Evaluate $\int_0^1 4x - 6\sqrt[3] {x^2}dx$

Analyze the question: We need the definite integral of 4x − 6x^(2/3) from 0 to 1.

Concept: Power rule for integration: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C, and for the cube root, ∛(x²) = x^(2/3).

Solution:

1. Rewrite the integrand: 4x − 6x^(2/3)

2. Integrate term by term: • ∫4x dx = 4·x²/2 = 2x² • ∫6x^(2/3) dx = 6 · x^(5/3)/(5/3) = 6 · (3/5) x^(5/3) = (18/5) x^(5/3)

3. So the antiderivative is: 2x² − (18/5) x^(5/3)

4. Apply the limits 0 to 1: • At x = 1: 2 − 18/5 = 10/5 − 18/5 = −8/5 • At x = 0: 0

5. Result: −8/5 − 0 = −8/5

Tip: When you see a root, convert it to a fractional power first (∛x² = x^(2/3)). Then add 1 to the power and divide by the new power. For limits 0 to 1, the answer is just the antiderivative with x = 1, since everything at 0 vanishes. Also, a quick sanity check: since 6x^(2/3) is larger than 4x on (0,1), the integrand is negative, so the answer must be negative. That immediately eliminates C and D.