Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{0}^{2} \frac{e^{x} d x}{3+e^{x}}$$

Short answer

The answer is \( ln(3 + e^{2}) - ln(4) \).

Step-by-step solution

Identify the substitution function

Let us define the substitution \( u = 3 + e^x \).It's derivative, \( du \), can be derived from the standard exponential function rule: \( du = e^x dx \).

Change the limits of integration

The integral is definite and goes from \( 0 \) to \( 2 \). These limits have to be adapted to the new variable \( u \). The limits for \( u \) would be \( u(0) = 3 + e^0 = 4 \) and \( u(2) = 3 + e^2 \). So the new limits of integration are from \( 4 \) to \( 3 + e^2 \).

Rewrite the integral

Replace \( e^x dx \) with \( du \) and \( 3 + e^x \) with \( u \) in the integral. Therefore, the integral becomes \( \int_{4}^{3 + e^{2}} \frac{du}{u} \).

Solve the integral

The integral \( \int_{4}^{3 + e^{2}} \frac{du}{u} \) represents the logarithm to the base \( e \). This makes the integral \( ln(u) \) evaluated from \( 4 \) to \( 3 + e^{2} \).

Evaluate the definite integral

Substitute the limits of integration into the result from the previous step to get the result: \( ln(3 + e^{2}) - ln(4) \).