Questions: ∫₀¹ e^(-x) · (4-e^x) dx =

Transcript text: $\int_{0}^{1} e^{-x} \cdot\left(4-e^{x}\right) d x=$

Solution

Solution Steps

To solve the integral $\int_{0}^{1} e^{-x} \cdot\left(4-e^{x}\right) dx$, we can split it into two separate integrals and solve each one individually. Specifically, we can rewrite it as $\int_{0}^{1} 4e^{-x} dx - \int_{0}^{1} e^{-x} \cdot e^{x} dx$. The second integral simplifies to $\int_{0}^{1} 1 dx$.

Step 1: Split the Integral

We start by splitting the integral $\int_{0}^{1} e^{-x} \cdot (4 - e^{x}) \, dx$ into two separate integrals: \[ \int_{0}^{1} 4e^{-x} \, dx - \int_{0}^{1} e^{-x} \cdot e^{x} \, dx \]

Step 2: Simplify the Second Integral

The second integral simplifies to: \[ \int_{0}^{1} 1 \, dx \]

Step 3: Evaluate Each Integral

Evaluate the first integral: \[ \int_{0}^{1} 4e^{-x} \, dx = 4 \left[ -e^{-x} \right]_{0}^{1} = 4 \left( -e^{-1} + 1 \right) = 4 - 4e^{-1} \]

Evaluate the second integral: \[ \int_{0}^{1} 1 \, dx = \left[ x \right]_{0}^{1} = 1 - 0 = 1 \]

Step 4: Combine the Results

Combine the results of the two integrals: \[ 4 - 4e^{-1} - 1 = 3 - 4e^{-1} \]