current objective find the area of a region bounded above by two different functions question determine the…

Explanation:

Step1: Determine the upper - function

On the interval $[1,4]$, we need to find which of $f(x)=x^{2}-8x + 16$ and $g(x)=4x - 4$ is greater. Let's find the difference $h(x)=f(x)-g(x)=x^{2}-8x + 16-(4x - 4)=x^{2}-12x + 20=(x - 2)(x - 10)$. For $x\in[1,4]$, when $x\in[1,2]$, $f(x)\geq g(x)$; when $x\in[2,4]$, $g(x)\geq f(x)$.

Step2: Set up the integral for the area

The area $A$ is given by $A=\int_{1}^{2}(x^{2}-8x + 16)dx+\int_{2}^{4}(4x - 4)dx$.

Step3: Integrate the first integral

$\int_{1}^{2}(x^{2}-8x + 16)dx=\left[\frac{1}{3}x^{3}-4x^{2}+16x\right]_{1}^{2}=\left(\frac{1}{3}(2)^{3}-4(2)^{2}+16(2)\right)-\left(\frac{1}{3}(1)^{3}-4(1)^{2}+16(1)\right)=\left(\frac{8}{3}-16 + 32\right)-\left(\frac{1}{3}-4 + 16\right)=\left(\frac{8}{3}+16\right)-\left(\frac{1}{3}+12\right)=\frac{8 + 48}{3}-\frac{1+36}{3}=\frac{56}{3}-\frac{37}{3}=\frac{19}{3}$.

Step4: Integrate the second integral

$\int_{2}^{4}(4x - 4)dx=\left[2x^{2}-4x\right]_{2}^{4}=(2(4)^{2}-4(4))-(2(2)^{2}-4(2))=(32 - 16)-(8 - 8)=16$.

Step5: Calculate the total area

$A=\frac{19}{3}+16=\frac{19+48}{3}=\frac{67}{3}$.

Answer:

$\frac{67}{3}$