Problem 35 - Calculus: Early Transcendental Functions

Question

Evaluate the definite integral. $$\int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x$$

Step-by-Step Solution

Answer

The definite integral from 0 to 2 of the given function is equal to: \( \arctan(e^2) - \frac{\pi}{4} \).

Step 1: Simplify the integrand

Firstly, let's set \( u = e^x \). By taking derivative both sides we have: \( du = e^x dx \). Now, we can rewrite our integral in terms of u, which is: \(\int_{1}^{e^2} \frac{u}{1+u^2} du\) where we have also changed our limits of integration by substituting x for u.

Step 2: Perform the integration

Looking at the simplified integral, it should be clear that it's a standard integral and recognised as the derivative of the inverse tangent function. Thus, the integral simplifies to \( \arctan(u)\). So, the integral from 1 to \( e^2 \) of our simplified function is: \( \arctan(e^2) - \arctan(1) \).

Step 3: Substituting the values

\(\arctan(1)\) is \( \frac{\pi}{4} \). Therefore, we have: \( \arctan(e^2) - \frac{\pi}{4} \). Now as \( e^2 > 1 \), \( \arctan(e^2) \) is between \( \frac{\pi}{2} \) and \( \frac{\pi}{4} \).

Key Concepts

Substitution Method

The substitution method is a common technique used in calculus, especially when working with integrals. It's like a tool to help us simplify tricky problems. Think of it as changing the characters in a story to make the plot easier to understand. Here's how it works:

• Identify a portion of the integral that can be substituted with another variable, called 'u'. This helps transform the integral into a simpler form.
• Next, compute the derivative of the substituted variable, 'du', to replace 'dx' in the original function.
• Don't forget to change the limits of integration when you substitute, based on your new variable 'u'.

In our exercise, we used the substitution method by setting \( u = e^x \), which turned the integral into a form easy to integrate: \( \int \frac{u}{1+u^2} du \). This step is crucial for simplifying and solving complex integrals, making them more manageable.

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arctan(x) \), play a significant role in calculus, especially in integration. They "undo" what the original trig functions do.In our example:

• The function \( \arctan(u) \) is the inverse of the tangent function and is used to evaluate the integral \( \int \frac{1}{1+u^2} du \).
• This is a standard result that comes up frequently, known by many calculus students, where the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \).

When we use \( \arctan(u) \), we're effectively 'reversing' part of the problem while still keeping it grounded in trigonometric reasoning. In our integral, after substitution had given us \( \frac{u}{1+u^2} \), realising that this was reminiscent of the derivative of an inverse trig function led to our solution.

Limits of Integration

The limits of integration dictate where the integration process starts and stops on the x-axis. When switching variables, as in substitution, these limits must be adjusted too. This maintains the relationship within the new context of the substituted variable.Here's how it affected our example:

• Originally, the integral's limits were from 0 to 2 for \( x \).
• Once \( u = e^x \) was introduced, the limits changed to reflect this new variable: \( u \) went from 1 to \( e^2 \).

By reassessing these bounds, we ensure that the area calculation remains consistent, even after transformation.This careful adjustment keeps the integrity of the integral aligning with its initial setup, allowing for accurate area measurement.