Evaluate \(\int_0^1 4x - 6\sqrt[3] {x^2}dx\)
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Correct Answer: Option B
Explanation:
\(\int_0^1 4x - 6^3\sqrt x^2\) dx
=\(\int_0^1 4x - 6x^{2/3}\) dx
=\((2x^2 - \frac{8x^{5/3}}{5})_0^1\)
=\((2(1)^2 - \frac{18(1)^{5/3}}{5}) - (2(0)^2 - \frac{18(0)^{5/3}}{5}\)
= - \(\frac{8}{5}\) - 0
- \(\frac{8}{5}\)
\(\int_0^1 4x - 6^3\sqrt x^2\) dx
=\(\int_0^1 4x - 6x^{2/3}\) dx
=\((2x^2 - \frac{8x^{5/3}}{5})_0^1\)
=\((2(1)^2 - \frac{18(1)^{5/3}}{5}) - (2(0)^2 - \frac{18(0)^{5/3}}{5}\)
= - \(\frac{8}{5}\) - 0
- \(\frac{8}{5}\)