Problem 364 - Calculus Volume 1

Question

In the following exercises, integrate using the indicated substitution. \(\int e^{2 x} \sqrt{1-e^{2 x}} d x ; u=e^{2 x}\)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{3}(1-e^{2x})^{3/2} + C\).

Step 1: Identify the Substitution

We are asked to use the substitution \(u = e^{2x}\). This substitution is indicated to simplify the integral. We will differentiate \(u = e^{2x}\) to express \(dx\) in terms of \(du\).

Step 2: Differentiate the Substitution

Differentiate both sides of \(u = e^{2x}\) with respect to \(x\) to find \(du\). We obtain: \(du = 2e^{2x}dx\). This implies that \(dx = \frac{du}{2e^{2x}}\).

Step 3: Rewrite the Integral

Substitute \(u = e^{2x}\) and \(dx = \frac{du}{2u}\) into the original integral. The integral \(\int e^{2x} \sqrt{1 - e^{2x}} \, dx\) becomes \(\int u \sqrt{1 - u} \cdot \frac{du}{2u}\).

Step 4: Simplify the Integral

Cancel \(u\) in the numerator and denominator to simplify: \(\int \sqrt{1-u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int \sqrt{1-u} \, du\).

Step 5: Integrate with Respect to u

We now integrate \(\frac{1}{2} \int \sqrt{1-u} \, du\). Let \(v = 1 - u\), then \(dv = -du\). This transforms the integral into \(-\frac{1}{2} \int \sqrt{v} \, dv\).

Step 6: Compute the Integral

Integrate \(\int \sqrt{v} \, dv = \int v^{1/2} \, dv = \frac{2}{3}v^{3/2} + C\). Therefore, \(-\frac{1}{2} \cdot \frac{2}{3}v^{3/2} = -\frac{1}{3}v^{3/2} + C\).

Step 7: Back Substitution

Replace \(v\) with \(1 - u\), which results in the expression \(-\frac{1}{3}(1-u)^{3/2} + C\).

Step 8: Substitute Back in terms of x

Since \(u = e^{2x}\), replace \(u\) back to get the final expression \(-\frac{1}{3}(1-e^{2x})^{3/2} + C\) in terms of \(x\).

Key Concepts

Substitution Method

The substitution method is a common technique used in integration, often simplifying complex expressions into more manageable forms. In this approach, we replace a complicated expression in the integral with a single variable, making it easier to evaluate. For instance, consider the integral \( \int e^{2x} \sqrt{1 - e^{2x}} \, dx \), where the substitution \( u = e^{2x} \) is suggested. This helps transform the integral to one involving \( u \), a simpler variable. The idea is to express \( dx \) in terms of \( du \), allowing the transformation of the whole integral.

• This involves differentiating the substitution, \( u = e^{2x} \), to find \( du \).
• As we differentiate, we get \( du = 2e^{2x}dx \), leading to \( dx = \frac{du}{2e^{2x}} \).
• The substitution simplifies the original integral by considerably reducing its complexity.

This method is useful, especially when dealing with chain rule derivatives or complicated polynomial expressions inside integrals.

Definite Integral

A definite integral computes the total accumulation of a quantity, often represented as the area under a curve between two bounds. While our problem didn't specify limits of integration, understanding definite integrals is crucial to tackle similar problems.

• The definite integral is represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits.
• Unlike indefinite integrals, definite integrals result in a numerical value, not an algebraic expression. This is because they calculate the net area.
• Using the substitution method within definite integrals requires careful modification of the limits according to the substitution used.

Grasping definite integrals ensures you can apply both conceptual and procedural understanding effectively during problem-solving in calculus.

Calculus Problems

Calculus problems often require a mix of strategic thinking and mathematical techniques to solve. These problems can range from finding the derivative or integral of functions to tackling complex applications in real-world contexts.

• Calculus splits into differential calculus, focusing on the rate of change, and integral calculus, centering on accumulation of quantities.
• Effective problem-solving in calculus involves understanding the relationships between functions, derivatives, and integrals.
• Integration techniques like substitution simplify solving integrals, making these problems more approachable.

For students, developing a strong foundational understanding of the core principles is key to tackling any calculus problems confidently.

Change of Variables

The change of variables is a powerful technique in calculus that involves substituting one variable for another to make integration easier. It's similar to substitution but focuses on changing not just the variable, but also changing the bounds if dealing with definite integrals.

• This helps in simplifying the relationship, transforming it to a basic form that's easier to manage.
• In problems like ours, the initial change is \( u = e^{2x} \), which subsequently leads us into easier integrals to solve.
• Changing variables requires you to also change expressions involving the limits or bounds if they exist.

Using a change of variables streamlines complex integrals, paving the way for effective problem-solving in more challenging calculus scenarios.