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 $y_1 = e^{-x}$ and $y_2=3 + x$. For $x\in[2,4]$, we know that $y_2=3 + x$ is greater than $y_1 = e^{-x}$ since $e^{-x}=\frac{1}{e^{x}}$ and for $x = 2,e^{-2}=\frac{1}{e^{2}}\approx0.14$ and $3 + 2=5$; for $x = 4,e^{-4}=\frac{1}{e^{4}}\approx0.02$ and $3+4 = 7$. The formula for the area $A$ between two curves $y = f(x)$ and $y = g(x)$ from $x=a$ to $x = b$ is $A=\int_{a}^{b}(f(x)-g(x))dx$ where $f(x)\geq g(x)$ on $[a,b]$. Here, $f(x)=3 + x$ and $g(x)=e^{-x}$, $a = 2$ and $b = 4$.

Step2: Set up the integral

The area $A$ is given by the integral $A=\int_{2}^{4}((3 + x)-e^{-x})dx$. We know that $\int(3 + x - e^{-x})dx=\int3dx+\int xdx-\int e^{-x}dx$. Using the integral rules $\int kdx=kx + C$ ($k$ is a constant), $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$) and $\int e^{-x}dx=-e^{-x}+C$, we have: $\int(3 + x - e^{-x})dx=3x+\frac{x^{2}}{2}+e^{-x}+C$.

Step3: Evaluate the definite integral

$A=\left[3x+\frac{x^{2}}{2}+e^{-x}\right]_{2}^{4}$ $=\left(3\times4+\frac{4^{2}}{2}+e^{-4}\right)-\left(3\times2+\frac{2^{2}}{2}+e^{-2}\right)$ $=(12 + 8+e^{-4})-(6 + 2+e^{-2})$ $=(20+e^{-4})-(8 + e^{-2})$ $=20+e^{-4}-8 - e^{-2}$ $=12+e^{-4}-e^{-2}$ $=12+\frac{1}{e^{4}}-\frac{1}{e^{2}}$ $\approx12 + 0.02 - 0.14$ $=11.88$

Answer:

$12+\frac{1}{e^{4}}-\frac{1}{e^{2}}\approx11.88$