To evaluate the definite integral of the function $\frac{e^{x+1}}{e^x}$ from 0 to 1, let's first simplify the integrand:

$\frac{e^{x+1}}{e^x} = e^{x+1 - x} = e^1 = e$

The integrand simplifies to a constant $e$. Now, we can compute the definite integral:

$\int_0^1 e \, dx$

Since $e$ is a constant, we can take it outside the integral:

$e \int_0^1 dx$

The integral of $dx$ from 0 to 1 is:

$\int_0^1 dx = [x]_0^1 = 1 - 0 = 1$

So, the definite integral becomes:

$e \times 1 = e$

Answer:

$\int_0^1 \frac{e^{x+1}}{e^x} \, dx = e$

Would you like more details or have any questions?

Related Questions:

1. What is the integral of $e^x$ from 0 to 1?
2. How would you evaluate the integral of $e^{2x}$ from 0 to 1?
3. What is the integral of $\frac{e^x}{x}$ from 1 to $e$?
4. How do you compute the integral of a constant?
5. How to evaluate definite integrals with different limits?

Tip:

When integrating exponential functions, always look for ways to simplify the exponent or apply properties of exponents to make the integration straightforward.