find the area of the region bounded by the graphs of the given equations. enter your answer in exact form or…

Explanation:

Step1: Determine the upper - lower functions

On the interval $[2,4]$, we need to determine which function is on the top and which is on the bottom. Let's compare $y_1 = e^{-0.4x}$ and $y_2=x^{2}+2$ at a value in the interval, say $x = 2$. $y_1(2)=e^{-0.4\times2}=e^{-0.8}\approx0.4493$ and $y_2(2)=2^{2}+2 = 6$. So, on the interval $[2,4]$, $y=x^{2}+2$ is the upper - function and $y = e^{-0.4x}$ is the lower - function.

Step2: Use the area formula

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, $f(x)=x^{2}+2$, $g(x)=e^{-0.4x}$, $a = 2$, and $b = 4$. So, $A=\int_{2}^{4}((x^{2}+2)-e^{-0.4x})dx=\int_{2}^{4}(x^{2}+2)dx-\int_{2}^{4}e^{-0.4x}dx$.

Step3: Integrate $x^{2}+2$

We know that $\int(x^{2}+2)dx=\frac{1}{3}x^{3}+2x+C$. Then $\int_{2}^{4}(x^{2}+2)dx=\left[\frac{1}{3}x^{3}+2x\right]_{2}^{4}=\left(\frac{1}{3}\times4^{3}+2\times4\right)-\left(\frac{1}{3}\times2^{3}+2\times2\right)=\left(\frac{64}{3}+8\right)-\left(\frac{8}{3}+4\right)=\frac{64 + 24}{3}-\frac{8 + 12}{3}=\frac{88}{3}-\frac{20}{3}=22\frac{2}{3}$.

Step4: Integrate $e^{-0.4x}$

Let $u=-0.4x$, then $du=-0.4dx$. When $x = 2$, $u=-0.8$; when $x = 4$, $u=-1.6$. $\int e^{-0.4x}dx=-\frac{1}{0.4}e^{-0.4x}+C = - 2.5e^{-0.4x}+C$. Then $\int_{2}^{4}e^{-0.4x}dx=-2.5\left[e^{-0.4x}\right]_{2}^{4}=-2.5(e^{-1.6}-e^{-0.8})=-2.5\left(\frac{1}{e^{1.6}}-\frac{1}{e^{0.8}}\right)\approx - 2.5(0.2019 - 0.4493)=-2.5\times(-0.2474)=0.6185$.

Step5: Calculate the area

$A = 22\frac{2}{3}-0.6185=\frac{68}{3}-0.6185\approx22.67 - 0.62=22.05$.

Answer:

$22.05$