terms of the numerator:\[\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\]

Answer 3

I think I got it
∫₋₈⁻¹ (x − x²) / (2x^(1/3)) dx = 1/2 ∫₋₈⁻¹ (x − x²) x^(−1/3) dx
= 1/2 ∫₋₈⁻¹ (x^(2/3) − x^(5/3)) dx
= 1/2 (3/5 x^(5/3) − 3/8 x^(8/3)) |₋₈⁻¹
= (3/10 x^(5/3) − 3/16 x^(8/3)) |₋₈⁻¹
= (3/10 (−1) − 3/16 (1)) − (3/10 (−32) − 3/16 (256))
= −3/10 − 3/16 (1) + 48/5 + 48
= 4569/80
= 57.113

Answer 4

Yep. That's correct!