3.4 find the area bounded by the parabola y = x^2 - 2x - 8, the x - axis and the lines x = 0 and x = 4…

Explanation:

Step1: Find the x - intercepts of the parabola

Set $y = x^{2}-2x - 8=0$. Factor the quadratic equation: $(x - 4)(x+2)=0$. So the x - intercepts are $x=-2$ and $x = 4$.

Step2: Determine the sign of the function on the interval $[0,4]$

Let $f(x)=x^{2}-2x - 8$. We can find $f(0)=0^{2}-2\times0 - 8=-8$ and $f(4)=4^{2}-2\times4 - 8=16 - 8 - 8 = 0$. The function $y=x^{2}-2x - 8$ is non - positive on the interval $[0,4]$.

Step3: Calculate the area using integration

The area $A$ between the curve $y = f(x)$, the x - axis and the lines $x = 0$ and $x = 4$ is given by $A=-\int_{0}^{4}(x^{2}-2x - 8)dx$. First, find the antiderivative of $x^{2}-2x - 8$. The antiderivative $F(x)=\frac{1}{3}x^{3}-x^{2}-8x$. Then, use the fundamental theorem of calculus: [ \begin{align*} A&=-\left[\frac{1}{3}x^{3}-x^{2}-8x\right]_{0}^{4}\ &=-\left(\frac{1}{3}(4)^{3}-(4)^{2}-8\times4\right)\ &=-\left(\frac{64}{3}-16 - 32\right)\ &=-\left(\frac{64}{3}-48\right)\ &=-\left(\frac{64 - 144}{3}\right)\ &=\frac{80}{3} \end{align*} ]

Answer:

$\frac{80}{3}$