Find the Area Between the Curves x=-1 , x=2 , y=3e^(3x) , y=2e^(3x)+1

Problem

(∫_−1^2)((3*e(3*x)−(2*e(3*x)+1))*d(x))

Solution

1. Identify the upper and lower functions by comparing (y_1)=3*e(3*x) and (y_2)=2*e(3*x)+1 on the interval [−1,2]

2. Set up the integral for the area A using the formula A=(∫_a^b)((ƒ(x)−g(x))*d(x))

3. Simplify the integrand by subtracting the functions.

3*e(3*x)−(2*e(3*x)+1)=e(3*x)−1

4. Integrate the simplified expression with respect to x

(∫_^)((e(3*x)−1)*d(x))=(e(3*x))/3−x

5. Evaluate the definite integral at the boundaries x=2 and x=−1

A=[(e(3*(2)))/3−2]−[(e(3*(−1)))/3−(−1)]

6. Simplify the numerical result to find the final area.

A=(e6)/3−2−(e(−3))/3−1

A=(e6−e(−3))/3−3

Final Answer

(∫_−1^2)((3*e(3*x)−(2*e(3*x)+1))*d(x))=(e6−e(−3)−9)/3

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