Question

In the following exercises, integrate using the indicated substitution. $$ \int \frac{x}{x-100} d x ; u=x-100 $$

Short answer

The integral evaluates to \( x - 100 + 100 \ln|x - 100| + C \).

Step-by-step solution

Identify the substitution

We are given the substitution: \( u = x - 100 \). This means that we will replace \( x - 100 \) with \( u \).

Differentiate to find du

Differentiate the substitution \( u = x - 100 \) with respect to \( x \). This gives us: \( \frac{du}{dx} = 1 \) or equivalently, \( du = dx \).

Express x in terms of u

Since \( u = x - 100 \), solve for \( x \) in terms of \( u \): \( x = u + 100 \).

Substitute into the integral

Replace \( x \) with \( u + 100 \) and \( x - 100 \) with \( u \) in the integral: \[ \int \frac{u + 100}{u} \, du \].

Simplify the integrand

Rewrite the integrand by dividing each term by \( u \): \( \int \left( 1 + \frac{100}{u} \right) \, du \).

Integrate term-by-term

Integrate each term separately: \[ \int 1 \, du + \int \frac{100}{u} \, du \]. The integrals are \( \, u + 100 \ln|u| + C \), where \( C \) is the constant of integration.

Back-substitute for x

Remember to substitute back for \( u = x - 100 \): \[ x - 100 + 100 \ln|x - 100| + C \].

Key concepts

Definite and Indefinite Integrals

In calculus, integrals are essential tools for understanding the area under curves. There are two main types of integrals: definite and indefinite integrals. Each serves different purposes and has its unique characteristics.

• Definite Integrals: A definite integral not only computes the area under a curve but also within a specific interval. It has upper and lower bounds, like from point 'a' to point 'b'. The notation for definite integrals is \( \int_a^b f(x) \, dx \). The result is a numerical value, representing the total area between the curve and x-axis.
• Indefinite Integrals: An indefinite integral finds the antiderivative of a function and does not have specified limits or bounds. This integral gives us a general function plus a constant \( C \), expressed as \( \int f(x) \, dx = F(x) + C \). The constant ensures all possible antiderivatives are covered.

Understanding these types is crucial because they provide solutions to different types of mathematical problems, from calculating total distance traveled to evaluating accumulated change.

Substitution Method

The substitution method is an integration technique used for simplifying integrals by transforming them into a more manageable form. This technique is particularly useful for integrals involving complex expressions.

1. Selecting the substitution: Choose a substitution \( u = g(x) \) that simplifies the integrand, focusing usually on making expressions easier to handle, such as turning a complicated fraction or product into a simple function.
2. Find the differential: Differentiate the substitution \( u = g(x) \) to find \( \frac{du}{dx} \), or \( du = g'(x) \, dx \). This step is vital for moving from the variable \( x \) to the variable \( u \).
3. Express the integral: Replace all parts of the integral with terms involving \( u \) to end up with an integral purely in terms of \( u \) and \( du \).
4. Integrate and back-substitute: Once simplified, perform the integration with respect to \( u \), and finally, substitute back the original variable \( x \) by reversing the initial substitution.

This method can turn a seemingly daunting integral into a straightforward problem, providing a clear path to the solution.

Integration Techniques

Integration can often be challenging, but learning specific techniques makes the process more accessible and systematic. Here are some common techniques:

• Basic Integration Rules: Knowing the antiderivatives of basic functions, such as power functions or exponential functions, is foundational. For example, the integral \( \int x^n \, dx \) is \( \frac{x^{n+1}}{n+1} + C \).
• Substitution: As discussed, the substitution method is a popular technique for handling integrals involving composite or more complex functions.
• Integration by Parts: This technique is particularly useful for integrals involving products of functions and stems from the product rule for differentiation. The formula used is \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Dealing with rational functions, this method involves breaking a fraction into simpler pieces that are easier to integrate.

These techniques are part of a broader toolkit essential for solving various integral problems. Mastering them helps in tackling even the most challenging integrals with confidence.