current objective find the area of a region bounded by two functions that cross question determine the area…

current objective find the area of a region bounded by two functions that cross question determine the area, in square units, of the region bounded by f(x)=−7x³−4x² + 2x−1 and g(x)=−6x³−12x² + 14x−1 over the interval 4,8. provide your answer below:

current objective find the area of a region bounded by two functions that cross question determine the area, in square units, of the region bounded by f(x)=−7x³−4x² + 2x−1 and g(x)=−6x³−12x² + 14x−1 over the interval 4,8. provide your answer below:

Answer

Explanation:

Step1: Find the difference of the functions

Let $h(x)=g(x)-f(x)=(-6x^{3}-12x^{2}+14x - 1)-(-7x^{3}-4x^{2}+2x - 1)=x^{3}-8x^{2}+12x$.

Step2: Use the definite - integral formula for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ on the interval $[a,b]$ is $A=\int_{a}^{b}|g(x)-f(x)|dx$. Here, on $[4,8]$, $g(x)-f(x)\geq0$, so $A=\int_{4}^{8}(x^{3}-8x^{2}+12x)dx$.

Step3: Integrate term - by - term

$\int(x^{3}-8x^{2}+12x)dx=\frac{1}{4}x^{4}-\frac{8}{3}x^{3}+6x^{2}+C$.

Step4: Evaluate the definite integral

$A=\left[\frac{1}{4}x^{4}-\frac{8}{3}x^{3}+6x^{2}\right]_{4}^{8}$ $=\left(\frac{1}{4}(8)^{4}-\frac{8}{3}(8)^{3}+6(8)^{2}\right)-\left(\frac{1}{4}(4)^{4}-\frac{8}{3}(4)^{3}+6(4)^{2}\right)$ $=\left(\frac{1}{4}\times4096-\frac{8}{3}\times512 + 6\times64\right)-\left(\frac{1}{4}\times256-\frac{8}{3}\times64+6\times16\right)$ $=(1024-\frac{4096}{3}+384)-(64-\frac{512}{3}+96)$ $=(1408-\frac{4096}{3})-(160-\frac{512}{3})$ $=1408-\frac{4096}{3}-160+\frac{512}{3}$ $=1248-\frac{3584}{3}$ $=\frac{3744 - 3584}{3}=\frac{160}{3}$.

Answer:

$\frac{160}{3}$