21. determine the area of the region bounded by the function y = 4x - x², the x - axis and y = 6.

Explanation:

Step1: Find intersection points

Set $4x - x^{2}=6$ and $4x - x^{2}=0$. For $4x - x^{2}=0$, we have $x(4 - x)=0$, so $x = 0$ or $x = 4$. For $4x - x^{2}=6$, rewrite it as $x^{2}-4x + 6=0$. The discriminant $\Delta=(-4)^{2}-4\times6=-8<0$, so no real - roots from this equation. We need to find where $y = 4x - x^{2}$ and $y = 0$ and calculate the area between the curves. The area $A=\int_{0}^{4}(4x - x^{2})dx$.

Step2: Apply integral formula

The antiderivative of $4x - x^{2}$ is $F(x)=4\times\frac{x^{2}}{2}-\frac{x^{3}}{3}=2x^{2}-\frac{x^{3}}{3}$.

Step3: Evaluate definite integral

$A=\left[2x^{2}-\frac{x^{3}}{3}\right]_{0}^{4}$. $A=(2\times4^{2}-\frac{4^{3}}{3})-(2\times0^{2}-\frac{0^{3}}{3})$. $A=(32-\frac{64}{3})-0$. $A=\frac{96 - 64}{3}=\frac{32}{3}$.

Answer:

$\frac{32}{3}$