Question

In the following exercises, solve for the antiderivative \(\int f\) of \(f\) with \(C=0\), then use a calculator to graph \(f\) and the antiderivative over the given interval \([a, b] .\) Identify a value of \(C\) such that adding \(C\) to the antiderivative recovers the definite integral \(F(x)=\int_{a}^{x} f(t) d t\). [T] \(\int \frac{e^{x}}{1+e^{2 x}} d x\) over \([-6,6]\)

Short answer

The antiderivative is \( \tan^{-1}(e^x) \) with \( C=0 \).

Step-by-step solution

Understand the Problem

We need to find the antiderivative of the function \( f(x) = \frac{e^x}{1+e^{2x}} \). This involves finding a function \( F(x) \) such that \( F'(x) = f(x) \) with an initial condition of \( C = 0 \). We will then find a constant \( C \) that, when added to \( F(x) \), results in the definite integral \( \int_{a}^{x} f(t) \, dt \) over the interval \([-6, 6]\).

Find the Antiderivative

To find the antiderivative \( F(x) \), observe that \( f(x) = \frac{e^x}{1+e^{2x}} \) can be rewritten such that it simplifies during integration. Let \( u = e^x \), then \( du = e^x \, dx \), changing our integral to:\[\int \frac{du}{1+u^2}\]This integral is the arctangent function, so \( F(x) = \tan^{-1}(e^x) + C \).

Evaluate the Definite Integral

The definite integral \( F(x) = \int_{-6}^{x} \frac{e^t}{1+e^{2t}} \, dt \) is evaluated using the antiderivative:\[F(x) = \tan^{-1}(e^x) - \tan^{-1}(e^{-6})\]Here, subtraction occurs because we want the integral from \(-6\) to \(x\). \( \tan^{-1}(e^{-6}) \) is a constant term that represents \( C \) plus any section of the curve evaluated at the lower bound \(-6\).

Identify the Correct Constant

The constant \( C \) that will allow us to use the indefinite integral form as a definite integral with bounds \([-6, 6]\) is given by solving:\[F(x) = \tan^{-1}(e^x) - \tan^{-1}(e^{-6})\]It confirms \( C = 0 \) effectively represents the initial condition in the context of total area from \(-6\) to \(x\).

Graph the Functions

Graph \( f(x) = \frac{e^x}{1+e^{2x}} \) and \( F(x) = \tan^{-1}(e^x) \) over the interval \([-6, 6]\). These graphs illustrate the function \( f(x) \) and how its area under the curve accumulates according to \( F(x) \). This final check shows the definite integral as a proper accumulation function of the area described by the antiderivative.

Key concepts

Definite Integrals

A definite integral gives us the area under the curve of a function on a specific interval. Unlike an indefinite integral, which results in a family of functions, a definite integral evaluates this result over a particular section. For example, in the case of \([a, b]\), it is defined as \( \int_{a}^{b} f(t) \, dt \).

To compute a definite integral, you evaluate the antiderivative at the upper limit and subtract the antiderivative evaluated at the lower limit:

• Calculate the antiderivative, \( F(x) \).
• Find \( F(b) - F(a) \).

This provides the net area between the curve of the function and the x-axis from \( a \) to \( b \). In our exercise, the interval \([-6, 6]\) ensures that the antiderivative computes the change over the specified range, depicting both positive and negative areas that the curve includes.

Arctangent Function

The arctangent function, written as \( \tan^{-1}(x) \), is a fundamental function in calculus, particularly when dealing with integrals involving the form \( \frac{1}{1+u^2} \). This kind of expression often appears when you simplify certain trigonometric integrals.

In our exercise's context, when we performed substitution with \( u = e^x \), we were left integrating \( \frac{du}{1+u^2} \), which directly results in \( \tan^{-1}(u) + C \).

• This reflects the process where we effectively reverse the trigonometric differentiation associated with the tangent function.
• Understanding the link between inverse trigonometric functions and integrals can simplify managing more complex expressions, as seen with transformations toward \( \tan^{-1} \) functions.

Substitution Method

The substitution method, also known as u-substitution, is one of the most essential techniques in integration. It simplifies integrals by changing variables, thus making the function easier to integrate. Here's how you apply this method:

• Select a substitution \( u = g(x) \) such that its derivative \( du \) replaces a part of the original integral.
• Rewrite the integral in terms of \( u \) and \( du \) rather than \( x \) and \( dx \).
• Finish the integration process, and then substitute back \( x \) for \( u \) in your final answer.

In our problem, setting \( u = e^x \) and thus \( du = e^x dx \) converted the issue into a more manageable form with \( \int \frac{du}{1+u^2} \), and ultimately leading us to an arctangent integral.
Recognizing when and how to properly substitute helps solve challenging integral problems smoothly and efficiently.

Integration Techniques

Integration techniques are the diverse methods used to find antiderivatives of functions. These techniques facilitate the process of calculating integrals that might not straightforwardly yield to basic rules. Here are some of the main techniques used:

• Basic Antiderivatives: Understanding the basic derivatives and their antiderivatives, like power rule \( \int x^n = \frac{x^{n+1}}{n+1} + C \), is foundational.
• Substitution: As shown with \( u = e^x \), it's vital for complex expressions.
• Integration by Parts: Often used when a product of functions exists, relying on \( \int u \, dv = uv - \int v \, du \).
• Partial Fractions: Decomposing a rational function into simpler fractions makes some integrals solvable.

Particularly focusing on substitution within our example, the technique permitted transforming a complex exponential integral into a simpler trigonometric one. Learning to choose the suitable technique determines the ease with which you solve integral problems.