Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the method of substitution to find each of the following indefinite integrals. $$ \int x^{2}\left(x^{3}+5\right)^{8} \exp \left[\left(x^{3}+5\right)^{9}\right] d x $$

Step-by-Step Solution

Answer

\( \frac{1}{27} \exp((x^{3}+5)^9) + C \)

Step 1: Identify the inner function for substitution

Let's start by identifying the inner function that would be useful for substitution. In the given integral \( \int x^{2}(x^{3}+5)^{8} \exp \left[ (x^{3}+5)^{9} \right] \, dx \), the inner function here is \( u = x^{3} + 5 \).

Step 2: Differentiate the inner function

Compute the derivative of \( u = x^{3} + 5 \). The derivative is \( \frac{du}{dx} = 3x^{2} \). This can be rearranged as \( du = 3x^{2} \, dx \) or \( x^{2} \, dx = \frac{1}{3} \, du \).

Step 3: Substitute and change variables

Replace \( x^{2} \, dx \) in the integral with \( \frac{1}{3} \, du \) and \( (x^{3} + 5) \) with \( u \). This transforms the integral into \( \int (u^{8}) \cdot \exp(u^{9}) \, \frac{1}{3} \, du \), which simplifies to \( \frac{1}{3} \int u^{8} \cdot \exp(u^{9}) \, du \).

Step 4: Integrate using substitution

Consider a second variable substitution for the exponential part: let \( v = u^{9} \). Then \( \frac{dv}{du} = 9u^{8} \), so \( du = \frac{1}{9u^{8}} \, dv \). Substitute these into the integral: \( \frac{1}{3} \int 1 \cdot \exp(v) \cdot \frac{1}{9} \, dv = \frac{1}{27} \int \exp(v) \, dv \).

Step 5: Evaluate the simple exponential integral

The integral of an exponential function \( \exp(v) \) is simply \( \exp(v) + C \), where \( C \) is the constant of integration. So, \( \frac{1}{27} \int \exp(v) \, dv = \frac{1}{27} \exp(v) + C \).

Step 6: Back-substitute the variables

Replace \( v \) back with \( u^{9} \) and then \( u \) back with \( x^{3} + 5 \): \( \frac{1}{27} \exp(u^{9}) = \frac{1}{27} \exp((x^{3}+5)^{9}) + C \).

Step 7: Write the final solution

The final indefinite integral of the function is \( \frac{1}{27} \exp((x^{3}+5)^{9}) + C \).

Key Concepts

Indefinite Integrals

Indefinite integrals are essential to calculus, allowing us to find functions whose derivatives are given functions. When you integrate a function without specifying the bounds, you perform an indefinite integral. This process determines a family of functions, usually represented as \( F(x) + C \), where \( C \) is a constant. The integration undoes the differentiation, reversing the process so we can understand the original function from its rate of change.

• The integral symbol \( \int \) signifies the start of this operation.
• Function inside the integral, known as the integrand, is what we integrate.
• Don't forget the constant \( C \), conveying the infinite number of solutions.

Understanding indefinite integrals is crucial because they form the basis for making sense of areas under curves and solving differential equations.

Substitution Method

The substitution method in calculus is a powerful tool for simplifying complex integrals. By transforming the integrand into a simpler form, this technique facilitates easier integration. In essence, you substitute a part of the integral with a new variable, reducing complexity.

• Hold onto the core idea: Identify an inner function within the integrand, usually denoted as \( u \).
• Differentiation plays a key role: Compute \( \frac{du}{dx} \) for replacing corresponding elements in the integral.
• This process is akin to reversing the chain rule from differentiation, making it more intuitive.

In the problem, we used substitution \( u = x^3 + 5 \) to simplify the original integral dramatically. The substitution method is particularly useful when dealing with nested functions, boosting your calculus toolkit.

Integration Techniques

Different integration techniques exist to handle the wide variety of integrals we encounter in calculus. There's no single method that works for everything, so understanding multiple techniques is beneficial.

• Substitution is just one of many methods. Others include integration by parts and partial fractions.
• These techniques bridge the gap between challenging integrals and simpler calculations.
• Choosing the right technique often depends on recognizing the form of the integrand.

The original exercise highlights substitution which streamlined the exponential integrand. Mastering various techniques enriches your mathematical problem-solving skills, making integration more approachable.

Exponential Functions

Exponential functions are one of the simplest yet most influential forms in calculus. The exponential function \( exp(x) \) represents constant growth or decay, appearing often in scientific and engineering problems.

• They involve the constant \( e \), approximately equal to 2.718, which serves as the base for natural logarithms.
• Exponential functions grow faster than polynomial functions as \( x \) increases.
• Integration of exponentials is usually straightforward, as both \( exp(x) \) and its integral are the same, with a subtle twist of a constant factor involved.

In the exercise, transforming the innermost exponential component using \( v = u^9 \) demonstrated their property of keeping form, simplifying the integration process. Being well-acquainted with exponential functions can yield insights into natural phenomena and complex calculus problems.