Find the area under the curve Y=4x-x square and the x-axis

\int (4x - x^2) \, dx
= \int 4x \, dx - \int x^2 \, dx
= 2x^2 - \frac{x^3}{3} + C

To find the area, evaluate this from \(x = 0\) to \(x = 4\).

So,

\left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}

Evaluating at upper limit \(4\):

\left( 2(4)^2 - \frac{(4)^3}{3} \right) = 2(16) - \frac{64}{3} = 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}

Evaluating at lower limit \(0\):

\left( 2(0)^2 - \frac{(0)^3}{3} \right) = 0

Difference between upper and lower limits:

\frac{32}{3} - 0 = \frac{32}{3}

Hence, the area under the curve is:

\frac{32}{3}