find the area of the shaded region.\nf(x)=x^{4}-8x^{3}+18x^{2}, g(x)=-25x + 102\nthe area is . (type an…

Explanation:

Step1: Determine the integral for area

The area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}|f(x)-g(x)|dx$. Here, we need to find the area between $f(x)=x^{4}-8x^{3}+18x^{2}$ and $g(x)=- 25x + 102$ from $x=-2$ to $x = 3$. So the area formula is $A=\int_{-2}^{3}[(x^{4}-8x^{3}+18x^{2})-(-25x + 102)]dx=\int_{-2}^{3}(x^{4}-8x^{3}+18x^{2}+25x - 102)dx$.

Step2: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have: $\int(x^{4}-8x^{3}+18x^{2}+25x - 102)dx=\frac{x^{5}}{5}-8\times\frac{x^{4}}{4}+18\times\frac{x^{3}}{3}+25\times\frac{x^{2}}{2}-102x+C=\frac{x^{5}}{5}-2x^{4}+6x^{3}+\frac{25x^{2}}{2}-102x+C$.

Step3: Evaluate the definite integral

$A=\left[\frac{x^{5}}{5}-2x^{4}+6x^{3}+\frac{25x^{2}}{2}-102x\right]_{-2}^{3}$ First, substitute $x = 3$: $\frac{3^{5}}{5}-2\times3^{4}+6\times3^{3}+\frac{25\times3^{2}}{2}-102\times3=\frac{243}{5}-162 + 162+\frac{225}{2}-306=\frac{243}{5}+\frac{225}{2}-306$. $=\frac{243\times2+225\times5}{10}-306=\frac{486 + 1125}{10}-306=\frac{1611}{10}-306=\frac{1611-3060}{10}=-\frac{1449}{10}$. Then substitute $x=-2$: $\frac{(-2)^{5}}{5}-2\times(-2)^{4}+6\times(-2)^{3}+\frac{25\times(-2)^{2}}{2}-102\times(-2)$ $=-\frac{32}{5}-32-48 + 50 + 204$ $=-\frac{32}{5}+174=\frac{-32 + 870}{5}=\frac{838}{5}$. Now, $A=\frac{1611}{10}-\frac{838}{5}=\frac{1611-1676}{10}=-\frac{65}{10}=-\frac{13}{2}$. But area is non - negative, so we take the absolute value.

Answer:

$\frac{13}{2}$