find the area between the curves ( y = e^{-0.3x} ) and ( y = 2.3x + 1 ) from ( x = 0 ) to ( x = 4 ).\narea =

Explanation:

Step1: Set up the integral for the area

The formula for the area 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). Here, (f(x)=e^{- 0.3x}), (g(x)=2.3x + 1), (a = 0), and (b = 4). Since (e^{-0.3x}\leq2.3x + 1) for (x\in[0,4]) (by checking the values at (x = 0): (e^{0}=1) and (2.3\times0+1=1); and as (x) increases, the linear function (y = 2.3x+1) grows while (y=e^{-0.3x}) decays), the area (A=\int_{0}^{4}[(2.3x + 1)-e^{-0.3x}]dx).

Step2: Integrate term - by - term

Use the integral rules (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), (\int e^{kx}dx=\frac{e^{kx}}{k}+C). (\int_{0}^{4}(2.3x + 1)dx-\int_{0}^{4}e^{-0.3x}dx) First integral: (\int_{0}^{4}(2.3x + 1)dx=\left[2.3\times\frac{x^{2}}{2}+x\right]{0}^{4}) [ \begin{align*} 2.3\times\frac{4^{2}}{2}+4-(2.3\times\frac{0^{2}}{2}+0)&=2.3\times8 + 4\ &=18.4+4\ &=22.4 \end{align*} ] Second integral: (\int{0}^{4}e^{-0.3x}dx=\left[\frac{e^{-0.3x}}{-0.3}\right]_{0}^{4}) [ \begin{align*} \frac{e^{-0.3\times4}}{-0.3}-\frac{e^{-0.3\times0}}{-0.3}&=\frac{1}{0.3}(1 - e^{-1.2})\ &\approx\frac{1}{0.3}(1-0.3012)\ &\approx\frac{0.6988}{0.3}\ &\approx2.3293 \end{align*} ]

Step3: Calculate the area

(A = 22.4-2.3293=20.0707)

Answer:

(20.07)