$$ \int _ { 0 } ^ { 1 } \frac { 2 e ^ { 8 x } - 2 } { e ^ { 4 x } } d x = \square $$

Question

UpStudy AI · Accepted Answer

To evaluate the integral:

$$
\int_{0}^{1} \frac{2e^{8x} - 2}{e^{4x}} \, dx
$$

Let's break it down step by step.

### Step 1: Simplify the Integrand

First, simplify the expression inside the integral:

$$
\frac{2e^{8x} - 2}{e^{4x}} = 2e^{8x} \cdot e^{-4x} - 2e^{-4x} = 2e^{4x} - 2e^{-4x}
$$

So, the integral becomes:

$$
\int_{0}^{1} \left(2e^{4x} - 2e^{-4x}\right) \, dx
$$

### Step 2: Split the Integral

Separate the integral into two parts:

$$
2\int_{0}^{1} e^{4x} \, dx - 2\int_{0}^{1} e^{-4x} \, dx
$$

### Step 3: Integrate Each Part

Recall that:

$$
\int e^{kx} \, dx = \frac{1}{k} e^{kx} + C
$$

Applying this:

1. **First Integral:**
$$
2\int e^{4x} \, dx = 2 \cdot \frac{1}{4} e^{4x} = \frac{1}{2} e^{4x}
$$

2. **Second Integral:**
$$
-2\int e^{-4x} \, dx = -2 \cdot \left( \frac{1}{-4} e^{-4x} \right) = \frac{1}{2} e^{-4x}
$$

### Step 4: Combine and Evaluate from 0 to 1

Combine the results:

$$
\frac{1}{2} e^{4x} + \frac{1}{2} e^{-4x} \Big|_{0}^{1}
$$

Evaluate at the bounds:

$$
\left( \frac{1}{2} e^{4(1)} + \frac{1}{2} e^{-4(1)} \right) - \left( \frac{1}{2} e^{0} + \frac{1}{2} e^{0} \right)
$$

Simplify:

$$
\left( \frac{e^{4} + e^{-4}}{2} \right) - \left( \frac{1 + 1}{2} \right) = \frac{e^{4} + e^{-4}}{2} - 1
$$

### Final Answer

$$
\int_{0}^{1} \frac{2e^{8x} - 2}{e^{4x}} \, dx = \frac{e^{4} + e^{-4}}{2} - 1
$$

Or, simplifying further:

$$
\frac{e^{4} + e^{-4} - 2}{2}
$$
$$ \int_{0}^{1} \frac{2e^{8x} - 2}{e^{4x}} \, dx = \frac{e^{4} + e^{-4} - 2}{2} $$

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