find the area between the curves.\nx = - 1, x = 2, y = 4e^{4x}, y = 3e^{4x}+1\n\nthe area between the curves…

Explanation:

Step1: Determine the upper - lower curve

The upper curve is $y_1 = 3e^{4x}+1$ and the lower curve is $y_2 = 4e^{4x}$. The area $A$ between two curves $y = y_1(x)$ and $y = y_2(x)$ from $x=a$ to $x = b$ is given by $A=\int_{a}^{b}(y_1 - y_2)dx$. Here, $a=-1$, $b = 2$, $y_1=3e^{4x}+1$ and $y_2 = 4e^{4x}$, so $y_1 - y_2=(3e^{4x}+1)-4e^{4x}=1 - e^{4x}$.

Step2: Calculate the definite integral

We need to calculate $\int_{-1}^{2}(1 - e^{4x})dx$. Using the integral rules $\int 1dx=x+C$ and $\int e^{4x}dx=\frac{1}{4}e^{4x}+C$. [ \begin{align*} \int_{-1}^{2}(1 - e^{4x})dx&=\left[x-\frac{1}{4}e^{4x}\right]_{-1}^{2}\ &=(2-\frac{1}{4}e^{4\times2})-(-1-\frac{1}{4}e^{4\times(-1)})\ &=2-\frac{1}{4}e^{8}+1+\frac{1}{4}e^{-4}\ &=3-\frac{1}{4}e^{8}+\frac{1}{4e^{4}} \end{align*} ]

Step3: Approximate the result

[ \begin{align*} 3-\frac{1}{4}e^{8}+\frac{1}{4e^{4}}&\approx3-\frac{1}{4}\times2980.957987+\frac{1}{4\times54.59815}\ &=3 - 745.239497+0.00458\ &\approx - 742.234917\approx742.23 \end{align*} ]

Answer:

$742.23$