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Chapter 5: Problem 4

Exer. \(1-8\) : Evaluate the integral using the given substitution, and express the answer in terms of \(x\). $$ \int \frac{5 x}{\sqrt{x^{2}-3}} d x ; \quad u=x^{2}-3 $$

Short Answer

Expert verified
The integral evaluates to \( 5\sqrt{x^2 - 3} + C \).

Step by step solution

01

Identify Substitution

The substitution provided is \( u = x^2 - 3 \). This suggests that we will use \( u \) to simplify the integral.
02

Differentiate Substitution

Differentiate the substitution equation with respect to \( x \) to find \( du \). \[ du = 2x \, dx \] This can be rearranged to express \( dx \): \[ dx = \frac{du}{2x} \]
03

Substitute in the Integral

Substitute \( u = x^2 - 3 \) and \( dx = \frac{du}{2x} \) into the integral: \[ \int \frac{5x}{\sqrt{x^2 - 3}} \, dx = \int \frac{5x}{\sqrt{u}} \cdot \frac{du}{2x} \]
04

Simplify the Integral

The \( x \) terms cancel out in the integrand, simplifying our expression:\[ \int \frac{5}{2\sqrt{u}} \, du \]
05

Evaluate the Integral

The integral \( \int \frac{1}{\sqrt{u}} \, du \) is recognized as a standard form, which integrates to \( 2\sqrt{u} \). Multiplying by constants:\[ \int \frac{5}{2\sqrt{u}} \, du = \frac{5}{2} \times 2\sqrt{u} = 5\sqrt{u} + C \]
06

Back-Substitute to Variable x

Replace \( u \) with \( x^2 - 3 \) as per the substitution:\[ 5\sqrt{x^2 - 3} + C \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique in calculus used to simplify the process of evaluating integrals. The main idea is to substitute a part of the integral with a single variable, usually denoted by \( u \), which makes the integral easier to evaluate.
  • First, choose a substitution \( u \) that will simplify the integrand. Typically, \( u \) is a function inside a composition, like an expression under a root or as an exponent.
  • Create a derivative equation based on the substitution. For instance, if \( u = x^2 - 3 \), differentiate both sides to get \( du = 2x \, dx \).
  • Solve for \( dx \) to replace it in the integral.
After substituting, the integral should become simpler to handle. You recalibrate the original variable at the end to express your antiderivative in the terms you started with. This method is particularly helpful in evaluating integrals involving complicated algebraic expressions.
Definite Integrals
Definite integrals allow us to calculate the area under a curve between two points on the \( x \)-axis. Unlike indefinite integrals, definite integrals provide a specific numerical value, which represents the cumulative area.
When using the substitution method with definite integrals, there are a few extra steps:
  • After substituting and before integrating, determine the new limits of integration based on your substitution variable \( u \).
  • Calculate the integral using these new limits. This means the antiderivative evaluations will now involve the transformed limits rather than the original ones.
  • The result is a numerical value, indicating the total area under the curve from one point to another.
This approach is highly useful in physics and engineering, where it often represents real-world quantities, like distance or probability.
Antiderivatives
An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function we started with. Finding antiderivatives involves undoing the process of differentiation, which is fundamental in solving integrals.
  • To find an antiderivative, you use integration techniques, like the substitution method, helping simplify expressions and solve them analytically.
  • Essentially, the integral gives us a function plus a constant \( C \), where \( C \) represents the integration constant.
  • Each unique \( C \) corresponds to a distinct function, all sharing the same slope at every point.
Remember, when solving any integral problem, you should always back-substitute your initial variable if you made a substitution. This provides your antiderivative in the correct form, aligned with your starting variables. This practice is crucial in ensuring that you've solved the problem accurately and thoroughly.

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