Problem 76 - Calculus Single Variable

Question

Calculate the integrals. $$ \int \frac{24 \exp (2 x)}{\sqrt{1-\exp (x)}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-16 (1 - \exp(x))^{3/2} + \frac{48}{5} (1 - \exp(x))^{5/2} + C\).

Step 1: Variable Substitution

Let's use a substitution to simplify the integral. Set \( u = 1 - \exp(x) \). Then, differentiate both sides with respect to \( x \), giving us \( du = -\exp(x) \, dx \) or \( dx = -\frac{du}{\exp(x)} \). Note that \( \exp(x) = 1 - u \). Substitute these into the integral.

Step 2: Substitute into Integral

Substitute \( u = 1 - \exp(x) \) and \( dx = -\frac{du}{\exp(x)}\) into the integral: \( \int \frac{24 \exp(2x)}{\sqrt{1-\exp(x)}} dx = \int \frac{24 (1-u)^2}{\sqrt{u}} \left( -\frac{du}{1-u} \right) \). This simplifies to \( -24 \int (1-u) \sqrt{u} \, du \).

Step 3: Simplify the Integrand

Distribute \( (1-u) \) in the integrand: \( -24 \int (\sqrt{u} - u\sqrt{u}) \, du \). This becomes \( -24 \left( \int u^{1/2} \, du - \int u^{3/2} \, du \right) \).

Step 4: Integrate Each Term

Integrate \( \int u^{1/2} \, du \) and \( \int u^{3/2} \, du \) separately: \( \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C_1 \) and \( \int u^{3/2} \, du = \frac{2}{5} u^{5/2} + C_2 \). Combine these to get: \( -24 \left( \frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2} \right) + C \).

Step 5: Simplify the Result

Distribute \( -24 \) across the terms: \( -24 \times \frac{2}{3} u^{3/2} + 24 \times \frac{2}{5} u^{5/2} + C \). This gives \( -16 u^{3/2} + \frac{48}{5} u^{5/2} + C \).

Step 6: Substitute Back

Substitute back \( u = 1 - \exp(x) \) into the expression: \( -16 (1 - \exp(x))^{3/2} + \frac{48}{5} (1 - \exp(x))^{5/2} + C \). This is the final answer.

Key Concepts

Variable Substitution

Variable substitution is a powerful technique to simplify complex integrals. By introducing a new variable, we can transform an integral into a more understandable form.
In our exercise, we set \( u = 1 - \exp(x) \). This new variable helps us rewrite expressions involving \( \exp(x) \) more conveniently. This process also requires us to convert the differential \( dx \) into \( du \), which involves differentiating \( u \) with respect to \( x \).

• The substitution yields \( du = -\exp(x) \, dx \).
• Replacing \( dx \) gives \( dx = -\frac{du}{\exp(x)} \).

With these transformations, the integral becomes much easier to handle. This method is particularly helpful when dealing with expressions that involve composite functions that are complex to integrate in their original form.

Exponential Functions

Exponential functions are fundamental in calculus, and they're often characterized by their base, usually the constant \( e \), the base of the natural logarithms. In this particular problem, \( \exp(x) \) is the function of interest.
The key property of exponential functions is that the rate of change is proportional to the function itself. This means that the derivative and the integral of an exponential function remain related to the function in a simple way. In our integral:

• We express \( \exp(x) \) using the newly defined variable \( u = 1 - \exp(x) \), making it more manageable.
• Substituting \( \exp(x) = 1-u \) further simplifies calculations.

This approach highlights the versatility and simplicity of exponential functions, especially within integrals where they frequently appear.

Definite Integral

A definite integral (

a to b
) is the calculation of the area under a curve between two limits, \( a \) and \( b \).
Even though our example involves an indefinite integral, this concept is critical in understanding broader integration applications. In the broader context, definite integrals:

• Provide numerical values, giving precise areas.
• Are computed using limits with the fundamental theorem of calculus that links derivatives and integrals.

By understanding variable substitutions and exponential functions, you prepare yourself for dealing with definite integrals by knowing how to handle parts of more complex integrands often encountered in these types of integrals.

Integration by Parts

Integration by parts is another crucial integration technique that often appears in calculus problems where standard methods, like substitution, don't work smoothly.
This technique is based on the product rule for differentiation and allows us to transform the integral of a product of functions into an easier form:

• The formula is \( \int u \, dv = uv - \int v \, du \).
• Choosing \( u \, \text{and}\, dv \) appropriately can simplify tough integrals.

Although not directly applied in this exercise, understanding integration by parts expands your toolkit, enabling you to solve a broader range of problems. It's most effective when dealing with polynomials multiplied by logarithmic or trigonometric functions.