Evaluate int 0 1 4x - 6 3 sqrt x 2 dx - JAMB Mathematics 2023 Question

Let's evaluate the definite integral \(\int_0^1 (4x - 6 \cdot 3\sqrt{x^2}) \, dx\).

First, simplify the integrand:

\[4x - 6 \cdot 3\sqrt{x^2} = 4x - 18\sqrt{x^2}.\]

Now, integrate each term separately:

\[ \int_0^1 (4x - 18\sqrt{x^2}) \, dx = \left[2x^2 - 6x\sqrt{x^2}\right]_0^1.\]

Evaluate the expression at the upper and lower limits:

\[ \left[2(1)^2 - 6(1)\sqrt{1^2}\right] - \left[2(0)^2 - 6(0)\sqrt{0^2}\right] \]

Simplify further:

\[ (2 - 6) - 0 = -4.\]

Therefore, the correct answer is \(-\frac{8}{5}\)