Find the points of intersection of the graphs of the functions 𝑦=𝑒𝑥
and 𝑦=𝑒−3𝑥
for 𝑥
between 𝑥=−2
and 𝑥=1
(enter your answer as a comma separated list)
Then find the area bounded by the curves 𝑦=𝑒𝑥
and 𝑦=𝑒−3𝑥
for 𝑥
between 𝑥=−2
and 𝑥=1

To solve the problem, we will first find the points of intersection of the graphs of the functions \(y = e^x\) and \(y = e^{-3x}\) in the interval \(x \in [-2, 1]\). Then, we will find the area bounded by these curves in this interval.

### 1. Finding Points of Intersection

The points of intersection occur where the two functions are equal:

\[
e^x = e^{-3x}
\]

To solve this equation, let's rewrite it:

\[
e^x = \frac{1}{e^{3x}}
\]

This implies:

\[
e^x \cdot e^{3x} = 1
\]

\[
e^{4x} = 1
\]

Since the exponential function is \(1\) when its exponent is \(0\):

\[
4x = 0 \implies x = 0
\]

Now, let's find the corresponding \(y\) value at \(x = 0\):

\[
y = e^0 = 1
\]

So, the point of intersection is \((0, 1)\).

### 2. Area Bounded by the Curves

Next, we need to find the area between the curves \(y = e^x\) and \(y = e^{-3x}\) over the interval \(x \in [-2, 1]\).

To find this area, we compute the integral of the absolute difference of the functions:

\[
\text{Area} = \int_{-2}^{1} |e^x - e^{-3x}| \, dx
\]

Since \(e^x > e^{-3x}\) in the interval \([-2, 0)\) and \(e^{-3x} > e^x\) in the interval \((0, 1]\), we can split the integral:

\[
\text{Area} = \int_{-2}^{0} (e^x - e^{-3x}) \, dx + \int_{0}^{1} (e^{-3x} - e^x) \, dx
\]

Let's compute each integral separately.

#### Integral from \(-2\) to \(0\):

\[
\int_{-2}^{0} (e^x - e^{-3x}) \, dx
\]

The integral of \(e^x\) is \(e^x\), and the integral of \(e^{-3x}\) is \(-\frac{1}{3} e^{-3x}\):

\[
\left[ e^x + \frac{1}{3}e^{-3x} \right]_{-2}^{0}
\]

Evaluating this:

\[
\left( e^0 + \frac{1}{3}e^{0} \right) - \left( e^{-2} + \frac{1}{3}e^{6} \right)
\]

\[
= (1 + \frac{1}{3}) - \left( e^{-2} + \frac{1}{3}e^{6} \right)
\]

\[
= \frac{4}{3} - \left( e^{-2} + \frac{1}{3}e^{6} \right)
\]

#### Integral from \(0\) to \(1\):

\[
\int_{0}^{1} (e^{-3x} - e^x) \, dx
\]

Integrating, we get:

\[
\left[ -\frac{1}{3} e^{-3x} - e^x \right]_{0}^{1}
\]

Evaluating this:

\[
\left( -\frac{1}{3} e^{-3 \cdot 1} - e^1 \right) - \left( -\frac{1}{3} e^{0} - e^0 \right)
\]

\[
= \left( -\frac{1}{3} e^{-3} - e \right) - \left( -\frac{1}{3} - 1 \right)
\]

\[
= \left( -\frac{1}{3e^3} - e \right) - \left( -\frac{1}{3} - 1 \right)
\]

Simplifying:

\[
= -\frac{1}{3e^3} - e + \frac{1}{3} + 1
\]

### 3. Putting it All Together

Now we combine the two integrals:

\[
\text{Area} = \left( \frac{4}{3} - e^{-2} - \frac{1}{3}e^{6} \right) + \left( -\frac{1}{3e^3} - e + \frac{1}{3} + 1 \right)
\]

Let's simplify this expression:

\[
\text{Area} = \frac{4}{3} - e^{-2} - \frac{1}{3}e^{6} - \frac{1}{3e^3} - e + \frac{1}{3} + 1
\]

### Final Answer:

- **Intersection Point**: \((0, 1)\)
- **Area Bounded by the Curves**:

\[
\text{Area} = \frac{5}{3} - e^{-2} - \frac{1}{3}e^{6} - \frac{1}{3e^3} - e
\]

### Do you want the exact numerical value for the area?

Feel free to ask if you have any questions or need further details!

#### Related Questions:

1. How do you find the area between two curves in general?
2. What is the significance of finding the intersection points between two curves?
3. How can you graph the functions \(y = e^x\) and \(y = e^{-3x}\) to visualize their intersection?
4. How does changing the interval of \(x\) affect the bounded area?
5. What is the importance of absolute value in the area calculation between curves?

#### Tip:

To ensure accurate calculations, especially when integrating, break down the problem into simpler parts and verify each step.

Learn how to find the points of intersection and calculate the area bounded by the exponential functions y = e^x and y = e^{-3x} over the interval [-2, 1]. This problem involves advanced concepts in calculus, including integration and exponential functions.

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